Differentiation

MATHS

Method of differentiation

First principle of differentiation 1.

The derivative of a given function f at a point x = a in its domain represent the slope of the tangent at that point, and it is defined as:

Limit f (a  h)f (a ) , provided the limit exists & is denoted by f (a). h0 h i.e. f (a) = Limit x a 2.

f ( x )f (a) , provided the limit exists. xa

If x and x + h belong to the domain of a function f defined by y = f(x), then

Limit f ( x  h)f ( x ) if it exists, is called the Derivative of f at x & is denoted by f (x) or dy . h0 dx h f ( x  h)f ( x ) h This method of differentiation is also called ab-initio method or first principle method. i.e., f (x) = Limit h0

Example # 1 : Find derivative of following functions by first principle with respect to x. (i) f(x) = x 2 (ii) f(x) = tan x (iii) f(x) = esinx Solution :

(i)

(ii)

f(x)

( x  h) 2  x 2 = hlim 0 h

f(x)

sin( x  h) sin x  tan( x  h)  tan x lim lim cos( x  h) cos x = h0 = h0 h h

= hlim 0

(iii)

f(x)

2xh  h2 = hlim = 2x. 0 h

sin( x  h – x ) = sec 2x. h cos x. cos( x  h)

e sin ( x  h)  e sin x = hlim 0 h





e sin ( x  h ) sin x  1  sin( x  h)  sin x  lim sin x   = h0 e h  sin( x  h)  sin x 

lim sin( x  h)  sin x = esin x lim h0 h

= esin x

h0

= esin x

h0

lim

xh x  xhx 2 cos  sin  2 2     h

 h  sin(h / 2)   sin x cos x cos x    2 h/2  = e  

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1

MATHS Derivative of some elementary functions : f(x)

f(x)

1.

xn

nx n – 1 ,

(x  R, n  R)

2.

ax

ax n a ,

a>0

3.

n |x|

1 x

4.

logax

1 x n a

5.

sin x

cos x

6.

cos x

– sin x

7.

sec x

sec x tan x

8.

cosec x

– cosec x cot x

9.

tan x

sec 2 x

10.

cot x

– cosec 2x

Basic theorems : Sum of two differentiable functions is always differentiable. Sum of two non-differentiable functions may be differentiable. There are certain basic theorems in differentiation: 1.

d (f ± g) = f(x) ± g(x) dx

2.

d d (k f(x)) = k f(x) dx dx

3.

d (f(x) . g(x)) = f(x) g(x) + g(x) f(x) dx

4.

d dx

5.

d (f(g(x))) = f(g(x)) g(x) dx

g( x ) f ( x )  f ( x ) g( x )  f ( x)    = g2 ( x )  g( x ) 

This rule is also called the chain rule of differentiation and can be written as dy dy dz = . dx dz dx

Note that an important inference obtained from the chain rule is that

dy dy dx =1= . dx dy dy 

1 dy = dx / dy dx

another way of expressing the same concept is by considering y = f(x) and x = g(y) as inverse functions of each other.


2

MATHS dy = f(x) dx



g(y) =

dx dy = g(y)

and

1 f ( x )

Example # 2 : Find the derivative of the following functions with respect to x.

Solution :

sin( 2 x  3)

(i)

f(x) =

(ii)

f(x) =

(i)

f(x) =

sin ( 2 x  3 )



f(x) =

d 1 d ( sin ( 2 x  3 ) ) = . (sin (2x + 3)) dx 2 sin (2x  3) dx

x

(iii)

1 x2

f(x) = x . sin x

(chain rule)

cos(2x  3) =

(ii)

f(x) =

sin (2x  3) x 1 x2

= (iii)



f(x)

=

(1  x 2 )  x(2x ) (1  x 2 )2

(Quotiant rule)

1 x2 (1  x 2 )2

f(x) = x sin x  f(x) = x. cos x + sin x

(Product rule)

Example # 3 : If f(x) = sin (x + tanx), then find value of f(0). Solution :

 f(x) = sin (x + tanx)  f(x) = cos (x + tanx) (1 + sec 2x) Hence, f(0) = 2

(chain rule)

Self Practice Problems : (1)

Find the derivative of following functions using first principle with respect to x. (i) f(x) = x sin x (ii) f(x) = sin2 x

(2)

f (5  t )  f (5  t ) If f(5) = 7, then find the value of tlim 0 2t

(3)

Differentiate the following functions with respect to x. (i)

(iii)

(1 + 3x 2) (2x 3 – 1)

