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DIFFERENTIATION
Let us take the graph of a function f . We know that the point ( x,f(x) ) is a point on the
graph. What line, if any, should be the tangent to the graph at this point?




We can take another point on ( x+h, f(x+h) ) on the graph and draw a secant line through
these two points. Keeping ( x,f(x) ) fixed, we move the other point closer to it (h tends to
zero from the left or from the right). The secant line tends to a limiting position. This is
“the tangent to the graph at the point ( x,f(x) )”. If this limit exists, that is defined to be the
slope of the tangent line at that fixed point ( x,f(x) ) on the graph of y = f (x).
                            DIFFERENTIABILITY &
                 THE DERIVATIVE OF A FUNCTION


Differentiation is the process of finding the derivative of a function.
A function f is said to be differentiable at x if and only if
                                      ffff+fffffffff
                                       `      a   ` a
                                         x fh @f x
                                        ffffffffffff
                                        ffffffffffff
                                       ffffffffffff
                                  lim               exists.
                                 hQ0         h
                                                                                 ` a
If this limit exists, it is called the derivative of f at x and is denoted by f. x .
                                                    x + f fff f
                                                    `      a     ` a
                                                   ffffffffffxff
                                                   ffffhf@ f fff
                                                 ffffffffffffff
                                                  fffffffffffff
                           Therefore, f. x = lim
                                        ` a
                                                                .
                                            hQ 0         h
                                                                                         ` a
If we draw a tangent to the graph at this point, the slope of the tangent is equal to f. x .




The derivative of a function y = f(x) with respect to x at a point x = x 0 is given by

                 x 0 + f ff x 0
                b      c     `   a
                           f
              ffffffhf@ffffff
               fffffffffffff
                ffffffff fffff
 f. x 0 = lim ffffffffffffff provided that this limit exists.
   ` a
         hQ0            h
This limit is also called the instantaneous rate of change of y with respect to x at x = x 0 .


                                                     ` a d f ` a dyf
                                                           ff
                                                           ff
                                                           f       ff
                                                                   ff
                                                                   f
The d/dx notation : If y = f(x), then we can write f. x = fff x = ff
                                                          dx      dx


The derivative of the function y =f(x) with respect to x may be indicated by any one of
                                                               ` a
              dff
              ff` a
               ff
              ff   dyf
                   ff
                    f
                    ff
                    ff         ` a                        dfffff
                                                           ffx f
                                                              f
                                                            ffff
                                                           fffff
the symbols    y ,    , y. , f. x , Dx , Dx f,                  . We can use any of these,
            dx     dx                                       dx
while solving problems.
   •   A function is said to be differentiable at a point x = x 0 if the derivative of the
       function exists at that point.
   •   Also, if a function is differentiable at every point in an interval, the function is
       said to be differentiable on that interval.
   •   If we say “a function (of x) is differentiable” without mentioning the interval, we
       mean that it is differentiable for every value of x in its domain.
Example :
                         f x = x 3 @ 3x
                             ` a
   1. The function                               is differentiable on the interval (-1,1) as it is
       differentiable at every point in that interval.
   2. The function f x = |x @ 2| is not differentiable on the interval (1,3) as it is not
                         ` a


       differentiable at x = 2 which lies in that interval.


Example : Find f’(x) for the function f x = x 2 + 4 .
                                             ` a


Solution :
                We formthe difference quotient
                                    +fffffffffffff4 f
                                             @ f2 + f
                                      hff+fffffffffff
                               B`      a2   C B      C
                ffff+fffffffff fffffffff
                                  xfffffffffffxfffff
                                           4ffff fffff
                  `     a  ` a
                    x fh @f x
                  ffffffffffff fff
                  ffffffffffff f
                 ffffffffffff fff f f
                                   =
                         h                     h
                              + 2xh + h + 4 @ x 2 @ 4   ff+ h
                            2          2                      2
                          xfffffffffffffffffffff 2xhfffff
                            fffffffffffffffffffff fffffff
                            ffffffffffffffffffff ffffff
                           fffffffffffffffffffff fffffff
                        =                             =         = 2x + h
                                       h                  h
                                        x + h @f x
                                             `        a    ` a
                                     fffffffffffff
                                      ffffffffffff
                                       ffffffffffff
                Therefore f. x = lim fffffffffffff lim 2x + h = 2x
                                                   =
                               ` a                       `   a
                                hQ0          h       hQ0




