# parc-der

Document Sample

```					                  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

(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)
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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