1 x 2

(ii)

(iv)

( x  1) ( x  2)( x  3) 1 x 1 x

(v)

cos 3 x sin x

(vi)

x ex sin x

(vii)

sin x 1  cos x

(viii)

n (sin x – cos x)


3

MATHS Answers :

(1)

(i)

(2)

7

(3)

(i)

x cosx + sinx

(ii)

2sin x cos x

6x (5x 3 + x – 1)

(ii)

( x  2) 2 ( x  3 ) 2

(iv)

(1  x )1 / 2 (1  x )3 / 2

 x 2  2x  1

x (iii)

1 x

1 2

(v)

cos 4 x – 3 cos 2x sin2x (vi)

ex ((sin x + cos x) x + sin x)

(vii)

1 x sec 2 2 2

cos x  sin x sin x  cos x

(viii)

Derivative of inverse trigonometric functions : 1.



–

dx = cos y dy



dy 1 = = dx cos y

dy = dx

1

, siny  ± 1

1  sin 2 y

1 1 x 2

, – 1 < x < 1.

In general |cos y| =

2.

  y and x = sin y 2 2

y = sin–1 x

y = tan–1x



1  sin 2 y . But here since –

x = tan y and –

  y 2 2

 |cos y| = cosy

  y 2 2

dx = sec 2y = 1 + tan2 y dy 

3.

dx 2 dy = 1 + x

y = sec –1x



dx dy = sec y tan y



dy 1 = dx 1 x2

(x  R)

 y  [0, ] –   and x = secy 2



1 dy = sec y. tan y , tan y  0 dx

1     = | sec y. tan y | , y   0,    ,   2   2  1 1 = | sec y || tan y | = | x | x2  1 

1 dy = dx | x | x2  1

x  (– , – 1)  (1, )


4

MATHS Results for the derivative of inverse trigonometric functions can be summarized as : f(x) f(x) 1 sin–1x ; |x| < 1 1 x 2 1 cos –1x ; |x| < 1 1 x 2 1 tan–1x ; xR 1 x2 1

cot–1x

xR

;

1 x2

1

sec –1 x

;

| x | x2  1

|x| > 1

1 cosec -1 x

; |x| > 1

| x | x2  1

Example # 4 : If f(x) = n (sin–1 x 2), then find f(x).

Solution :

f(x)

=

1 (sin 1 x 2 )

2x

1

.

. 2x =

1  ( x 2 )2

(sin 1 x 2 ) 1  x 4

Example # 5 : If f(x) = 2x sec –1x – cosec –1(x) , then find f(–2). Solution :

f(x) = 2 sec –1(x) +

2x | x | x2  1

Hence, f(–2) = 2.sec –1(– 2) –

f(–2) =

+

1 | x | x2  1

2 3

1 +

2 3

4 3 – 3 2

Logarithmic differentiation : The process of taking logarithm of the function first and then differentiate is called Logarithmic differentiation. It is often useful in situations when (i) (ii)

a function is the product or quotient of a number of functions OR a function is of the form [f(x)]g(x) where f & g are both derivable,

Example # 6 : If y = (sin x)n x, find Solution :

dy dx

n y = n x . n (sin x)

1 dy cos x 1 y dx = x n (sin x) + n x. sin x 

 n sin x   dy  cot x n x  = (sin x)n x  x dx  


5

MATHS Example # 7 : If y =

Solution :



x1/ 2 (1  2x )2 / 3 ( 2  3 x )3 / 4 (3  4 x ) 4 / 5 y=

, then find

dy . dx

x1/ 2 (1  2x )2 / 3 ( 2  3 x )3 / 4 (3  4 x ) 4 / 5

taking loge on both side n y =

2 4 1 3 n x + n (1 – 2x) – n (2 – 3x) – n (3 – 4x) 3 5 2 4



1 dy 9 16 4 1 = – + + y dx 4 (2  3 x ) 5 (3  4 x ) 3(1  2x ) 2x



 1  4 9 16 dy = y  2x  3 (1  2 x )  4(2  3 x )  5 (3  4 x )  dx  

Implicit differentiation : If f(x, y) = 0, is an implicit function i.e. y can't be expressed explicitly as a function of x then in order to find dy/dx, we differentiate each term w.r.t. x regarding y as a function of x and then collect terms in dy/dx. Example # 8 : If x 3 + y3 = 3xy, then find Solution :

