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					                  The partial derivatives
                                 r´
                        Robert Maˇık
                      February 20, 2006




⊳⊳   ⊳   ⊲   ⊲⊲                             c Robert Maˇık, 2006 ×
                                                       r´
Contents
     x2 + xy + y 3      . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
     (x + y)e−x . .     . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
     x + y2
              . . . .   . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
      y−1
            y
     arctg      . . .   . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
           x
       1 − x2 − y 2     . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
                 2
     (x + y)ex −y
       2
                        . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
       2   2
     ex +y . . . .      . . . . . . . . . . . . . . . . . . . . . . . . . . . 88




⊳⊳   ⊳   ⊲   ⊲⊲                                                  c Robert Maˇık, 2006 ×
                                                                            r´
                                                                       ′
Remark. In all the following exercises the mixed partial derivatives (zx )′
                                                                          y
      ′ ′
and (zy )x are identical and we make no difference between them.




⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                    ′
                   zx = 2x + 1 · y + 0 = 2x + y
                   zy = 0 + x · 1 + 3y 2 = x + 3y 2
                    ′

                   ′′
                  zxx = (2x + y)′ = 2 · 1 + 0 = 2
                                x
                   ′′
                  zxy = (2x + y)′ = 0 + 1 = 1
                                y
                  zyy = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                   ′′
                                   y




⊳⊳   ⊳   ⊲   ⊲⊲                                            c Robert Maˇık, 2006 ×
                                                                      r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                       ′
                      zx = 2x + 1 · y + 0 = 2x + y
                  zy = 0 + x · 1 + 3y 2 = x + 3y 2
                     ′

We differentiate the sum (x2 + xy = 2 3 ) with = 2
                 z ′′ = (2x + y)′ + y · 1 + 0 respect to x.
                      xx            x
                     ′′            ′
                   zxy = as the function 1 one
     • x2 is differentiated(2x + y)y = 0 + of= 1 variable.
                   z ′′ = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                                     y
     • The variable yy in the expression xy is considered to be a constant
                     y
       factor and we use the constant multiple rule

                                  (xy)′ = y(x)′ .
                                      x       x

         The derivative of x with respect to x is the usual derivative.
     • Term y 3 does not involve the variable x. Hence this term is treated
       to be constant and the derivative is zero.

⊳⊳   ⊳   ⊲   ⊲⊲                                               c Robert Maˇık, 2006 ×
                                                                         r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                    ′
                   zx = 2x + 1 · y + 0 = 2x + y
                   zy = 0 + x · 1 + 3y 2 = x + 3y 2
                    ′

                   ′′
                  zxx = (2x + y)′ = 2 · 1 + 0 = 2
                                x
                   ′′
                  zxy = (2x + y)′ = 0 + 1 = 1
                                y
                  zyy = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                   ′′
                                   y




We simplify.
⊳⊳   ⊳   ⊲   ⊲⊲                                            c Robert Maˇık, 2006 ×
                                                                      r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                       ′
                      zx = 2x + 1 · y + 0 = 2x + y
                      zy = 0 + x · 1 + 3y 2 = x + 3y 2
                       ′

                  ′′
                 zxx = (2x + y)′ = 2 · 1 + 0 = 2
                               x
                  ′′
                 zxy = (2x 2 y)′ = 0 3 1 = 1
                           + y       +
We differentiate the sum (x + xy + y ) with respect to y.
                              2 ′
                 zyy = (x + 3y )y = 0 + 3 · 2y 1 = 6y
                  ′′
      2
  • x is differentiated as a constant, since it does not involve the
     variable y.
     • The variable x in the expression xy is considered to be a constant
       factor and we use the constant multiple rule (xy)′ = x(y)′ . The
                                                         y        y
         derivative of y with respect to y is the usual derivative.
     • Term y 3 is differentiated as one-variable function.

⊳⊳   ⊳   ⊲   ⊲⊲                                                c Robert Maˇık, 2006 ×
                                                                          r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                    ′
                   zx = 2x + 1 · y + 0 = 2x + y
                   zy = 0 + x · 1 + 3y 2 = x + 3y 2
                    ′

                   ′′
                  zxx = (2x + y)′ = 2 · 1 + 0 = 2
                                x
                   ′′
                  zxy = (2x + y)′ = 0 + 1 = 1
                                y
                  zyy = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                   ′′
                                   y




We simplify.
⊳⊳   ⊳   ⊲   ⊲⊲                                            c Robert Maˇık, 2006 ×
                                                                      r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                    ′
                   zx = 2x + 1 · y + 0 = 2x + y
                   zy = 0 + x · 1 + 3y 2 = x + 3y 2
                    ′

                   ′′
                  zxx = (2x + y)′ = 2 · 1 + 0 = 2
                                x
                   ′′
                  zxy = (2x + y)′ = 0 + 1 = 1
                                y
                  zyy = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                   ′′
                                   y




                 ′
We differentiate zx with respect to x
⊳⊳   ⊳   ⊲   ⊲⊲                                            c Robert Maˇık, 2006 ×
                                                                      r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                    ′
                   zx = 2x + 1 · y + 0 = 2x + y
                   zy = 0 + x · 1 + 3y 2 = x + 3y 2
                    ′

                   ′′
                  zxx = (2x + y)′ = 2 · 1 + 0 = 2
                                x
                   ′′
                  zxy = (2x + y)′ = 0 + 1 = 1
                                y
                  zyy = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                   ′′
                                   y




We use the sum rule, the constant multiple rule and the rule for the
derivative of constant function.
⊳⊳   ⊳   ⊲   ⊲⊲                                            c Robert Maˇık, 2006 ×
                                                                      r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                    ′
                   zx = 2x + 1 · y + 0 = 2x + y
                   zy = 0 + x · 1 + 3y 2 = x + 3y 2
                    ′

                   ′′
                  zxx = (2x + y)′ = 2 · 1 + 0 = 2
                                x
                   ′′
                  zxy = (2x + y)′ = 0 + 1 = 1
                                y
                  zyy = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                   ′′
                                   y




We simplify.
⊳⊳   ⊳   ⊲   ⊲⊲                                            c Robert Maˇık, 2006 ×
                                                                      r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                    ′
                   zx = 2x + 1 · y + 0 = 2x + y
                   zy = 0 + x · 1 + 3y 2 = x + 3y 2
                    ′

                   ′′
                  zxx = (2x + y)′ = 2 · 1 + 0 = 2
                                x
                   ′′
                  zxy = (2x + y)′ = 0 + 1 = 1
                                y
                  zyy = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                   ′′
                                   y




        ′′                  ′
To find zxy we differentiate zx with respect to y.
⊳⊳   ⊳   ⊲   ⊲⊲                                            c Robert Maˇık, 2006 ×
                                                                      r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                    ′
                   zx = 2x + 1 · y + 0 = 2x + y
                   zy = 0 + x · 1 + 3y 2 = x + 3y 2
                    ′

                   ′′
                  zxx = (2x + y)′ = 2 · 1 + 0 = 2
                                x
                   ′′
                  zxy = (2x + y)′ = 0 + 1 = 1
                                y
                  zyy = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                   ′′
                                   y




We use the sum rule. Since x is treated to be a constant, (2x) is
constant as well.
⊳⊳   ⊳   ⊲   ⊲⊲                                            c Robert Maˇık, 2006 ×
                                                                      r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                    ′
                   zx = 2x + 1 · y + 0 = 2x + y
                   zy = 0 + x · 1 + 3y 2 = x + 3y 2
                    ′

                   ′′
                  zxx = (2x + y)′ = 2 · 1 + 0 = 2
                                x
                   ′′
                  zxy = (2x + y)′ = 0 + 1 = 1
                                y
                  zyy = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                   ′′
                                   y




We simplify.
⊳⊳   ⊳   ⊲   ⊲⊲                                            c Robert Maˇık, 2006 ×
                                                                      r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                    ′
                   zx = 2x + 1 · y + 0 = 2x + y
                   zy = 0 + x · 1 + 3y 2 = x + 3y 2
                    ′

                   ′′
                  zxx = (2x + y)′ = 2 · 1 + 0 = 2
                                x
                   ′′
                  zxy = (2x + y)′ = 0 + 1 = 1
                                y
                  zyy = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                   ′′
                                   y




        ′′                  ′
To find zyy we differentiate zy with respect to y.
⊳⊳   ⊳   ⊲   ⊲⊲                                            c Robert Maˇık, 2006 ×
                                                                      r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                    ′
                   zx = 2x + 1 · y + 0 = 2x + y
                   zy = 0 + x · 1 + 3y 2 = x + 3y 2
                    ′

                   ′′
                  zxx = (2x + y)′ = 2 · 1 + 0 = 2
                                x
                   ′′
                  zxy = (2x + y)′ = 0 + 1 = 1
                                y
                  zyy = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                   ′′
                                   y




We use the sum rule, the derivative of constant function, the constant
multiple rule and the power rule.
⊳⊳   ⊳   ⊲   ⊲⊲                                            c Robert Maˇık, 2006 ×
                                                                      r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                    ′
                   zx = 2x + 1 · y + 0 = 2x + y
                   zy = 0 + x · 1 + 3y 2 = x + 3y 2
                    ′

                   ′′
                  zxx = (2x + y)′ = 2 · 1 + 0 = 2
                                x
                   ′′
                  zxy = (2x + y)′ = 0 + 1 = 1
                                y
                  zyy = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                   ′′
                                   y




We simplify.
⊳⊳   ⊳   ⊲   ⊲⊲                                            c Robert Maˇık, 2006 ×
                                                                      r´
Find derivatives of z(x, y) = x2 + xy + y 3 up to the 2nd order.

                    ′
                   zx = 2x + 1 · y + 0 = 2x + y
                   zy = 0 + x · 1 + 3y 2 = x + 3y 2
                    ′

                   ′′
                  zxx = (2x + y)′ = 2 · 1 + 0 = 2
                                x
                   ′′
                  zxy = (2x + y)′ = 0 + 1 = 1
                                y
                  zyy = (x + 3y 2 )′ = 0 + 3 · 2y 1 = 6y
                   ′′
                                   y




All derivatives up to the second order have been found.
⊳⊳   ⊳   ⊲   ⊲⊲                                            c Robert Maˇık, 2006 ×
                                                                      r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
          ′′
         zxy = (e−x )′ = −e−x
                     x
          ′′
         zyy = 0




⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
          ′′
         zxy = (e−x )′ = −e−x
                     x
        ′′
       zyy = 0
     • The function consists from the product of two factors
        z = (x + y) · e−x .