Example : Find f. @ 3 and f. @ 1 given that f x = x 2
                    `    a           `   a                ` a


Solution :
                                               @ f + h fff @ 3
                                                         `           a     `   a
                                              ffffffffff fffff
                                            ffff3 ffff@fffffff
                                              fffffffffffffff
                By definition f. @ 3 = lim ffffffffffffffff
                               `       a
                                      hQ0              h
                                              f3 + h @ @ 3
                                           `         a2 `   a2
                                             @ffffffffffffff
                                             ffffffffffffff
                                             ffffffffffffff
                                            fffffffffffffff
                                     = lim
                                      hQ0            h c
                                               @ 6h + h ff
                                           b
                                                         2
                                             9 fffffffff@ff
                                            ffffffffffffff
                                             ffffffffffffff
                                             fffffffffff f
                                              f
                                                           9
                                     = lim
                                      hQ0           h
                                               6h + h
                                                      2
                                           @fffffff
                                             ffffffff
                                             fffffff
                                            ffffffff
                                     = lim
                                      hQ0       h a
                                     = lim @ 6 + h
                                           `
                                               hQ0
                                           =@6
Similarly
                                            @ f + h fff @ 3
                                                         `           a    `    a
                                           ffffffffff fffff
                                         ffff1 ffff@fffffff
                                           fffffffffffffff
                            f. @ 1 = lim ffffffffffffffff
                              `        a
                                    hQ0             h
                                           f1 + f @ fff
                                         `        a2 `   a2
                                           ffffffffff 1 ff
                                          fffffhffff@ffff
                                          @fffffffffffff
                                           ffffffffffffff
                                   = lim
                                     hQ0          h
                                   = lim @ 2 + h
                                         `        a
                                               hQ0
                                           =@2

                              ` a                                                  ` a
Alternative method : Find f. x from f(x) first, and to put x=-3 and x=-1 in f. x to get
   `   a       `   a
f. @ 3 and f. @ 1 as shown below


                                          ffff+fffffffff
                                                         `       a       ` a
                                             x fh @f x
                                            ffffffffffff
                                            ffffffffffff
                                           ffffffffffff
                              f. x = lim
                                   ` a
                                    hQ0            h
                                           xf+ffffffff
                                         `       a2     2
                                             fh @x
                                           ffffffffff
                                            ffffffffff
                                           ffffffffff
                                   = lim
                                     hQ0         h
                                                       +fffffff
                                               +fffffffffffff
                                         b                 c
                                             2            2
                                                         h @ f2
                                                 2xhfffffffff
                                           xfffffffffffffxff
                                           fffffff
                                            fff
                                            f
                                           fffffff f
                                   = lim
                                     hQ0              h
                                            ff+ h
                                                    2
                                         2xhfffff
                                           fffffff
                                            ffffff
                                           fffffff
                                   = lim
                                     hQ0       h
                                   = lim 2x + h
                                         `         a
                                               hQ0
                                           = 2x

                   Therefore f. @ 3 = 2 A @ 3 = @ 6
                                   `       a         `       a

                     and f. @ 1 = 2 A @ 1 = @ 2
                           `    a    `    a




Example : Find the derivative of f x = x 3 + 5 at the point (2,13).
                                       ` a
Solution :
         f x = x3 + 5
             ` a

                                 x + h @f x
                               `     a        ` a
                             fffffffffffff
                   =            ffffffffffff
                               ffffffffffff
                         lim fffffffffffff
             ` a
        f. x
                        hQ0            h
                            B`                C B      C
                                     hff+fffffffffff
                                               @ f3 + f
                                   +fffffffffffff5 f
                                      a3
                                            5ffff fffff
                                xfffffffffffxfffff
                   =           fff
                                f
                              fff f f
                         lim fffffffff
                        hQ0                 h
                                             + fff + f + f fffffff
                                   +fffffff3xhfffhffffffffffff
                             b                            c
                                 3       2          2  3
                                           hfffffffffffffff x 3 @ f
                                     3xffffffffffffffff@ffffff
                               xfffffffffffffffffff5fffffff5f
                   =           fff
                                f
                              fffff
                         lim ffffffff
                        hQ0                         h
                                f h + 3xh + h
                                 2           2    3
                             3xfffffffffffff
                   =           fffffffffffff
                              fffffffffffff
                         lim ffffffffffffff
                        hQ0             h
                   =     lim 3x 2 + 3xh + h
                            b                    c
                                                2
                        hQ0

