Calculus

DIFFERENTIATION-II 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 derivative of

2 xf +f 5x @ 9f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f by the limit method will be very long. So, many useful 3 x @x + 2

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 a du f f f f f f f cu = c A f dx dx ` df f f f f f f f

Scaler multiple of a function :

B df f f f f f f f

[

dx

c A f x = c f. x ` aC

` a

` a df du f f f f f f f f f f f f f dv f f f f f f f Sum and Difference Rule : f uFv = f F f dx dx dx B ` aC ` a ` a df f f f f f f f ` a [ f x0 F g x = f. x F g. x dx

Product Rule : [

` df f f f f f f f

a dv du f f f f f f f f f f f f f f uv = u A f + vA f dx dx dx B ` a ` aC ` a ` a ` a ` a d f f f f f f f f f x A g x = g x A f. x + f x A g. x dx

we can also extend this to product of three functions as ` a dw dv du df f f f f f f f f f f f f f f f f f f f f f f f f f f f f f uvw = uv A f + uw A f + vw A f dx dx dx dx Quotient Rule : [

d e df f f f f f f f uf f f f

dx v

=

H ` aI xf d f f f f f f f f f f f f f f f f f f f f fJ ff ` aK

dx

g x

du dx

dv dx

f f f f f f f f f f f f f f f f f f f @ uf Af vf Af f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f

v2 =

` a

` a

` a

` a

Af f. xf @ ff xf Af g. xf gf xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f B ` aC2

g x

DIFFERENTIATION OF x n AND ` a df f f f f f f f

c =0

dx

` df f f f f f f f

dx

that is, if

`

c CONSTANT

a

f x = c, then f. x = 0 ` a

` a

xn = n A xn@ 1 a

using this we can write the derivative of another two common functions ` a df f f f f f f f x =1 dx f g df f f f f f f f 1f f f f

1f f f f f f =@ f dx x x2

1f f 3 1f f f2 Example : Differentiate y = f x @ f x + x @3 6 4 Answer : dy df f f f f f f f f f f f f f f f 1f f f 3 1f f f f = x @ x2 + x @ 3 dx dx 6 4 f

=

f g df f f f f f f f 1f f f3

dx 6

x

@

f g df f f f f f f f 1f f f f2

dx 4

x

g

+

using the sum @ difference rule ` a df f f f f f f f

dx

x @

b c 1 ` a ` a ` a 1f f f f f f f = A 3x 2 @ A 2x + 1 @ 0 6 4 2 xf f f f f f f xf f f f = @ +1 2 2

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

Answer :

cb

` a df f f f f f f f

dx

c

3

b cb c dy df f f f f f f f f f f f f f f 2 = f x @ 2x + 3 x 3 @ x 2 + 2 using the product rule dx dx b c d b c b c d b c f f f f f f f 2 f f f f f f f 3 x @ 2x + 3 + x 2 @ 2x + 3 A f x @ x2 + 2 = x3@ x2 + 2 A f dx dx b c` cb c a b 2 3 2 2 = x @ x + 2 A 2x @ 2 + x @ 2x + 3 A 3x @ 2x D

E

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

`

c

ab

`

c

3 xf +f 1f f f f f f f f f f f f f f f f Example : Differentiate y = f 3 x @1

Answer : 3 dy df +f 1f f f f f f f f f f f f f f f xf f f f f f f f f f f f f f f f f = f dx dx x 3 @ 1

f

=

=

g

d dx

using the quotient rule d dx

f f f f f f f f f 3 f f f f f f f f f 3 3 3 xf @ 1f Af xf +f 1f @ xf +f 1f Af xf @ 1f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f

b

c

b

b

c

b

c2

c

b

c

x3@ 1

3 2 3 2 xf @ 1f Af 3x +f 0f @ xf + 1f Af 3x @ 0f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f

b

cb

b

c

b

c2

x3@ 1

5 2 5 2 3x @ 3x @ 3x @ 3x f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = f b c2 x3@ 1 2 6x f f f f f f f f f f f f f f f f f f f f f f f f f f f = @ bf c2 3 x @1

cb

c

DIFFERENTIATION OF TRIGONOMETRIC FUNCTIONS ` df f f f f f f f

dx

` df f f f f f f f

dx

a

dx

a

dx

a

dx

a

cot x = @csc 2 x

` df f f f f f f f

sec x = sec x tan x

Example : Find y.