dy . dx

Differentiating both sides w.r.t.x, we get 3x 2 + 3y2

dy dy = 3x + 3y dx dx

y  x2 dy = 2 y x dx Note that above result holds only for points where y2 – x  0 Example # 9 : If x y + yx = 2 then find Solution :

dy dx

u+v=2 du dv + =0 dx dx where u = x y  n u = y n x



........(i) & &

v = yx n v = x n y



1 du dy y = + n x & u dx dx x

x dy 1 dv = n y + y dx v dx



dy  y du  & = x y   n x dx  dx x

 x dy  dv = yx  n y  y dx  dx  

Now, equation (i) becomes

 x dy  dy  y   + y x  n y  x y   n x y dx  = 0. dx  x 



 x y y  y n y  x .  dy x  =– dx  y x  x n x  y x .  y 


6

MATHS Self Practice Problems (4)

(5)

Differentiate the following functions with respect to x. (i)

y = sec –1 (x 2)

(iii)

1  y = 1   x 

(v)

y = (ln x)x + (x)sin x

Find (i) (iii)

(ii)

 1 x   y = tan–1   1 x 

(iv)

y = ex

(ii)

x 2/3 + y2/3 = a2/3

x

dy if dx y = cos (x + y) x = y n (x – y)

(6)

If x y = ex – y, then prove that

(7)

If

x

n x dy = . dx (1  n x )2

x a x dy = log , then prove that =2– . xy xy y dx

Answers :

(4)

(i)

2

(ii)

x x 1 4

1  (iii) 1   x 

x

  1 1  n 1     x 1  x   

1 1 x2 x

(iv) x x. e x (nx + 1)

  1   sin x   cos x nx  (v)  n (nx )   nx   (n x)x + x sinx     x    sin( x  y ) (i) 1  sin( x  y )

(5)

y (ii) –   x

1/ 3

(iii)

y( x  2 y ) x( x  y )  y 2

Differentiation using substitution : In certain situations as mentioned below, substitution simplifies differentiation. For each of the following expression, appropriate substitution is as follows (i)

x 2  a2

 x = a tan  , where 

  << 2 2

or

x = a cot , where 0 <  < 

(ii)

a2  x 2



x = a sin  , where 

   2 2

or

x = a cos , where 0    

(iii)

x 2  a2

  x = a sec , where  [0 –   2

or

(iv)

xa ax



   x = a cosec , where   ,  – {0}  2 2

x = a cos , where 0    


7

MATHS  1  x 2  1   Example # 10 : Differentiate y = tan–1   with respect to x. x   Solution :

Let

  

x = tan  , where    2 , 2  – {0}         | sec  | sec      ,    2 2  

 | sec  | 1   y = tan–1   tan  



 1  cos    y = tan–1   sin  



  y = tan–1  tan  2  



y=

 2



y=

1 tan–1 x 2

Example # 11 : Find Solution :

     1  tan (tan x )  x for x    ,    2 2   1 dy = 2(1  x 2 ) dx



dy , where y = tan–1 dx

 1 x  1 x   1 x  1 x 

Let

x = cos , where  [0, ]



 1  cos   1  cos    y = tan–1    1  cos   1  cos  



y = tan–1

    2 cos  2 sin  2 2      2 cos  2 sin  2 2 

  



y = tan–1



y=

 .  

1  cos  

  1  tan    1  tan 

2 cos

  2   2

 1 – cos –1x 4 2

    but for   0,  , 2 2  2

  – 4 2



y=



1 dy = dx 2 1 x2

2 cos

as 

   2 cos  2 2

     –  4 4 2 4

 2x  Example # 12 : If f(x) = sin–1   , then find  1 x2 

(i)

f(2)

(ii)

 1 f   2

(iii)

f(1)


8

MATHS Solution :

x = tan , 

where –

  << 2 2

y = sin–1 (sin 2)

  ,  2      2 2    ,   2  y =  2 2 2   (   2) ,    2     2 2   2  1 x  2  2 f(x) =  1  x  2   1  x 2





   2 tan 1 x x 1  1 2 tan x 1 x  1 f(x) =   (   2 tan 1 x ) x  1 

x 1 1 x  1 x  1

2 5

(ii)

 1 8 f   = 2 5  

(i)

f(2) = –

(iii)

f(1+) = – 1 and f(1–) = + 1  f(1) does not exist.

Aliter Above problem can also be solved without any substution also, but in a little tedious way.  2x   f(x) = sin–1   1 x2 



1

f(x) = 1

.