                   = 3x   2



        Therefore f. 2 = 3.2                  = 12
                           ` a           2




        Hence the derivative of f x = x 3 + 5 at the point 2,13 is 12 A
                                             ` a                     b     c




                                    COMPARISON OF
              DIFFERENTIABILITY AND CONTINUITY

                       If a function f is differentiable x, it is continuous at x.


   But if a function is continuous at x, it may or may not be differentiable at x.
Example :
   1. The function f x = x 3 @ 3x is differentiable on its whole domain (which is R).
                              ` a


       As differentiability implies continuity, it is continuous at every point.
   2. f(x) = |x| is continuous at any real value of x. But as continuity does not imply
       differentiability, we have to check its differentiability at every at every point. We
       find that at x=0, it is not differentiable though continuous.
                                2
                          ` a xfffff
                                  @4
   3. The function              ffff
                                fffff
                         f x = fffff is not continuous at             x = 2 . As continuity is
                               x @2
       necessary for differentiability, it is not differentiable at that point.



                   DIFFERENTIATION FORMULAS

We can use the limit definition of derivative to find the derivative of any function, but
this application may be cumbersome and very long at times. For example, finding the
                  f+fff@ 9 f
                  2
                    f 5x fff
                xfffffffff
                  fffffffff
                 fffffffff
derivative of                by the limit method will be very long. So, many useful
                 x @x + 2
                   3


differentiation formulas are first proven using the definition of derivative, and then these
are used directly in solving problems.


                           DIFFERENTIATION OF
                    COMBINATION OF FUNCTIONS

The functions to be differentiated are usually combination of two or more functions. Here
we discuss about the combination of two functions by addition, subtraction,
multiplication, division only and also scalar multiples of functions. (Differentiation of
composite functions are discussed separately in Chain Rule).
                                                           ` a       ` a   ` a
In the following formulas u , v and w, or f x , g x , h x are functions of x and are
differentiable, and c is a constant number

                                                 dff
                                                 ff`
                                                 ff
                                                  ff        duf
                                                              ff
                                                              ff
                                                    cu = c A fff
                                                      a
Scaler multiple of a function :
                                                 dx          dx
                                            dff
                                            ff
                                             ff
                                            ff
                                                 c A f x = c f. x
                                              B        ` aC          ` a
                                 [
                                            dx
                            dff        duf dvf
                            ff
                             ff         ff ff
                                        ff ff
                                         f
Sum and Difference Rule : ff u F v = ffF ff f
                               `   a
                           dx          dx dx
                         dff
                         ff
                          ff
                         ff ` a
                              f x0 F g x = f. x F g. x
                           B          ` aC    ` a   ` a
                   [
                        dx
                    dff
                     ff
                    ff`
                    ff            dvf     duf
                                   ff
                                   ff
                                   f
                        uv = u A ff+ v A ff ff
                                            f
                                           ff
                           a
Product Rule :
                   dx             dx      dx
                    dff ` a ` aC ` a ` a ` a ` a
                    ff
                   ffff
                        f x A g x = g x A f. x + f x A g. x
                      B
           [
                   dx
we can also extend this to product of three functions as
                 dff
                 ff
                  ff
                ff`               dwf         dvf        duf
                                   ff
                                   ff
                                    f
                     uvw = uv A fff uw A ff + vw A ff
                                       +       f
                                               ff
                                               ff         f
                                                          ff
                                                          ff
                         a
                dx                dx          dx         dx
                                                 du           dv
                        dff f
                        ff uf
                          d e
                        ff f
                         ff f
                                               A fff@ffffff
                                                 ff
                                                  f
                                                  ff
                                                 ff
                                             vffffffffffff
                                                      u A fff
                                               f fff
                                               fffffffffff
                                              ffffffffffff
                                                   f
                                                          ff
                                                           ff
                                                           ff
                                                 dx           dx
Quotient Rule :                        =
                        dx v                          v2
                           H ` aI                 ` a      ` a       ` a    ` a
                               x
                        dff fffff
                         ff fff
                        ff ffff
                        ffJ ffff                   ffffffff@ f ffffffx f
                                                 gfffff.ffffffffffg.ffff
                                                    x A f x fff x A ff f
                                                   ffffffffffffffffffff
                                                  ffffffffffffffffffff
               [              ` aK          =                 B ` aC2
                        dx       g x
                                                               g x




