# Differentiation - PDF by examville

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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
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

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

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

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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