     • Both factors involve the variable x and hence we differentiate by
       the product rule.

⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
          ′′
         zxy = (e−x )′ = −e−x
                     x
          ′′
         zyy = 0




The usual rules are employed and the variable y is treated as a
constant.
⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
          ′′
         zxy = (e−x )′ = −e−x
                     x
          ′′
         zyy = 0




We take out the repeating factor e−x .
⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
        ′′
       zxy = (e−x )′ = −e−x
                   x
       zyy differentiate with respect to y. The function is a product of
     • We = 0
        ′′

       two factors z = (x + y) · e−x .

     • The green expression does not involve the variable and it is con-
       sidered to be constant. Hence we have a constant multiple of the
       function (x + y) and work with the constant multiple rule.

⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
          ′′
         zxy = (e−x )′ = −e−x
                     x
          ′′
         zyy = 0




We use the sum rule, the variable x is considered to be a constant
parameter.
⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
          ′′
         zxy = (e−x )′ = −e−x
                     x
          ′′
         zyy = 0




We simplify.
⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
          ′′
         zxy = (e−x )′ = −e−x
                     x
        ′′
       zyy = 0
               ′′                                      ′
     • To find zxx we differentiate the first derivative zx with respect to
       x.
     • The variable x is involved in both factors and we have to use the
       product rule.

⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
          ′′
         zxy = (e−x )′ = −e−x
                     x
          ′′
         zyy = 0




We take out the common factor.
⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
          ′′
         zxy = (e−x )′ = −e−x
                     x
          ′′
         zyy = 0




We simplify.
⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
          ′′
         zxy = (e−x )′ = −e−x
                     x
          ′′
         zyy = 0



                                                           ′
To find the mixed derivative we find either the derivative (zx )′ or
                                                              y
  ′ ′
(zy )x . The second possibility seems to be easier.
⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
          ′′
         zxy = (e−x )′ = −e−x
                     x
          ′′
         zyy = 0

Since the function is a function of one variable, the partial derivative
becomes to be the usual derivative. We use the chain rule as follows.

                       (e−x )′ = e−x (−x)′ = e−x (−1)

⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
          ′′
         zxy = (e−x )′ = −e−x
                     x
          ′′
         zyy = 0



                        ′′                  ′
To find the derivative zyy we differentiate zy with respect to y.
                                                          ′          ′
However, the variable y is missing in the expression for zy . Hence zy is
constant and its derivative is zero.
⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
Find derivatives of z(x, y) = (x + y)e−x up to the 2nd order.

           ′
          zx = (x + y)′ · e−x + (x + y) · (e−x )′
                      x                         x
             = (1 + 0)e−x + (x + y) · e−x · (−1) = e−x · (1 − x − y)
           ′
          zy = (x + y)′ · e−x = (0 + 1)e−x = e−x
                      y
          ′′
         zxx = e−x · (−1) · (1 − x − y) + e−x (0 − 1 − 0)
             = e−x (−1 + x + y − 1) = e−x (x + y − 2)
          ′′
         zxy = (e−x )′ = −e−x
                     x
          ′′
         zyy = 0




⊳⊳   ⊳   ⊲   ⊲⊲                                             c Robert Maˇık, 2006 ×
                                                                       r´
                                   x + y2
Find derivatives of z(x, y) =             up to the 2nd order.
                                   y−1

                           ′      1               1
                          zx =       · (1 + 0) =
                                 y−1             y−1

 ′        1     ′   y 2 − 2y − x         ′′           ′′          1
zx =         , zy =              ,      zxx = 0,     zxy = −
         y−1           (y − 1)2                                (y − 1)2

                    ′
                       (x + y 2 )′ (y − 1) − (x + y 2 )(y − 1)′
                                 y                            y
                   zy =
                                       (y − 1)2
                       (0 + 2y)(y − 1) − (x + y 2 )(1 − 0)
                     =
                                      (y − 1)2
                           2
                       2y − 2y − (x + y 2 )
                     =
                              (y − 1)2
                       y 2 − 2y − x
                     =
                          (y − 1)2
⊳⊳   ⊳    ⊲   ⊲⊲                                                  c Robert Maˇık, 2006 ×
                                                                             r´
                                x + y2
Find derivatives of z(x, y) =          up to the 2nd order.
                                y−1

                       ′      1               1
                      zx =       · (1 + 0) =
                             y−1             y−1

 ′      1       ′    y 2 − 2y − x        ′′           ′′          1
zx =       , zy =               2
                                    , zxx = 0, zxy = −
      y−1               (y − 1)                               (y − 1)2
    • In order to differentiate the function with respect to x we write
                      (x + y 2 )′ (y − 1) − (x + y 2 )(y −11)′
                                y                            y· (x + y 2 ) .
      the function as the product of two factors:
                 ′
               zy =
                                      (y − 1) 2          y−1

                     1(0 + 2y)(y − 1) − (x + y 2 )(1 − 0)
    • The factor =                   involve
                           does not (y − 1)2the variable x and it is a con-
                   y−1
                          2 use the constant multiple rule.
      stant multiple.2y − 2y − (x + y 2 )
                       We
                   =
                             (y − 1)2
    • It remains to differentiate the sum (x + y 2 ) by the sum rule.
                      y 2 − 2y − x
                   =
                         (y − 1)2
⊳⊳   ⊳   ⊲   ⊲⊲                                               c Robert Maˇık, 2006 ×
                                                                         r´
                                   x + y2
Find derivatives of z(x, y) =             up to the 2nd order.
                                   y−1

                           ′      1               1
                          zx =       · (1 + 0) =
                                 y−1             y−1

 ′        1     ′   y 2 − 2y − x         ′′           ′′          1
zx =         , zy =              ,      zxx = 0,     zxy = −
         y−1           (y − 1)2                                (y − 1)2

                    ′
                       (x + y 2 )′ (y − 1) − (x + y 2 )(y − 1)′
                                 y                            y
                   zy =
                                       (y − 1)2
                       (0 + 2y)(y − 1) − (x + y 2 )(1 − 0)
                     =
                                      (y − 1)2
                           2
                       2y − 2y − (x + y 2 )
                     =
                              (y − 1)2
We simplify.           y 2 − 2y − x
                     =
                          (y − 1)2
⊳⊳   ⊳    ⊲   ⊲⊲                                                  c Robert Maˇık, 2006 ×
                                                                             r´
                                x + y2
Find derivatives of z(x, y) =          up to the 2nd order.
                                y−1

 ′        1     ′   y 2 − 2y − x      ′′          ′′          1
zx =         , zy =              ,   zxx = 0,    zxy = −
         y−1           (y − 1)2                            (y − 1)2

                   (x + y 2 )′ (y − 1) − (x + y 2 )(y − 1)′
                    ′        y                            y
                   zy =
                                   (y − 1)2
                   (0 + 2y)(y − 1) − (x + y 2 )(1 − 0)
                =
                                  (y − 1)2
                   2y 2 − 2y − (x + y 2 )
                =
        ′                (y − 1)2
To find zy we have to use the quotient rule, since the variable y is in
                    2
both numerator and y − 2y − x Hence we differentiate
                = denominator.
                     (y − 1)2
                                 x + y2
                                  y−1
                    ′′
                   zxx = 0
⊳⊳   ⊳    ⊲   ⊲⊲                                      1       c Robert Maˇık, 2006 ×
                                                                         r´
                                 x + y2
Find derivatives of z(x, y) =           up to the 2nd order.
                                 y−1

 ′      1     ′   y 2 − 2y − x          ′′           ′′          1
zx =       , zy =              ,       zxx = 0,     zxy = −
       y−1           (y − 1)2                                 (y − 1)2

                 ′   (x + y 2 )′ (y − 1) − (x + y 2 )(y − 1)′
                               y                            y
                zy =
                                     (y − 1)2
                     (0 + 2y)(y − 1) − (x + y 2 )(1 − 0)
                   =
                                    (y − 1)2
                     2y 2 − 2y − (x + y 2 )
                   =
                            (y − 1)2
                     y 2 − 2y − x
                   =
                        (y − 1)2
We evaluate the derivative of the numerator and denominator. To do
                 sum rule, constant rule and power rule.
this we use the ′′
              zxx = 0
⊳⊳  ⊳  ⊲  ⊲⊲                                         1   c Robert Maˇık, 2006 ×
                                                                    r´
                                   x + y2
Find derivatives of z(x, y) =             up to the 2nd order.
                                   y−1

 ′        1     ′   y 2 − 2y − x          ′′           ′′          1
zx =         , zy =              ,       zxx = 0,     zxy = −
         y−1           (y − 1)2                                 (y − 1)2

                    ′   (x + y 2 )′ (y − 1) − (x + y 2 )(y − 1)′
                                  y                            y
                   zy =
                                        (y − 1)2
                        (0 + 2y)(y − 1) − (x + y 2 )(1 − 0)
                      =
                                       (y − 1)2
                        2y 2 − 2y − (x + y 2 )
                      =
                               (y − 1)2
                        y 2 − 2y − x
                      =
                           (y − 1)2


We simplify.        ′′
                   zxx = 0
⊳⊳   ⊳    ⊲   ⊲⊲                                           1       c Robert Maˇık, 2006 ×
                                                                              r´
                                  x + y2
Find derivatives of z(x, y) =            up to the 2nd order.
                                  y−1

 ′        1     ′   y 2 − 2y − x         ′′           ′′          1
zx =         , zy =              ,      zxx = 0,     zxy = −
         y−1           (y − 1)2                                (y − 1)2

                    ′  (x + y 2 )′ (y − 1) − (x + y 2 )(y − 1)′
                                 y                            y
                   zy =
                                       (y − 1)2
                       (0 + 2y)(y − 1) − (x + y 2 )(1 − 0)
                     =
                                      (y − 1)2
                       2y 2 − 2y − (x + y 2 )
                     =
                              (y − 1)2
                       y 2 − 2y − x
                     =
                          (y − 1)2


We simplify even more.
               ′′
             zxx = 0
⊳⊳   ⊳    ⊲   ⊲⊲                                          1       c Robert Maˇık, 2006 ×
                                                                             r´
                                  x + y2
Find derivatives of z(x, y) =            up to the 2nd order.
                                  y−1

 ′        1     ′   y 2 − 2y − x         ′′         ′′           1
zx =         , zy =              ,      zxx = 0,   zxy = −
         y−1           (y − 1)2                               (y − 1)2
                    ′′
                   zxx = 0
                                                           1
                    ′′
                   zxy = −1 · (y − 1)−2 · (1 − 0) = −
                                                        (y − 1)2