        DIFFERENTIATION OF x n AND
                                                                                  `            a
                                                                                  c CONSTANT


                   dff
                   ff
                    ff
                   ff` a
                         c =0                              f x = c, then f. x = 0
                                                              ` a                 ` a
                                           that is, if
                   dx
                   dff
                   ff
                    ff
                   ff`
                         xn = n A xn@ 1
                             a
                   dx
        using this we can write the derivative of another two common functions
                        dff
                        ff
                         ff
                       ff` a
                            x =1
                       dx
                             ff 1f
                             dff f    1f
                               f g
                              ff f
                             ff f     ff
                                       ff
                                  = @ ff
                             dx x    x2
                           1f      1f
                             f      ff
Example : Differentiate y = fx 3 @ fx 2 + x @ 3
                           6       4


Answer :
                dyf ff 1f 3 1f 2
                ff d f f
                           f                               g
                 ff ff
                 f   f
                 ff ff f     f
                             ff
                   =     x @ x + x @3                               using the sum @ difference rule
                dx dx 6     4
                       dff f
                       ff 1f 3             ff 1f 2
                                           dff f               dff        dff
                         f     g             f     g
                        ff
                       ff f                 ff f
                                           ff f                ff` a
                                                               ff
                                                                ff        ff
                                                                           ff
                                                                          ff` a
                   =           x       @          x        +        x @        3
                       dx 6                dx 4                dx         dx
                    1fb c 1f ` a ` a ` a
                     ff      fff
                   = A 3x 2 @ A 2x + 1 @ 0
                    6        4
                      2
                    xff xf
                      ff f
                      f
                     ff f  f
                   = @ +1
                     2 2


Example : Differentiate y = x 2 @ 2x + 3 x 3 @ x 2 + 2
                                   b                  cb                c




Answer :
               dyf ff b 2
               ff d f
                           D                                        E
                ff ff
                f
                ff ff
                  = f x @ 2x + 3 x 3 @ x 2 + 2
                                       cb          c
                                                             using the product rule
               dx dx
                                 c d b                            c d b
                                     ff
                                     ff
                                      ff                              ff
                                                                      ff
                                                                       ff
                  = x 3 @ x 2 + 2 A ff x 2 @ 2x + 3 + x 2 @ 2x + 3 A ff x 3 @ x 2 + 2
                   b                               c b                               c
                                    dx                               dx
                  = x @ x + 2 A 2x @ 2 + x @ 2x + 3 A 3x @ 2x
                   b             c`        a b 2          cb         c
                      3     2                                  2



                  = 2 x @ 1 x 3 @ x 2 + 2 + x 3x @ 2 x 2 @ 2x + 3
                       `       ab                 c            `        ab          c




                              f+ff
                              3
                            xfffff
                                f1
                             fffff
Example : Differentiate y = fffff
                             3
                            x @1


Answer :
                      ff ff f+ff
                              3
                     dyf ff xfffff
                     ff d f fffff
                                1
                                   f          g
                      f
                      ff ff fffff
                        = f 3f                                   using the quotient rule
                     dx dx x @ 1
                                             d                                        d
                                   1 A fffx 3 + 1 @ x 3 + 1 A fffx 3 @ 1
                              b          c        b             c       b         c        b       c
                              x f@ffffffffffffffffffffffffffffff
                                3
                               f ff
                                       ff
                                        f
                                        ff
                                        f
                              ffffffffffffffffffffffffffffffffff
                               fff fffffffffffffffffffffffffffff
                              ffffffffffffffffffffffffffffffffff
                                                              ff
                                                              ff
                                                               f
                                                               f
                                             dx                                       dx
                          =                                b                c2
                                                               x3@ 1
         
								
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