4x 2

(1  x 2 )2

thus

2(1  x 2 ) (1  x 2 )2

(1  x ) 2

.

| 1 x 2 |

 2 , |x|1  2 1  x f(x) =   2  , | x|1  1  x 2

Example # 13 : If

Solution :

.

(1  x 2 )

2 f(x) =

(1  x 2 )2

(1  x 2 )2

(1  x 2 ) =

2{(1  x 2 )  2x 2 }

Put

1 x 2 +

1 y 2 = a(x – y), then prove that

x = sin where –

     2 2

dy = dx

1 y 2 1 x 2

.

and

    . 2 2 cos + cos  = a (sin – sin) y = sin where –

 

         cos   = 2a cos   sin   2cos   2   2   2   2 


9

MATHS    =a cot   2 



  –  = 2 cot–1 (a)  sin–1 x – sin–1 y = 2 cot–1a differentiating w.r.t to x.

1 1 x

2

1 y

2

dy =0 dx

1 y 2

dy = dx



1

–

–  + 2 cot–1a –  + 2 cot–1a

or or

1 x 2

Aliter Using implicit differentiation

x 1 x





dy = dx

y 2

–

1 y

dy   dy  = a 1  dx  dx 

2

 y   a  1 y2 

  dy =a+   dx

x 1 x 2

dy = dx

a

x 1 x2 y 1 y2

1 x 2  1 y 2 x  xy 1 x 2 1 x2  1 y2 y  xy 1 y2

(1  x 2 )  (1  x 2 )(1  y 2 )  x 2  xy dy = . dx (1  x 2 )(1  y 2 )  (1  y 2 )  xy  y 2 

dy = dx



a

1 y 2 1 x 2

1  (1  x 2 )(1  y 2 )  xy

1 y2 1 x2

=

1  (1  x 2 )(1  y 2 )  xy

1 y2

.

1 x2

Hence proved

Parametric differentiation : dy dy / d If y = f() & x = g() where is a parameter, then dx  dx / d . Example # 14 : If x= a cos 3t and y = a sin3t, then find the value of

Solution :

dy . dx

 3a sin 2 t cos t dy dy / dt = = = – tan t 3a cos 2 t sin t dx dx / dt

Example # 15 : If y = a cos t and x = a (t – sint), then find the value of

Solution :

dy a sin t = dx a(1  cos t )



dy dx

t

 2

dy  at t = . dx 2

= – 1.


10

MATHS Derivative of one function with respect to another function : Let y = f(x); z = g(x) then

dy dy / dx f '(x)   . dz dz / dx g' (x)

Example # 16 : Find derivative of y = n x with respect to z = ex. Solution :

1 dy dy / dx = = dz dz / dx xex

Self Practice Problems :

(8)

Find

dy when dx

(i)

x = a (cos t + t sin t)

and

y = a (sin t – t cos t)

(ii)

 1 t2    x=a  2   1 t 

and

 2t   y=b   1 t 2 

sin–1

 x2   4 4  x a

 2 xa 2 dy  , then prove that = .  dx x 4  a4 

(9)

If y =

(10)

 2x  dy 2  If y = tan–1  = (| x |  1) 2 , then prove that dx  1 x  1 x2

(11)

If u = sin (m cos –1x) and v = cos (m sin–1 x), then prove that

Answer :

(8)

(i)

tan t

du = dv

1  u2 1 v 2

.

( t 2  1)b 2at

(ii)

Higher order derivatives : Let a function y = f(x) be defined on an open interval (a, b). It’s derivative, if it exists in (a, b) is a certain function f (x) [or (dy/dx) or y ] & is called the first derivative of y w. r. t. x. If it happens that the first derivative has a derivative in (a, b) then this derivative is called the second derivative of y w. r. t. x & is denoted by f (x) or (d2y/dx 2) or y . While the first derivative denotes slope of the graph, the second derivative denotes it's concavity. Similarly, the 3rd order derivative of y w. r. t. x, if it exists, is defined by

d3 y dx 3



d  d2 y  dx  dx 2  , it is also

denoted by f (x) or y . It must be carefully noted that in case of parametric functions although

dy dy / dt = dx dx / dt

but

d2 y dx

2



d2 y / dt 2 2

dx / dt

2

rather

d2 y dx

2

=

d  dy / dt    dx  dx / dt 

which on applying chain rule can be resolved as


11

MATHS

d2 y dx 2

d2 y dx 2

d  dy / dt  dt  . = dt  dx / dt  dx

=

d2 y



dx 2

=

 dx d2 y dy d2 x   .   .  dt dt 2 dt dt 2     dx     dt 

2

.

dt dx

 dx d2 y dy d2 x  .  . 2   dt dt 2   dt dt  dx     dt 

3

Example # 17 : If y = x 3 n x, then find yand y Solution :

y = 3x 2 n x + x 3

1 = 3x 2 n x + x 2 x

y = 6x n x + 3x 2 .