           (2y − 2)(y − 1)2 − (y 2 − 2y − x) · 2 · (y − 1) · (1 − 0)
          ′′
         zyy =
                                     (y − 1)4
                            2      2
                     (y − 1) − (y − 2y − x)
         = 2(y − 1)
                              (y − 1)4
              x+1
         =2
             (y − 1)3
The first derivatives have been found.
⊳⊳   ⊳    ⊲   ⊲⊲                                                   c Robert Maˇık, 2006 ×
                                                                              r´
                                  x + y2
Find derivatives of z(x, y) =            up to the 2nd order.
                                  y−1

 ′        1     ′   y 2 − 2y − x         ′′         ′′           1
zx =         , zy =              ,      zxx = 0,   zxy = −
         y−1           (y − 1)2                               (y − 1)2
                    ′′
                   zxx = 0
                                                           1
                    ′′
                   zxy = −1 · (y − 1)−2 · (1 − 0) = −
                                                        (y − 1)2


          ′′ (2y − 2)(y − 1)2 − (y 2 − 2y − x) · 2 · (y − 1) · (1 − 0)
         zyy =
                                     (y − 1)4
                             2      2′
                       differentiate − 2y − x)
     • To find zxx we(y − 1) − (y zx with respect to x.
                ′′
          = 2(y − 1)
              ′
                               (y − 1)4
     • Since zx does not involve the variable x, it is treated as a constant
                x+1
       and= 2 derivative is zero by the constant rule.
           the
              (y − 1)3

⊳⊳   ⊳    ⊲   ⊲⊲                                                   c Robert Maˇık, 2006 ×
                                                                              r´
                                  x + y2
Find derivatives of z(x, y) =            up to the 2nd order.
                                  y−1

 ′        1     ′   y 2 − 2y − x         ′′         ′′           1
zx =         , zy =              ,      zxx = 0,   zxy = −
         y−1           (y − 1)2                               (y − 1)2
                    ′′
                   zxx = 0
                                                           1
                    ′′
                   zxy = −1 · (y − 1)−2 · (1 − 0) = −
                                                        (y − 1)2


        ′′   (2y − 2)(y − 1)2 − (y 2 − 2y − x) · 2 · (y − 1) · (1 − 0)
      zyy =
                 ′′                   (y − 1)4
                                     z′
     • To find zxy we differentiate 2 x with respect to y.
                             2
                      (y − 1) − (y − 2y − x)
           = the expression for z ′ does not involve the variable x, it
     • Since 2(y − 1)          (y − 1)4
                                   x
       is a one-variable function and the partial derivative is the usual
                x+1
           =2
       derivative.− 1)3
               (y

⊳⊳   ⊳    ⊲   ⊲⊲                                                   c Robert Maˇık, 2006 ×
                                                                              r´
                                  x + y2
Find derivatives of z(x, y) =            up to the 2nd order.
                                  y−1

 ′        1     ′   y 2 − 2y − x         ′′         ′′           1
zx =         , zy =              ,      zxx = 0,   zxy = −
         y−1           (y − 1)2                               (y − 1)2
                    ′′
                   zxx = 0
                                                           1
                    ′′
                   zxy = −1 · (y − 1)−2 · (1 − 0) = −
                                                        (y − 1)2


          ′′(2y − 2)(y − 1)2 − (y 2 − 2y − x) · 2 · (y − 1) · (1 − 0)
         zyy =
                                    (y − 1)4
                           2      2
                    (y − 1) − (y − 2y − x)
        = 2(y − 1)
                             (y − 1)4
              x+1
        =2
We simplify. (y − 1)3

⊳⊳   ⊳    ⊲   ⊲⊲                                                   c Robert Maˇık, 2006 ×
                                                                              r´
                                x + y2
Find derivatives of z(x, y) =          up to the 2nd order.
                                y−1

 ′        1     ′   y 2 − 2y − x      ′′          ′′          1
zx =         , zy =              ,   zxx = 0,    zxy = −
         y−1           (y − 1)2                            (y − 1)2

          ′′ (2y − 2)(y − 1)2 − (y 2 − 2y − x) · 2 · (y − 1) · (1 − 0)
         zyy =
                                       (y − 1)4
                              2      2
                       (y − 1) − (y − 2y − x)
           = 2(y − 1)
                                (y − 1)4
                x+1
           =2                                2
              (y′′− we3 differentiate z ′ = y − 2y − x with respect to y.
     • To find zyy 1)                   y
                                              (y − 1)2
       Since y is in both numerator and denominator, we use the quotient
       rule.
     • The expression (y − 1)2 is differentiated by the chain rule.

⊳⊳   ⊳    ⊲   ⊲⊲                                              c Robert Maˇık, 2006 ×
                                                                         r´
                                   x + y2
Find derivatives of z(x, y) =             up to the 2nd order.
                                   y−1

 ′        1     ′   y 2 − 2y − x          ′′          ′′          1
zx =         , zy =              ,       zxx = 0,    zxy = −
         y−1           (y − 1)2                                (y − 1)2

          ′′    (2y − 2)(y − 1)2 − (y 2 − 2y − x) · 2 · (y − 1) · (1 − 0)
         zyy =
                                         (y − 1)4
                                2      2
                         (y − 1) − (y − 2y − x)
              = 2(y − 1)
                                  (y − 1)4
                   x+1
              =2
                 (y − 1)3




We take out the common factor 2(y − 1)
⊳⊳   ⊳    ⊲   ⊲⊲                                                 c Robert Maˇık, 2006 ×
                                                                            r´
                                   x + y2
Find derivatives of z(x, y) =             up to the 2nd order.
                                   y−1

 ′        1     ′   y 2 − 2y − x          ′′          ′′          1
zx =         , zy =              ,       zxx = 0,    zxy = −
         y−1           (y − 1)2                                (y − 1)2

          ′′    (2y − 2)(y − 1)2 − (y 2 − 2y − x) · 2 · (y − 1) · (1 − 0)
         zyy =
                                         (y − 1)4
                                2      2
                         (y − 1) − (y − 2y − x)
              = 2(y − 1)
                                  (y − 1)4
                   x+1
              =2
                 (y − 1)3




We simplify the numerator by expanding the power of the sum and
adding the corresponding terms.
⊳⊳   ⊳    ⊲   ⊲⊲                                                 c Robert Maˇık, 2006 ×
                                                                            r´
                                   x + y2
Find derivatives of z(x, y) =             up to the 2nd order.
                                   y−1

 ′        1     ′   y 2 − 2y − x          ′′          ′′          1
zx =         , zy =              ,       zxx = 0,    zxy = −
         y−1           (y − 1)2                                (y − 1)2

          ′′    (2y − 2)(y − 1)2 − (y 2 − 2y − x) · 2 · (y − 1) · (1 − 0)
         zyy =
                                         (y − 1)4
                                2      2
                         (y − 1) − (y − 2y − x)
              = 2(y − 1)
                                  (y − 1)4
                   x+1
              =2
                 (y − 1)3




All derivatives up to the second order have been found.
⊳⊳   ⊳    ⊲   ⊲⊲                                                 c Robert Maˇık, 2006 ×
                                                                            r´
                                              y
Find derivatives of z(x, y) = arctg             up to the 2nd order.
                                              x

            ′        1                                 x2     y      y
           zx =          y2
                              · y · (−1)x−2 = −             ·   =− 2
                  1+                              x2   + y 2 x2   x + y2
                         x2

            ′        1            1       x2   1    x
           zy =          y2
                              ·     ·1= 2     · = 2
                                             2 x
                  1+              x    x +y      x + y2
                         x2


 ′             y               ′        x
zx = −               ,        zy =            ,
            x2 + y 2                 x2 + y 2

                                                       2xy
     zxx = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) =
      ′′
                                                    (x2 + y 2 )2
      ′′       1 · (x2 + y 2 ) − y · (0 + 2y)      x2 − y 2        y 2 − x2
     zxy   =−                                 =− 2            = 2
                         (x2 + y 2 )2             (x + y 2 )2    (x + y 2 )2
                                                       2xy
      ′′
     zyy   = x · (−1) · (x2 + y 2 )−2 · (0 + 2y) = − 2
⊳⊳    ⊳    ⊲ ⊲⊲
                                                    (x + y 2 )2 c Robert Maˇ´ık, 2006 ×
                                                                           r
                                            y
Find derivatives of z(x, y) = arctg           up to the 2nd order.
                                            x

            ′       1                                 x2     y      y
           zx =         y2
                             · y · (−1)x−2 = −             ·   =− 2
                  1+                             x2   + y 2 x2   x + y2
                        x2

            ′       1 1         x2    1      x
           zy =  y2
                        ·1= 2·      · = 2
                                   2 x
            1 + x2 x         x +y         x + y2
   • We differentiate the arctg(·) function by the rule
          y                  x
zx = − arctg f2(x) z= = 21 2 ·, f ′ (x) (formula for arctangent and
 ′                  ′ ′
                ,     y
       x2 + y             x f+(x)
                        1+ 2 y
         the chain rule).
                                                       2xy
      ′′
     zxx  = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) = 2
                          y
      • The expression                              (x + y 2 )or the constant
                            is differentiated as the product 2
                     2    x
                          2
               1 · (x + y ) − y · (0 + 2y)        x2 − y 2         y 2 − x2
      ′′
          =−
     zxyfactor and the power 2 2             = y = · 2
                                function, i.e. − (x2y+ x−12 .= (x2 + y 2 )2
                       (x2 + y )               x       y )
                                                       2xy
     zyy = x · (−1) · (x2 + y 2 )−2 · (0 + 2y) = − 2
      ′′

⊳⊳    ⊳   ⊲  ⊲⊲
                                                    (x + y 2 )2 c Robert Maˇ´ık, 2006 ×
                                                                           r
                                          y
Find derivatives of z(x, y) = arctg         up to the 2nd order.
                                          x

         ′        1                                 x2     y      y
        zx =          y2
                           · y · (−1)x−2 = −             ·   =− 2
               1+                              x2   + y 2 x2   x + y2
                      x2

         ′        1            1       x2   1    x
        zy =          y2
                           ·     ·1= 2     · = 2
                                          2 x
               1+              x    x +y      x + y2
                      x2

              y                    x
 zx simplify. Among zy = 2we use,
We = −
   ′
                    , others,
                          ′
           x2 + y 2            x + y2
                         1             x2            x2
                             2 =               = 2 2xy
      ′′              1 (xy
     zxx = −y · (−1) ·+ x2+ y 2x2 1 (2x + 0) = + y 2 2
                            2             y2
                                  )−2 · + x2     x       2