cos x = @ sin x

` df f f f f f f f

tan x = sec 2 x

` df f f f f f f f

dx

` df f f f f f f f

sin x = cos x

a

csc x = @ csc x cot x a

y = x 2 sec x

if

Solution : y. =

b df f f f f f f f

c

x 2 sec x

dx b c ` a df df f f f f f f f 2 f f f f f f f = x A sec x + x 2 A sec x dx dx 2 = 2x sec x + x sec x tan x

Example : ` a πf f f f f Find f. if f x = 5 secx + tan x A 3 d

e

What is the slope of the tangent on the curve at x =

πf f f f f ? 3

Solution : ` a ` a df f f f f f f f. x = f 5 sec x + tan x dx = 5 sec x tan x + sec 2 x d e πf πf πf f f f f f f πf f f f f f f f. f = 5 sec f tan f + sec 2 f 3 3 3 3 w w w w w w

= 5 A 2 A p3 + 2 w w w w w w = 4 + 10 p3

2

d e b w w w w w wc πf πf f f f f f f f The slope of the tangent on the curve at x = is f. f or 4 + 10 p3 A 3 3

THE CHAIN RULE For composite functions, finding the derivative directly is not possible. If we use the limit method from the definition, it will become very cumbersome. So we make it easy by using the chain rule. For example, finding the derivative of the function f x = cos x 2 A ` a

directly by above methods is not possible. If we assume two functions g x = cos x and h x = x 2 then we can write f x = gοh x = g h x . ` a

` a

` a

In such a case where f x = g h x ` a

b ` ac

The Chain Rule :

b

c` a

B ` aC

it can be proven that f. x = g. h x A h. x . ` a

If y = f u and u = g x then ` a

` a

dy dy f f f f f f f f f f f f f f f du f f f f f f f = f A f dx du dx or B C B C ` a du df df f f f f f f f ` a f f f f f f ` a du f f f f f f f f f f f f f f f u = f f u A f = f. u A f dx dx du dx

Using the chain rule we get @ df f f f f f f f

A du f f f f f f f un = n un@ 1 f dx dx

which is very useful to find the derivatives of some common functions. dy f f f f f f f Example : Find f by the chain rule given that dx uf +f 1f f f f f f f f f f f f f y= f and u = x 3 A u@1

Solution :

B ` aC

` a

uf +f 1f f f f f f f f f f f f f y= f using the quotient rule u @` 1 a ` a uf @ 1f Af 1f @ uf + 1f Af 1f dy 2f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = = @`f ` a a2 2 dx u@1 u@1 du f f f f f f f f = 3x 2 dx

and

H

I

2 dy 2f 6f xf f f f f f f f dy f f f f f f f du f f f f f f f J f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f KA 3x 2 = @ f So f = f A f = @` f a2 b c2 dx du dx u@1 x2@ 1

Example : Find the derivative of

f

g@ 2

1f f f f x+ x

Solution : g@ 2

f df f f f f f f f

1f f f f x+ dx x

g@ 3

1f f f f = @2 x + x f

f df f f f f f f f

g@ 3f

1f 1f f f f f f f x+ = @2 x + dx x x g

f

c2

b

Example : Find the derivative of 3x x 2 @ 1 Solution : F df f f f f f f f

dx

b

c2 G

3x x 2 @ 1

c2

= x2@ 1 b

` df f f f f f f f

dx

c2

= 3 x2@ 1 b

c2

= 3 x2@ 1 b

3x + 3x a

b df f f f f f f f

dx

+ 3x A 2 x 2 @ 1 A 2x b

c

+ 12 x 2 x 2 @ 1 b

= 3 x 2 @ 1 x 2 @ 1 + 4x 2 b

cb

= 3 x 2 @ 1 5x 2 @ 1 b

c2

x2@ 1

cb

c

c

c

Extending the chain rule we can also write dy f f f f f f f f = dx dy f f f f f f f f = dx

dy f f f f f f f f A du dy f f f f f f f f A du

du f f f f f f f f A dv du f f f f f f f f A dv

dv f f f f f f f f or dx dv f f f f f f f f f dw f f f f f f f f f A dw dx

and so on A

1@

1f f f f f f f 2 x

g