1 + 2x = 6x n x + 5x x

y = 6 n x + 11

x

 1 Example # 18 : If y =   , then find y(1) x Solution :

Now taking loge both sides, we get n y = – x n x when x = 1, then y = 1 n y = – x n x 

y y = – (1 + n x)



y = – y (1 + n x)



y = y (1 + n x)2 –

......(i)

again diff. w.r.t. to x, y = – y(1 + n x) – y . 

1 x

y(1) = 0

Example # 19 : If x = t + 1 and y = t2 + t3, then find

Solution :

dy = 2t + 3t2 ; dt





y (using (i)) x

d2 y dx 2

.

dx =1 dt

dy = 2t + 3t2 dx

d2 y dx

2

d2 y dx 2

=

d dt (2t + 3t2) . dt dx

= 2 + 6t.


12

MATHS Example # 20 : Find second order derivative of y= sin x with respect to z = ex. Solution :

cos x dy dy / dx = = dz dz / dx ex



d2 y



dz

= d2 y dz

2

2

=

d  cos x    dz  e x 

 e x sin x  cos x e x x 2

(e )

=–



.

d2 y dx

2

=

d  cos x  dx  . dx  e x  dz

1 ex

(sin x  cos x ) e2x

Example # 21 : y = f(x) and x =g(y) are inverse functions of each other, then express g(y) and g(y) in terms of derivative of f(x). dx dy Solution : = f(x) and dy = g(y) dx 1  g(y) = ...........(i) f ( x ) again differentiating w.r.t. to y

d g(y) = dy

=

d dx

=–



 1     f ( x ) 

 1  dx   .  f ( x )  dy

f ( x ) f ( x )2

. g(y)

f ( x ) g(y) = –

f ( x )3

.........(ii)

which can also be remembered as

d2 y d x dx 2 2 = – 3 . dy  dy     dx  2

Example # 22 : y = sin (sinx) then prove that y + (tanx) y + y cos 2x = 0 Solution :

Such expression can be easily proved using implicit differentiation  y = cos (sin x) cos x  sec x.y = cos (sin x) again differentiating w.r.t x, we get secx y + y sec x tan x = – sin (sin x) cos x  y + y tan x = – y cos 2 x  y +(tanx) y + y cos 2x = 0

Self Practice Problems :

(12)

d2 y n x If y = , then find the value of . x dx 2


13

MATHS cos 2

(13)

Prove that y = x + tan x satisfies the differential equation

(14)

If x = a (cos +  sin ) and y = a(sin  –  cos), then find the value of

(15)

Find the second order derivative of nx with respect to sin x.

(16)

If y =

e– x

Answers :

(A cos x + B sin x), then prove that

(12)

2n x  3 x

(14)

3

d2 y dx 2

+2.

sec 3  a

x

d2 y dx 2

– 2y + 2x = 0.

d2 y dx 2

.

dy + 2y = 0. dx

(15)

x sin x  cos x x 2 cos3 x

Derivative of a determinant :

If F(x) =

f (x)

g( x )

h( x )

l( x )

m( x )

n( x )

u( x )

v( x )

w( x)

, where f, g, h, l, m, n, u, v, w are differentiable functions of x, then

f ' ( x ) g' ( x ) h' ( x ) F (x) =

l( x )

m( x )

n( x )

u( x )

v( x )

w( x)

f (x) +

g( x )

h( x )

l' ( x ) m' ( x ) n' ( x ) u( x )

v( x )

w( x )

+

f ( x)

g( x )

h( x )

l( x )

m( x )

n( x )

u' ( x ) v' ( x ) w ' ( x )

L’ hospital’s rule : If f(x) & g(x) are functions of x such that: (i)

Limit f(x) = x a

 = Limit g(x) x a

OR (ii)

Limit f(x) = 0 = Limit g(x), both f(x) and g(x) are continuous at x = a, both f(x) and x a

x a

g(x) are differentiable at x = a and both f (x) and g (x) are continuous at x = a, then

Limit x a

f (x) f ' (x) Limit g( x ) = x a g' ( x )


14

Differentiation

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