                                                   (x2 + y )
                    2     2                        2      2        2          2
and ′′ = − 1 · (x + y ) − y · (0 + 2y) = − x − y = y − x
     zxy
                       (x2 + y 2 )2 x−2 =  1      2      2 2   (x2 + y 2 )2
                                             . (x + y )
                                          x2            2xy
     zyy = x · (−1) · (x2 + y 2 )−2 · (0 + 2y) = − 2
      ′′

⊳⊳    ⊳  ⊲  ⊲⊲
                                                   (x + y 2 )2 c Robert Maˇ´ık, 2006 ×
                                                                          r
                                           y
Find derivatives of z(x, y) = arctg          up to the 2nd order.
                                           x

         ′        1                                 x2     y      y
        zx =          y2
                           · y · (−1)x−2 = −             ·   =− 2
               1+                              x2   + y 2 x2   x + y2
                      x2

         ′        1            1       x2   1    x
        zy =          y2
                           ·     ·1= 2     · = 2
                                          2 x
               1+              x    x +y      x + y2
                      x2


 ′          y               ′        x
zx = −            ,        zy =            ,
         x2 + y 2                 x2 + y 2

                                                  2xy
   zxx = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) =
    ′′
                                               (x2 + y 2 )2
    ′′     1 · (x2 + y 2 ) − y · (0 + 2y)     x2 − y 2        y 2 − x2
   zxy = −                                 =− 2          = 2
                     (x2 + y 2 )2            (x + y 2 )2    (x + y 2 )2
                                          2
We multiply the fractions. 2 −2 term x cancels. 2xy
                             The
   zyy = x · (−1) · (x2 + y ) · (0 + 2y) = − 2
    ′′

⊳⊳  ⊳  ⊲ ⊲⊲
                                               (x + y 2 )2 c Robert Maˇ´ık, 2006 ×
                                                                      r
                                            y
Find derivatives of z(x, y) = arctg           up to the 2nd order.
                                            x

         ′         1                                 x2     y      y
        zx =           y2
                            · y · (−1)x−2 = −             ·   =− 2
                1+                              x2   + y 2 x2   x + y2
                       x2

          ′        1            1       x2   1    x
         zy =          y2
                            ·     ·1= 2     · = 2
                                           2 x
                1+              x    x +y      x + y2
                       x2


 ′           y               ′        x
zx = −             ,        zy =            ,
          x2 + y 2                 x2 + y 2
                          ′         1
Formula (arctg(f (x)))2 = 2 −2 2          f ′ (x) is used and the expression
                                                        2xy
     ′′
   zxx = −y · (−1) · (x + y )  1 + f ·(x) + 0) =
                                       (2x
 y                                                   (x2 + y 2 )2
   is treated as a2product or the constant factor and2 the power 2
 x ′′        1 · (x + y 2 ) − y · (0 + 2y)         x2 − y           y2 − x
   zxy = −                                  =− 2               = 2
                 y   1
                     (x2 + y 2 )2                 (x + y 2 )2     (x + y 2 )2
function, i.e.     = ·y .
                 x   x2                                 2xy
   zyy = x · (−1) · (x + y 2 )−2 · (0 + 2y) = − 2
     ′′

⊳⊳   ⊳  ⊲  ⊲⊲
                                                     (x + y 2 )2 c Robert Maˇ´ık, 2006 ×
                                                                            r
                                           y
Find derivatives of z(x, y) = arctg          up to the 2nd order.
                                           x

         ′        1                                 x2     y      y
        zx =          y2
                           · y · (−1)x−2 = −             ·   =− 2
               1+                              x2   + y 2 x2   x + y2
                      x2

         ′        1            1       x2   1    x
        zy =          y2
                           ·     ·1= 2     · = 2
                                          2 x
               1+              x    x +y      x + y2
                      x2


 ′          y               ′        x
zx = −            ,        zy =            ,
         x2 + y 2                 x2 + y 2

                                                   2xy
   zxx = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) =
    ′′
                                                (x2 + y 2 )2
    ′′     1 · (x2 + y 2 ) − y · (0 + 2y)      x2 − y 2        y 2 − x2
   zxy = −                                =− 2            = 2
                     (x2 + y 2 )2             (x + y 2 )2    (x + y 2 )2
We simplify.                                       2xy
   zyy = x · (−1) · (x2 + y 2 )−2 · (0 + 2y) = − 2
    ′′

⊳⊳  ⊳  ⊲ ⊲⊲
                                                (x + y 2 )2 c Robert Maˇ´ık, 2006 ×
                                                                       r
                                          y
Find derivatives of z(x, y) = arctg         up to the 2nd order.
                                          x

         ′       1                                 x2     y      y
        zx =         y2
                          · y · (−1)x−2 = −             ·   =− 2
               1+                             x2   + y 2 x2   x + y2
                     x2

         ′       1            1       x2   1    x
        zy =         y2
                          ·     ·1= 2     · = 2
                                         2 x
               1+             x    x +y      x + y2
                     x2


 ′          y              ′        x
zx = −            ,       zy =            ,
         x2 + y 2                x2 + y 2

                                                 2xy
   zxx = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) =
    ′′
                                              (x2 + y 2 )2
    ′′     1 · (x2 + y 2 ) − y · (0 + 2y)    x2 − y 2        y 2 − x2
   zxy = −                                =− 2          = 2
                     (x2 + y 2 )2           (x + y 2 )2    (x + y 2 )2
We multiply and cancel x. 2 −2                   2xy
   zyy = x · (−1) · (x2 + y ) · (0 + 2y) = − 2
    ′′

⊳⊳  ⊳  ⊲ ⊲⊲
                                              (x + y 2 )2 c Robert Maˇ´ık, 2006 ×
                                                                     r
                                            y
Find derivatives of z(x, y) = arctg           up to the 2nd order.
                                            x
 ′                y          ′        x
zx = −                  ,   zy =            ,
               x2 + y 2            x2 + y 2

                                                       2xy
     zxx = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) =
      ′′
                                                       + y 2 )2
                                                      (x2
                     2     2
      ′′       1 · (x + y ) − y · (0 + 2y)          2
                                                   x − y2        y 2 − x2
     zxy   =−                                 =− 2            = 2
                         (x2 + y 2 )2             (x + y 2 )2   (x + y 2 )2
                                                       2xy
      ′′
     zyy   = x · (−1) · (x2 + y 2 )−2 · (0 + 2y) = − 2
                                                    (x + y 2 )2




The first derivatives are known.
⊳⊳    ⊳    ⊲    ⊲⊲                                               c Robert Maˇık, 2006 ×
                                                                            r´
                                            y
Find derivatives of z(x, y) = arctg           up to the 2nd order.
                                            x
 ′                y          ′        x
zx = −                  ,   zy =            ,
               x2 + y 2            x2 + y 2

                                                       2xy
     zxx = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) =
      ′′
                                                       + y 2 )2
                                                      (x2
                     2     2
      ′′       1 · (x + y ) − y · (0 + 2y)          2
                                                   x − y2        y 2 − x2
     zxy   =−                                 =− 2            = 2
                         (x2 + y 2 )2             (x + y 2 )2   (x + y 2 )2
                                                       2xy
      ′′
     zyy   = x · (−1) · (x2 + y 2 )−2 · (0 + 2y) = − 2
                                                    (x + y 2 )2



We differentiate zx = −y · (x2 + y 2 )−1 with respect to x. The factor
                   ′

(−y) is a constant multiple and the constant multiple rule is folowed
by the chain rule for (x2 + y 2 )−1 .
⊳⊳    ⊳    ⊲    ⊲⊲                                               c Robert Maˇık, 2006 ×
                                                                            r´
                                            y
Find derivatives of z(x, y) = arctg           up to the 2nd order.
                                            x
 ′                y          ′        x
zx = −                  ,   zy =            ,
               x2 + y 2            x2 + y 2

                                                       2xy
     zxx = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) =
      ′′
                                                       + y 2 )2
                                                      (x2
                     2     2
      ′′       1 · (x + y ) − y · (0 + 2y)          2
                                                   x − y2        y 2 − x2
     zxy   =−                                 =− 2            = 2
                         (x2 + y 2 )2             (x + y 2 )2   (x + y 2 )2
                                                       2xy
      ′′
     zyy   = x · (−1) · (x2 + y 2 )−2 · (0 + 2y) = − 2
                                                    (x + y 2 )2




We simplify.
⊳⊳    ⊳    ⊲    ⊲⊲                                               c Robert Maˇık, 2006 ×
                                                                            r´
                                            y
Find derivatives of z(x, y) = arctg           up to the 2nd order.
                                            x
 ′                y          ′        x
zx = −                  ,   zy =            ,
               x2 + y 2            x2 + y 2

                                                       2xy
     zxx = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) =
      ′′
                                                       + y 2 )2
                                                       (x2
                     2     2
      ′′       1 · (x + y ) − y · (0 + 2y)          2
                                                   x − y2        y 2 − x2
     zxy   =−                                 =− 2            = 2
                         (x2 + y 2 )2             (x + y 2 )2   (x + y 2 )2
                                                       2xy
      ′′
     zyy   = x · (−1) · (x2 + y 2 )−2 · (0 + 2y) = − 2
                                                    (x + y 2 )2




                 ′                  y
We differentiate zx = −                   with respect to y by the quotient rule.
                               x2   + y2
⊳⊳    ⊳    ⊲    ⊲⊲                                                c Robert Maˇık, 2006 ×
                                                                             r´
                                            y
Find derivatives of z(x, y) = arctg           up to the 2nd order.
                                            x
 ′                y          ′        x
zx = −                  ,   zy =            ,
               x2 + y 2            x2 + y 2

                                                       2xy
     zxx = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) =
      ′′
                                                       + y 2 )2
                                                      (x2
                     2     2
      ′′       1 · (x + y ) − y · (0 + 2y)          2
                                                   x − y2        y 2 − x2
     zxy   =−                                 =− 2            = 2
                         (x2 + y 2 )2             (x + y 2 )2   (x + y 2 )2
                                                       2xy
      ′′
     zyy   = x · (−1) · (x2 + y 2 )−2 · (0 + 2y) = − 2
                                                    (x + y 2 )2




We simplify the numerator.
⊳⊳    ⊳    ⊲    ⊲⊲                                               c Robert Maˇık, 2006 ×
                                                                            r´
                                            y
Find derivatives of z(x, y) = arctg           up to the 2nd order.
                                            x
 ′                y          ′        x
zx = −                  ,   zy =            ,
               x2 + y 2            x2 + y 2

                                                       2xy
     zxx = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) =
      ′′
                                                       + y 2 )2
                                                      (x2
                     2     2
      ′′       1 · (x + y ) − y · (0 + 2y)          2
                                                   x − y2        y 2 − x2
     zxy   =−                                 =− 2            = 2
                         (x2 + y 2 )2             (x + y 2 )2   (x + y 2 )2
                                                       2xy
      ′′
     zyy   = x · (−1) · (x2 + y 2 )−2 · (0 + 2y) = − 2
                                                    (x + y 2 )2




We multiply the fraction by −1 which stays in the front of the fraction.
⊳⊳    ⊳    ⊲    ⊲⊲                                               c Robert Maˇık, 2006 ×
                                                                            r´
                                            y
Find derivatives of z(x, y) = arctg           up to the 2nd order.
                                            x
 ′                y          ′        x
zx = −                  ,   zy =            ,
               x2 + y 2            x2 + y 2

                                                       2xy
     zxx = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) =
      ′′
                                                       + y 2 )2
                                                      (x2
                     2     2
      ′′       1 · (x + y ) − y · (0 + 2y)          2
                                                   x − y2        y 2 − x2
     zxy   =−                                 =− 2            = 2
                         (x2 + y 2 )2             (x + y 2 )2   (x + y 2 )2
                                                       2xy
      ′′
     zyy   = x · (−1) · (x2 + y 2 )−2 · (0 + 2y) = − 2
                                                    (x + y 2 )2



We differentiate zy = x · (x2 + y 2 )−1 with respect to y, treating x as a
                 ′

constant and (x2 + y 2 )−1 as a power function with inside function
(x2 + y 2 ).
⊳⊳    ⊳    ⊲    ⊲⊲                                               c Robert Maˇık, 2006 ×
                                                                            r´
                                            y
Find derivatives of z(x, y) = arctg           up to the 2nd order.
                                            x
 ′                y          ′        x
zx = −                  ,   zy =            ,
               x2 + y 2            x2 + y 2

                                                       2xy
     zxx = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) =
      ′′
                                                       + y 2 )2
                                                      (x2
                     2     2
      ′′       1 · (x + y ) − y · (0 + 2y)          2
                                                   x − y2        y 2 − x2
     zxy   =−                                 =− 2            = 2
                         (x2 + y 2 )2             (x + y 2 )2   (x + y 2 )2
                                                       2xy
      ′′
     zyy   = x · (−1) · (x2 + y 2 )−2 · (0 + 2y) = − 2
                                                    (x + y 2 )2




We simplify.
⊳⊳    ⊳    ⊲    ⊲⊲                                               c Robert Maˇık, 2006 ×
                                                                            r´
                                            y
Find derivatives of z(x, y) = arctg           up to the 2nd order.
                                            x
 ′                y          ′        x
zx = −                  ,   zy =            ,
               x2 + y 2            x2 + y 2

                                                       2xy
     zxx = −y · (−1) · (x2 + y 2 )−2 · (2x + 0) =
      ′′
                                                       + y 2 )2
                                                      (x2
                     2     2
      ′′       1 · (x + y ) − y · (0 + 2y)          2
                                                   x − y2        y 2 − x2
     zxy   =−                                 =− 2            = 2
                         (x2 + y 2 )2             (x + y 2 )2   (x + y 2 )2
                                                       2xy
      ′′
     zyy   = x · (−1) · (x2 + y 2 )−2 · (0 + 2y) = − 2
                                                    (x + y 2 )2




All derivatives have been found.
⊳⊳    ⊳    ⊲    ⊲⊲                                               c Robert Maˇık, 2006 ×
                                                                            r´
Find derivatives of z(x, y) =         1 − x2 − y 2 up to the 2nd order.

                   1                                         x
                ′
               zx =  (1 − x2 − y 2 )−1/2 (−2x) = −
                   2                                     1 − x2 − y 2
                   1                                         y
               zx = (1 − x2 − y 2 )−1/2 (−2y) = −
                ′
                   2                                     1 − x2 − y 2

 ′                   x                                   ′               y
zx = −                                                  zy = −
               1−    x2   −   y2                                   1 − x2 − y 2

      ′′             1 − x2 − y 2 − x · 1 · (1 − x2 − y 2 )−1/2 (−2x)
                    1·                   2
     zxx = −
                                     1 − x2 − y 2
                (1 − x2 − y 2 ) + x2          y2 − 1
             =−                      =
                 (1 − x2 − y 2 )3/2     (1 − x2 − y 2 )3/2
                     1                                            xy
      ′′
     zxy     = −x −      (1 − x2 − y 2 )−3/2 (−2y) = −
                     2                                   (1 − x  2 − y 2 )3/2
⊳⊳   ⊳   ⊲     ⊲⊲         2                                       c Robert Maˇık, 2006 ×
                                                                             r´
Find derivatives of z(x, y) =    1 − x2 − y 2 up to the 2nd order.

                 1                                      x
              ′
             zx =  (1 − x2 − y 2 )−1/2 (−2x) = −
                 2                                  1 − x2 − y 2
                 1                                      y
             zx = (1 − x2 − y 2 )−1/2 (−2y) = −
              ′
                 2                                 1 − x2 − y 2

 ′                x                                 ′               y
zx = −                                             zy = −
             1−   x2   −   y2                                1 − x2 − y 2

           1 · 1 − x2 − y 2 − x · 1 · (1 − x2 − y 2 )−1/2 (−2x)
   zxx = −power rule with power 12 and the chain rule
    ′′
We use the                        1 −2x2 − y 2
                 2     2        2
           (1 − x − y ) + x ′             y2 − 1
       =−                         = ′
            (1 − x2 −(g(x)) = f (g(x))− g ′2 )3/2
                     f y 2 )3/2      (1 − x2 · y (x)
                1                                           xy
    ′′
   zxy = −x −       (1 − x2 − y 2 )−3/2 (−2y) = −
                2                                    (1 − x2 − y 2 )3/2
⊳⊳   ⊳   ⊲   ⊲⊲        2                                     c Robert Maˇık, 2006 ×
                                                                        r´
Find derivatives of z(x, y) =     1 − x2 − y 2 up to the 2nd order.

                 1                                       x
              ′
             zx =  (1 − x2 − y 2 )−1/2 (−2x) = −
                 2                                   1 − x2 − y 2
                 1                                       y
             zx = (1 − x2 − y 2 )−1/2 (−2y) = −
              ′
                 2                                   1 − x2 − y 2

 ′                 x                                 ′               y
zx = −                                              zy = −
             1−    x2   −   y2                                 1 − x2 − y 2

      ′′         1 − x2 − y 2 − x · 1 · (1 − x2 − y 2 )−1/2 (−2x)
                  1·                 2
     zxx = −
                                 1 − x2 − y 2
            (1 − x2 − y 2 ) + x2          y2 − 1
       =−                        =
             (1 − x2 − y 2 )3/2     (1 − x2 − y 2 )3/2
                 1                                            xy
We zxy = −x −
   simplify
    ′′
                     (1 − x2 − y 2 )−3/2 (−2y) = −
                 2                                   (1 − x  2 − y 2 )3/2
⊳⊳   ⊳   ⊲   ⊲⊲         2                                     c Robert Maˇık, 2006 ×
                                                                         r´
Find derivatives of z(x, y) =    1 − x2 − y 2 up to the 2nd order.

                 1                                      x
              ′
             zx =  (1 − x2 − y 2 )−1/2 (−2x) = −
                 2                                  1 − x2 − y 2
                 1                                      y
             zx = (1 − x2 − y 2 )−1/2 (−2y) = −
              ′
                 2                                 1 − x2 − y 2

 ′                x                                 ′               y
zx = −                                             zy = −
             1−   x2   −   y2                                1 − x2 − y 2

           1 · 1 − x2 − y 2 − x · 1 · (1 − x2 − y 2 )−1/2 (−2x)
   zxx = −power rule with power 12 and the chain rule
    ′′
We use the                        1 −2x2 − y 2
                 2     2        2
           (1 − x − y ) + x ′             y2 − 1
       =−                         = ′
            (1 − x2 −(g(x)) = f (g(x))− g ′2 )3/2
                     f y 2 )3/2      (1 − x2 · y (x)
                1                                           xy
    ′′
   zxy = −x −       (1 − x2 − y 2 )−3/2 (−2y) = −
                2                                    (1 − x2 − y 2 )3/2
⊳⊳   ⊳   ⊲   ⊲⊲        2                                     c Robert Maˇık, 2006 ×
                                                                        r´
Find derivatives of z(x, y) =     1 − x2 − y 2 up to the 2nd order.

                 1                                        x
              ′
             zx =  (1 − x2 − y 2 )−1/2 (−2x) = −
                 2                                    1 − x2 − y 2
                 1                                        y
             zx = (1 − x2 − y 2 )−1/2 (−2y) = −
              ′
                 2                                    1 − x2 − y 2

 ′                 x                                  ′               y
zx = −                                               zy = −
             1−    x2   −   y2                                  1 − x2 − y 2

      ′′          1 − x2 − y 2 − x · 1 · (1 − x2 − y 2 )−1/2 (−2x)
                  1·                  2
     zxx = −
                                  1 − x2 − y 2
             (1 − x2 − y 2 ) + x2          y2 − 1
       =−                         =
              (1 − x2 − y 2 )3/2     (1 − x2 − y 2 )3/2
                  1                                            xy
We zxy = −x −
   simplify.
    ′′
                      (1 − x2 − y 2 )−3/2 (−2y) = −
                  2                                   (1 − x  2 − y 2 )3/2
⊳⊳   ⊳   ⊲   ⊲⊲         2                                      c Robert Maˇık, 2006 ×
                                                                          r´
Find derivatives of z(x, y) =       1 − x2 − y 2 up to the 2nd order.

 ′                 x                                        ′              y
zx = −                                                     zy = −
             1 − x2 − y 2                                           1 − x2 − y 2

      ′′         1 − x2 − y 2 − x · 1 · (1 − x2 − y 2 )−1/2 (−2x)
                  1·                2
     zxx = −
                                1 − x2 − y 2
           (1 − x2 − y 2 ) + x2          y2 − 1
       =−                       =
            (1 − x2 − y 2 )3/2     (1 − x2 − y 2 )3/2
                 1                                           xy
    ′′
   zxy = −x −       (1 − x2 − y 2 )−3/2 (−2y) = −
                 2                                   (1 − x2 − y 2 )3/2
    ′′        x2 − 1
We zyy = (1 − x2 − rule3/2
   use the quotient y 2 )

                            u   ′       u′ · v − u · v ′
                                    =
                            v                  v2

⊳⊳   ⊳   ⊲   ⊲⊲                                                     c Robert Maˇık, 2006 ×
                                                                               r´
Find derivatives of z(x, y) =         1 − x2 − y 2 up to the 2nd order.

 ′                   x                                  ′               y
zx = −                                                 zy = −
               1 − x2 − y 2                                       1 − x2 − y 2

      ′′            1·1 − x2 − y 2 − x · 1 · (1 − x2 − y 2 )−1/2 (−2x)
                                          2
     zxx = −
                                      1 − x2 − y 2
                 (1 − x2 − y 2 ) + x2          y2 − 1
             =−                       =
                  (1 − x2 − y 2 )3/2     (1 − x2 − y 2 )3/2
                      1                                           xy
      ′′
     zxy     = −x −       (1 − x2 − y 2 )−3/2 (−2y) = −
                      2                                   (1 − x2 − y 2 )3/2
      ′′            x2 − 1
     zyy     =
               (1 − x2 − y 2 )3/2

We multiply both numerator and denominator by the expression
 1 − x2 − y 2 . This removes the composite fraction.
⊳⊳   ⊳   ⊲     ⊲⊲                                                c Robert Maˇık, 2006 ×
                                                                            r´
Find derivatives of z(x, y) =         1 − x2 − y 2 up to the 2nd order.

 ′                   x                                  ′               y
zx = −                                                 zy = −
               1 − x2 − y 2                                       1 − x2 − y 2

      ′′            1·1 − x2 − y 2 − x · 1 · (1 − x2 − y 2 )−1/2 (−2x)
                                          2
     zxx = −
                                      1 − x2 − y 2
                 (1 − x2 − y 2 ) + x2          y2 − 1
             =−                       =
                  (1 − x2 − y 2 )3/2     (1 − x2 − y 2 )3/2
                      1                                           xy
      ′′
     zxy     = −x −       (1 − x2 − y 2 )−3/2 (−2y) = −
                      2                                   (1 − x2 − y 2 )3/2
      ′′            x2 − 1
     zyy     =
               (1 − x2 − y 2 )3/2


We simplify.
⊳⊳   ⊳   ⊲     ⊲⊲                                                c Robert Maˇık, 2006 ×
                                                                            r´
Find derivatives of z(x, y) =    1 − x2 − y 2 up to the 2nd order.

 ′                 x                                ′               y
zx = −                                             zy = −
             1 − x2 − y 2                                    1 − x2 − y 2

      ′′         1 − x2 − y 2 − x · 1 · (1 − x2 − y 2 )−1/2 (−2x)
                  1·                 2
     zxx = −
                                 1 − x2 − y 2
            (1 − x2 − y 2 ) + x2          y2 − 1
       =−                        =
             (1 − x2 − y 2 )3/2     (1 − x2 − y 2 )3/2
                 1                                           xy
    ′′
   zxy = −x −        (1 − x2 − y 2 )−3/2 (−2y) = −
                 2                                   (1 − x2 − y 2 )3/2
    ′′         x2 − 1
   zyy =        2     2
          (1 derivative 3/2
We write the − x − y )with respect to x in the form of
zx = −x · (1 − x2 − y 2 )−1/2 , treat x as a constant (we differentiate
 ′

with respect to y) and use the constant multiple rule and the chain
rule.
⊳⊳   ⊳   ⊲   ⊲⊲                                              c Robert Maˇık, 2006 ×
                                                                        r´
Find derivatives of z(x, y) =         1 − x2 − y 2 up to the 2nd order.

 ′                   x                                  ′               y
zx = −                                                 zy = −
               1 − x2 − y 2                                       1 − x2 − y 2

      ′′            1·1 − x2 − y 2 − x · 1 · (1 − x2 − y 2 )−1/2 (−2x)
                                          2
     zxx = −
                                      1 − x2 − y 2
                 (1 − x2 − y 2 ) + x2          y2 − 1
             =−                       =
                  (1 − x2 − y 2 )3/2     (1 − x2 − y 2 )3/2
                      1                                           xy
      ′′
     zxy     = −x −       (1 − x2 − y 2 )−3/2 (−2y) = −
                      2                                   (1 − x2 − y 2 )3/2
      ′′            x2 − 1
     zyy     =
               (1 − x2 − y 2 )3/2


We simplify.
⊳⊳   ⊳   ⊲     ⊲⊲                                                c Robert Maˇık, 2006 ×
                                                                            r´
Find derivatives of z(x, y) =         1 − x2 − y 2 up to the 2nd order.

 ′                   x                                  ′               y
zx = −                                                 zy = −
               1 − x2 − y 2                                       1 − x2 − y 2

      ′′            1·1 − x2 − y 2 − x · 1 · (1 − x2 − y 2 )−1/2 (−2x)
                                          2
     zxx = −
                                      1 − x2 − y 2
                 (1 − x2 − y 2 ) + x2          y2 − 1
             =−                       =
                  (1 − x2 − y 2 )3/2     (1 − x2 − y 2 )3/2
                      1                                           xy
      ′′
     zxy     = −x −       (1 − x2 − y 2 )−3/2 (−2y) = −
                      2                                   (1 − x2 − y 2 )3/2
      ′′            x2 − 1
     zyy     =
               (1 − x2 − y 2 )3/2


                  ′′                ′′
An evaluation of zyy is similar to zxx .
⊳⊳   ⊳   ⊲     ⊲⊲                                                c Robert Maˇık, 2006 ×
                                                                            r´
                                                                      2
Find derivatives of z(x, y) = (x2 + y)ex                                  −y
                                                                               up to the 2nd order.

                        2                                   2                              2
     zx = 2x · ex
      ′                     −y
                                 + (x2 + y) · ex                −y
                                                                     (2x) = 2xex               −y
                                                                                                    (x2 + y + 1)
                    2                                   2                         2
     zy = 1 · ex
      ′                 −y
                             + (x2 + y) · ex                −y
                                                                 (−1) = ex            −y
                                                                                           (1 − x2 − y 2 )


                    2                                                            2
         zxx = ex
          ′′            −y
                             (2x) · (2x3 + 2xy + 2x) + ex                            −y
                                                                                          (6x2 + 2y + 2)
                        2
             = 2ex          −y
                                 (2x4 + 2x2 y + 2x2 + 3x2 + y + 1)
                        2
             = 2ex          −y
                                 (2x4 + 2x2 y + 5x2 + y + 1)


                                  2                                                  2
                  zxy = ex
                   ′′                 −1
                                           (2x) · (1 − x2 − y) + ex                      −y
                                                                                               · (−2x)
                                       2
                        = 2xex             −y
                                                (1 − x2 − y − 1)
                                           2

⊳⊳   ⊳   ⊲   ⊲⊲
                        = −2xex                −y
                                                    (x2 + y)                                           c Robert Maˇık, 2006 ×
                                                                                                                  r´
                                                                 2
Find derivatives of z(x, y) = (x2 + y)ex                             −y
                                                                          up to the 2nd order.

                        2                              2                              2
     zx = 2x · ex
      ′                     −y
                                 + (x2 + y) · ex           −y
                                                                (2x) = 2xex               −y
                                                                                               (x2 + y + 1)
                    2                              2                         2
     zy = 1 · ex
      ′                 −y
                             + (x2 + y) · ex           −y
                                                            (−1) = ex            −y
                                                                                      (1 − x2 − y 2 )


                    2                                                       2
         zxx = ex
          ′′            −y
                             (2x) · (2x3 + 2xy + 2x) + ex                       −y
                                                                                     (6x2 + 2y + 2)
                        2
             = 2ex          −y
                                 (2x4 + 2x2 y + 2x2 + 3x2 + y + 1)
                        2
             = 2ex          −y
                                 (2x4 + 2x2 y + 5x2 + y + 1)


                                  2                                             2
                    x −1
             ′′
Product rulezxy = e      (2x) · (1 − x2 − y) + ex                                   −y
                                                                                          · (−2x)
                       2        ′    ′         ′
                = 2xex (u · v) =2u ·y − 1) · v
                         −y
                            (1 − x − v + u
                                      2

⊳⊳   ⊳   ⊲   ⊲⊲
                        = −2xex           −y
                                               (x2 + y)                                           c Robert Maˇık, 2006 ×
                                                                                                             r´
                                                                2
Find derivatives of z(x, y) = (x2 + y)ex                            −y
                                                                         up to the 2nd order.

                      2                               2                              2
    zx = 2x · ex
     ′                    −y
                               + (x2 + y) · ex            −y
                                                               (2x) = 2xex               −y
                                                                                              (x2 + y + 1)
                  2                               2                         2
    zy = 1 · ex
     ′                −y
                           + (x2 + y) · ex            −y
                                                           (−1) = ex            −y
                                                                                     (1 − x2 − y 2 )


                 2                                                         2
      zxx = ex
       ′′            −y
                           (2x) · (2x3 + 2xy + 2x) + ex                        −y
                                                                                    (6x2 + 2y + 2)
                      2
          = 2ex           −y
                               (2x4 + 2x2 y + 2x2 + 3x2 + y + 1)
                      2
          = 2ex           −y
                               (2x4 + 2x2 y + 5x2 + y + 1)


                                2                                              2
            zxy = ex
             ′′                     −1
                                         (2x) · (1 − x2 − y) + ex                  −y
                                                                                         · (−2x)
                                     2
                = 2xex −y (1 − x2 − y − 1)
                                         2
We simplify by taking out the factor 2xex −y .
                        2

⊳⊳ ⊳  ⊲  ⊲⊲
                = −2xex −y (x2 + y)                                                              c Robert Maˇık, 2006 ×
                                                                                                            r´
                                                                 2
Find derivatives of z(x, y) = (x2 + y)ex                             −y
                                                                          up to the 2nd order.

                        2                              2                              2
     zx = 2x · ex
      ′                     −y
                                 + (x2 + y) · ex           −y
                                                                (2x) = 2xex               −y
                                                                                               (x2 + y + 1)
                    2                              2                         2
     zy = 1 · ex
      ′                 −y
                             + (x2 + y) · ex           −y
                                                            (−1) = ex            −y
                                                                                      (1 − x2 − y 2 )


                    2                                                       2
         zxx = ex
          ′′            −y
                             (2x) · (2x3 + 2xy + 2x) + ex                       −y
                                                                                     (6x2 + 2y + 2)
                        2
             = 2ex          −y
                                 (2x4 + 2x2 y + 2x2 + 3x2 + y + 1)
                        2
             = 2ex          −y
                                 (2x4 + 2x2 y + 5x2 + y + 1)


We use the product rule
                    2                              2
           zxy = ex −1 (2x) · (1 − x2 − y) + ex −y · (−2x)
            ′′

                      2(u · v)′ = u′ · v + u · v ′
               = 2xex −y (1 − x2 − y − 1)
                                      2

⊳⊳   ⊳   ⊲   ⊲⊲
                        = −2xex           −y
                                               (x2 + y)                                           c Robert Maˇık, 2006 ×
                                                                                                             r´
                                                                2
Find derivatives of z(x, y) = (x2 + y)ex                            −y
                                                                         up to the 2nd order.

                      2                               2                              2
    zx = 2x · ex
     ′                    −y
                               + (x2 + y) · ex            −y
                                                               (2x) = 2xex               −y
                                                                                              (x2 + y + 1)
                  2                               2                         2
    zy = 1 · ex
     ′                −y
                           + (x2 + y) · ex            −y
                                                           (−1) = ex            −y
                                                                                     (1 − x2 − y 2 )


                 2                                                         2
      zxx = ex
       ′′            −y
                           (2x) · (2x3 + 2xy + 2x) + ex                        −y
                                                                                    (6x2 + 2y + 2)
                      2
          = 2ex           −y
                               (2x4 + 2x2 y + 2x2 + 3x2 + y + 1)
                      2
          = 2ex           −y
                               (2x4 + 2x2 y + 5x2 + y + 1)


                                2                                              2
            zxy = ex
             ′′                     −1
                                         (2x) · (1 − x2 − y) + ex                  −y
                                                                                         · (−2x)
                                     2
                = 2xex −y (1 − x2 − y − 1)
                                       2
We simplify by taking out the factor ex −y .
                        2

⊳⊳ ⊳  ⊲  ⊲⊲
                = −2xex −y (x2 + y)                                                              c Robert Maˇık, 2006 ×
                                                                                                            r´
                                                                2
Find derivatives of z(x, y) = (x2 + y)ex                            −y
                                                                         up to the 2nd order.

                        2                             2                              2
     zx = 2x · ex
      ′                     −y
                                 + (x2 + y) · ex          −y
                                                               (2x) = 2xex               −y
                                                                                              (x2 + y + 1)
                    2                             2                         2
     zy = 1 · ex
      ′                 −y
                             + (x2 + y) · ex          −y
                                                           (−1) = ex            −y
                                                                                     (1 − x2 − y 2 )


                    2                                                      2
         zxx = ex
          ′′            −y
                             (2x) · (2x3 + 2xy + 2x) + ex                      −y
                                                                                    (6x2 + 2y + 2)
                        2
         = 2ex −y (2x4 + 2x2 y + 2x2 + 3x2 + y + 1)
                          ′
We write the derivative zx in the form
              x2 −y
         = 2e           4
                    (2x + 2x2 y + 5x2 + y + 1)
                            2
                     zx = ex −y · 2x3 + 2xy + 2x
                      ′


and use the product rule
                    2                               2
            zxy = ex −1 (2x) · (1 − x2 − y) + ex −y · (−2x)
             ′′

                       2(u · v)′ = u′ · v + u · v ′
                = 2xex −y (1 − x2 − y − 1)
                                     2

⊳⊳   ⊳   ⊲   ⊲⊲
                        = −2xex          −y
                                              (x2 + y)                                           c Robert Maˇık, 2006 ×
                                                                                                            r´
                                                                2
Find derivatives of z(x, y) = (x2 + y)ex                            −y
                                                                         up to the 2nd order.

                      2                               2                              2
    zx = 2x · ex
     ′                    −y
                               + (x2 + y) · ex            −y
                                                               (2x) = 2xex               −y
                                                                                              (x2 + y + 1)
                  2                               2                         2
    zy = 1 · ex
     ′                −y
                           + (x2 + y) · ex            −y
                                                           (−1) = ex            −y
                                                                                     (1 − x2 − y 2 )


                 2                                                         2
      zxx = ex
       ′′            −y
                           (2x) · (2x3 + 2xy + 2x) + ex                        −y
                                                                                    (6x2 + 2y + 2)
                      2
          = 2ex           −y
                               (2x4 + 2x2 y + 2x2 + 3x2 + y + 1)
                      2
          = 2ex           −y
                               (2x4 + 2x2 y + 5x2 + y + 1)


                                2                                              2
            zxy = ex
             ′′                     −1
                                         (2x) · (1 − x2 − y) + ex                  −y
                                                                                         · (−2x)
                                     2
                = 2xex −y (1 − x2 − y − 1)
                                         2
We factorize, the repeating factor is 2ex −y .
                          2

⊳⊳ ⊳  ⊲  ⊲⊲
                = −2xex −y (x2 + y)                                                              c Robert Maˇık, 2006 ×
                                                                                                            r´
                                                                2
Find derivatives of z(x, y) = (x2 + y)ex                            −y
                                                                         up to the 2nd order.

                      2                               2                              2
    zx = 2x · ex
     ′                    −y
                               + (x2 + y) · ex            −y
                                                               (2x) = 2xex               −y
                                                                                              (x2 + y + 1)
                  2                               2                         2
    zy = 1 · ex
     ′                −y
                           + (x2 + y) · ex            −y
                                                           (−1) = ex            −y
                                                                                     (1 − x2 − y 2 )


                2                                                          2
     zxx = ex
      ′′            −y
                           (2x) · (2x3 + 2xy + 2x) + ex                        −y
                                                                                    (6x2 + 2y + 2)
                      2
         = 2ex            −y
                               (2x4 + 2x2 y + 2x2 + 3x2 + y + 1)
                      2
         = 2ex            −y
                               (2x4 + 2x2 y + 5x2 + y + 1)


                                2                                              2
           zxy = ex
            ′′                      −1
                                         (2x) · (1 − x2 − y) + ex                  −y
                                                                                         · (−2x)
                                     2
                = 2xex −y (1 − x2 − y − 1)
We simplify in the parenthesis.
                         2

⊳⊳ ⊳  ⊲  ⊲⊲
                = −2xex −y (x2 + y)                                                              c Robert Maˇık, 2006 ×
                                                                                                            r´
                                                                            2
Find derivatives of z(x, y) = (x2 + y)ex                                        −y
                                                                                     up to the 2nd order.

                        2                                         2                              2
     zx = 2x · ex
      ′                     −y
                                 + (x2 + y) · ex                      −y
                                                                           (2x) = 2xex               −y
                                                                                                          (x2 + y + 1)
                    2                                        2                          2
     zy = 1 · ex
      ′                 −y
                             + (x2 + y) · ex                     −y
                                                                       (−1) = ex            −y
                                                                                                 (1 − x2 − y 2 )


                                 2                                                      2
                  zxy = ex
                   ′′                −1
                                          (2x) · (1 − x2 − y) + ex                          −y
                                                                                                     · (−2x)
                                      2
                        = 2xex            −y
                                                  (1 − x2 − y − 1)
                                          2

We start with           = −2xex               −y
                                                   (x2 + y)
                                                        2
                                      zy = ex
                                       ′                    −y
                                                                 (1 − x2 − y 2 )
and differentiate with respect to x by the product rule

                  zyy = ex
                   ′′            2
                                      (u
                                     −y     · v)′ = u′ · v + u v ′ 2
                                          (−1) · (1 − x2 − y)·+ ex −y · (−1)
                                              2
                        = (−1)ex                  −y
                                                       (2 − x2 − y)
⊳⊳   ⊳   ⊲   ⊲⊲                                                                                              c Robert Maˇık, 2006 ×
                                                                                                                        r´
                                                                     2
Find derivatives of z(x, y) = (x2 + y)ex                                 −y
                                                                              up to the 2nd order.

                        2                                  2                              2
     zx = 2x · ex
      ′                     −y
                                 + (x2 + y) · ex               −y
                                                                    (2x) = 2xex               −y
                                                                                                   (x2 + y + 1)
                    2                                  2                         2
     zy = 1 · ex
      ′                 −y
                             + (x2 + y) · ex               −y
                                                                (−1) = ex            −y
                                                                                          (1 − x2 − y 2 )


                                 2                                               2
                  zxy = ex
                   ′′                −1
                                          (2x) · (1 − x2 − y) + ex                   −y
                                                                                              · (−2x)
                                      2
                        = 2xex            −y
                                               (1 − x2 − y − 1)
                                          2
                        = −2xex               −y
                                                   (x2 + y)




                                 2                                                   2
            zyy = ex −y (−1) · (1 − x2 − 2y) + ex
              ′′                                                                         −y
                                                                                              · (−1)
We factorize, the repeating factor is 2xex −y .
                          2
                 = (−1)ex −y (2 − x2 − y)
⊳⊳   ⊳   ⊲   ⊲⊲                                                                                       c Robert Maˇık, 2006 ×
                                                                                                                 r´
                                                                     2
Find derivatives of z(x, y) = (x2 + y)ex                                 −y
                                                                              up to the 2nd order.

                        2                                  2                              2
     zx = 2x · ex
      ′                     −y
                                 + (x2 + y) · ex               −y
                                                                    (2x) = 2xex               −y
                                                                                                   (x2 + y + 1)
                    2                                  2                         2
     zy = 1 · ex
      ′                 −y
                             + (x2 + y) · ex               −y
                                                                (−1) = ex            −y
                                                                                          (1 − x2 − y 2 )


                                 2                                               2
                  zxy = ex
                   ′′                −1
                                          (2x) · (1 − x2 − y) + ex                   −y
                                                                                              · (−2x)
                                      2
                        = 2xex            −y
                                               (1 − x2 − y − 1)
                                          2
                        = −2xex               −y
                                                   (x2 + y)




                                 2                                                   2
                  zyy = ex
                   ′′                −y
                                          (−1) · (1 − x2 − y) + ex                       −y
                                                                                              · (−1)
We simplify.
                                          x2 −y            2
                        = (−1)e                    (2 − x − y)
⊳⊳   ⊳   ⊲   ⊲⊲                                                                                       c Robert Maˇık, 2006 ×
                                                                                                                 r´
                                                                         2
Find derivatives of z(x, y) = (x2 + y)ex                                     −y
                                                                                  up to the 2nd order.

                        2                                      2                              2
     zx = 2x · ex
      ′                     −y
                                 + (x2 + y) · ex                   −y
                                                                        (2x) = 2xex               −y
                                                                                                       (x2 + y + 1)
                    2                                     2                          2
     zy = 1 · ex
      ′                 −y
                             + (x2 + y) · ex                  −y
                                                                    (−1) = ex            −y
                                                                                              (1 − x2 − y 2 )


                                 2                                                       2
                  zyy = ex
                   ′′                −y
                                          (−1) · (1 − x2 − y) + ex                           −y
                                                                                                  · (−1)
                                           2
        ′′     = (−1)ex                        −y
                                                    (2 − x2 − y)
To find zyy we start with
                                                     2
                                     zy = ex
                                      ′                  −y
                                                              (1 − x2 − y 2 )

and use the product rule

                                      (u · v)′ = u′ · v + u · v ′ .

⊳⊳   ⊳   ⊲   ⊲⊲                                                                                           c Robert Maˇık, 2006 ×
                                                                                                                     r´
                                                                       2
Find derivatives of z(x, y) = (x2 + y)ex                                   −y
                                                                                up to the 2nd order.

                        2                                    2                              2
     zx = 2x · ex
      ′                     −y
                                 + (x2 + y) · ex                 −y
                                                                      (2x) = 2xex               −y
                                                                                                     (x2 + y + 1)
                    2                                    2                         2
     zy = 1 · ex
      ′                 −y
                             + (x2 + y) · ex                 −y
                                                                  (−1) = ex            −y
                                                                                            (1 − x2 − y 2 )


                                 2                                                     2
                  zyy = ex
                   ′′                −y
                                          (−1) · (1 − x2 − y) + ex                         −y
                                                                                                · (−1)
                                           2
                        = (−1)ex               −y
                                                    (2 − x2 − y)




                                                    2
The repeating factor is (−1)ex                          −y


⊳⊳   ⊳   ⊲   ⊲⊲                                                                                         c Robert Maˇık, 2006 ×
                                                                                                                   r´
                                                       2
                                                           +y 2
Find derivatives of z(x, y) = ex                                  up to the 2nd order.

                         2
                             +y 2
              zx = ex
               ′
                                      · 2x
               ′        x2 +y 2
              zy   =e                 · 2y
                         2        2                          2
                                                                 +y 2               2
                                                                                        +y 2
             zxx = ex
              ′′             +y
                                      2x · 2x + ex                      · 2 = 2ex              (1 + 2x2 )
                                      2        2                    2        2
             zxy = 2x · ex
              ′′                          +y
                                                   2y = 4xyex           +y

                         2
                             +y 2                           2
                                                                +y 2                2
                                                                                        +y 2
             zyy = ex
              ′′
                                      2y · 2y + ex                      · 2 = 2ex              (1 + 2y 2 )




⊳⊳   ⊳   ⊲   ⊲⊲                                                                                      c Robert Maˇık, 2006 ×
                                                                                                                r´
                                                          2
                                                              +y 2
Find derivatives of z(x, y) = ex                                     up to the 2nd order.

                         2
                             +y 2
              zx = ex
               ′
                                      · 2x
               ′        x2 +y 2
              zy   =e                 · 2y
                         2        2                             2
                                                                    +y 2               2
                                                                                           +y 2
             zxx = ex
              ′′             +y
                                      2x · 2x + ex                         · 2 = 2ex              (1 + 2x2 )
                                      2        2                       2        2
             zxy = 2x · ex
              ′′                          +y
                                                   2y = 4xyex              +y

                         2
                             +y 2                              2
                                                                   +y 2                2
                                                                                           +y 2
             zyy = ex
              ′′
                                      2y · 2y + ex                         · 2 = 2ex              (1 + 2y 2 )




The chain rule                                        ′
                                  f (g(x))                = f ′ (g(x)) · g ′ (x).


⊳⊳   ⊳   ⊲   ⊲⊲                                                                                         c Robert Maˇık, 2006 ×
                                                                                                                   r´
                                                          2
                                                              +y 2
Find derivatives of z(x, y) = ex                                     up to the 2nd order.

                         2
                             +y 2
              zx = ex
               ′
                                      · 2x
               ′        x2 +y 2
              zy   =e                 · 2y
                         2        2                             2
                                                                    +y 2               2
                                                                                           +y 2
             zxx = ex
              ′′             +y
                                      2x · 2x + ex                         · 2 = 2ex              (1 + 2x2 )
                                      2        2                       2        2
             zxy = 2x · ex
              ′′                          +y
                                                   2y = 4xyex              +y

                         2
                             +y 2                              2
                                                                   +y 2                2
                                                                                           +y 2
             zyy = ex
              ′′
                                      2y · 2y + ex                         · 2 = 2ex              (1 + 2y 2 )




The chain rule                                        ′
                                  f (g(x))                = f ′ (g(x)) · g ′ (x).


⊳⊳   ⊳   ⊲   ⊲⊲                                                                                         c Robert Maˇık, 2006 ×
                                                                                                                   r´
                                                       2
                                                           +y 2
Find derivatives of z(x, y) = ex                                  up to the 2nd order.

                         2
                             +y 2
              zx = ex
               ′
                                      · 2x
               ′        x2 +y 2
              zy   =e                 · 2y
                         2        2                          2
                                                                 +y 2               2
                                                                                        +y 2
             zxx = ex
              ′′             +y
                                      2x · 2x + ex                      · 2 = 2ex              (1 + 2x2 )
                                      2        2                    2        2
             zxy = 2x · ex
              ′′                          +y
                                                   2y = 4xyex           +y

                         2
                             +y 2                           2
                                                                +y 2                2
                                                                                        +y 2
             zyy = ex
              ′′
                                      2y · 2y + ex                      · 2 = 2ex              (1 + 2y 2 )




The product rule
                                          (u · v)′ = u′ · v + u · v ′ .

⊳⊳   ⊳   ⊲   ⊲⊲                                                                                      c Robert Maˇık, 2006 ×
                                                                                                                r´
                                                       2
                                                           +y 2
Find derivatives of z(x, y) = ex                                  up to the 2nd order.

                         2
                             +y 2
              zx = ex
               ′
                                      · 2x
               ′        x2 +y 2
              zy   =e                 · 2y
                         2        2                          2
                                                                 +y 2               2
                                                                                        +y 2
             zxx = ex
              ′′             +y
                                      2x · 2x + ex                      · 2 = 2ex              (1 + 2x2 )
                                      2        2                    2        2
             zxy = 2x · ex
              ′′                          +y
                                                   2y = 4xyex           +y

                         2
                             +y 2                           2
                                                                +y 2                2
                                                                                        +y 2
             zyy = ex
              ′′
                                      2y · 2y + ex                      · 2 = 2ex              (1 + 2y 2 )




Simplifying.
⊳⊳   ⊳   ⊲   ⊲⊲                                                                                      c Robert Maˇık, 2006 ×
                                                                                                                r´
                                                       2
                                                           +y 2
Find derivatives of z(x, y) = ex                                  up to the 2nd order.

                         2
                             +y 2
              zx = ex
               ′
                                      · 2x
               ′        x2 +y 2
              zy   =e                 · 2y
                         2        2                          2
                                                                 +y 2               2
                                                                                        +y 2
             zxx = ex
              ′′             +y
                                      2x · 2x + ex                      · 2 = 2ex              (1 + 2x2 )
                                      2        2                    2        2
             zxy = 2x · ex
              ′′                          +y
                                                   2y = 4xyex           +y

                         2
                             +y 2                           2
                                                                +y 2                2
                                                                                        +y 2
             zyy = ex
              ′′
                                      2y · 2y + ex                      · 2 = 2ex              (1 + 2y 2 )




The constant multiple rule.
⊳⊳   ⊳   ⊲   ⊲⊲                                                                                      c Robert Maˇık, 2006 ×
                                                                                                                r´
                                                       2
                                                           +y 2
Find derivatives of z(x, y) = ex                                  up to the 2nd order.

                         2
                             +y 2
              zx = ex
               ′
                                      · 2x
               ′        x2 +y 2
              zy   =e                 · 2y
                         2        2                          2
                                                                 +y 2               2
                                                                                        +y 2
             zxx = ex
              ′′             +y
                                      2x · 2x + ex                      · 2 = 2ex              (1 + 2x2 )
                                      2        2                    2        2
             zxy = 2x · ex
              ′′                          +y
                                                   2y = 4xyex           +y

                         2
                             +y 2                           2
                                                                +y 2                2
                                                                                        +y 2
             zyy = ex
              ′′
                                      2y · 2y + ex                      · 2 = 2ex              (1 + 2y 2 )




Simplifying.
⊳⊳   ⊳   ⊲   ⊲⊲                                                                                      c Robert Maˇık, 2006 ×
                                                                                                                r´
                                                       2
                                                           +y 2
Find derivatives of z(x, y) = ex                                  up to the 2nd order.

                         2
                             +y 2
              zx = ex
               ′
                                      · 2x
               ′        x2 +y 2
              zy   =e                 · 2y
                         2        2                          2
                                                                 +y 2               2
                                                                                        +y 2
             zxx = ex
              ′′             +y
                                      2x · 2x + ex                      · 2 = 2ex              (1 + 2x2 )
                                      2        2                    2        2
             zxy = 2x · ex
              ′′                          +y
                                                   2y = 4xyex           +y

                         2
                             +y 2                           2
                                                                +y 2                2
                                                                                        +y 2
             zyy = ex
              ′′
                                      2y · 2y + ex                      · 2 = 2ex              (1 + 2y 2 )




The product rule
                                          (u · v)′ = u′ · v + u · v ′ .

⊳⊳   ⊳   ⊲   ⊲⊲                                                                                      c Robert Maˇık, 2006 ×
                                                                                                                r´
                                                       2
                                                           +y 2
Find derivatives of z(x, y) = ex                                  up to the 2nd order.

                         2
                             +y 2
              zx = ex
               ′
                                      · 2x
               ′        x2 +y 2
              zy   =e                 · 2y
                         2        2                          2
                                                                 +y 2               2
                                                                                        +y 2
             zxx = ex
              ′′             +y
                                      2x · 2x + ex                      · 2 = 2ex              (1 + 2x2 )
                                      2        2                    2        2
             zxy = 2x · ex
              ′′                          +y
                                                   2y = 4xyex           +y

                         2
                             +y 2                           2
                                                                +y 2                2
                                                                                        +y 2
             zyy = ex
              ′′
                                      2y · 2y + ex                      · 2 = 2ex              (1 + 2y 2 )




Simplifying.
⊳⊳   ⊳   ⊲   ⊲⊲                                                                                      c Robert Maˇık, 2006 ×
                                                                                                                r´
That’s all . . .




⊳⊳   ⊳   ⊲   ⊲⊲    c Robert Maˇık, 2006 ×
                              r´

				
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