chapter-28- differentiation

Class XI www.vedantu.com RS Aggarwal Solutions

CHAPTER-28- Differentiation

Exercise 28 A

1. Solution :

(i) x-3

Differentiating w.r.t x,

(

)

(ii) √


(

)

2. Solution :

(i)



(

)

(ii)

√


(

)

(iii)

√


3. Solution :

(i) 3x-5

Differentiating with respect to x,

( ) (

)

(ii) ⁄



(

)

(iii) 6. √


4. Solution :

(i) 6x5 + 4x3 – 3x2 + 2x – 7


(6x5 + 4x3 – 3x2 + 2x – 7) = 30x5-1 + 12x3-1 – 6x2-1 + 2x1-1 + 0 =

30x4 + 12x2 – 6x1 + 2x

(ii) ⁄

√ √


( ⁄

√ √

)

= 5×

=


(iii) ax3 + bx2 + cx + d, where a, b, c, d are constants


(ax3 + bx2 + cx + d) = 3ax3-1 + 2bx2-1 + cx1-1+ d×0

= 3ax2 + 2bx + c

5. Solution :

(i) 4x3+3.2x + 6.√

= 4x3+3.2x 6


= 4.3x3-1 + 3.logn(2).2x + 6×

+ 5 ×- cosec2x

= 12x2 + 3.logn(2).2x -3

- 5 cosec2x

(ii)

√

√ ⁄


⁄


(

⁄

)

(

)

( )

(

)

( )

6. Solution :

(i) 4 cot -

cos x +

-

+

+ 9

= 4 cot -

cos x + – + + 9



(

)

(

( ) (

( )

)

( )

(ii) -5 tan x + 4 tan x cos x – 3 cot x sec x + 2sec x – 13

= -5 tan x + 4 sinx – 3 cosecx+ 2sec x – 13


(-5 tan x + 4 sinx – 3 cosecx+ 2sec x – 13)

= -5 sec2x + 4cosx -3(- cosecx cotx) + 2 secx tanx – 0

= -5 sec2x + 4cosx + 3 cosecx cotx + 2 secx tanx

7. Solution :

(i) (2x + 3) (3x – 5)


( )( )

( )

( ) ( )

( )

= (2x + 3)(3x1-1+0) + (3x – 5)(2x1-1+0) = 6x + 9 + 6x -10

= 12x -1

(ii) x(1 + x)3

( )

( ) ( )

( )

= x×3×(1 + x)2 + (1 + x)3(1) = (1 + x)2(3x+x+1)

= (1 + x)2(4x+1)

(iii) .√

/ .√

/= (x1/2 + x-1)(x – x-1/2 )

(x1/2 + x-1)(x – x-1/2 )

= (x1/2 + x-1)

(x – x-1/2 ) + (x – x-1/2 )

(x1/2 + x-1)

= (x1/2 + x-1)(1+

x-3/2) + (x – x-1/2 )(

-1/2 – x-2)

= x1/2 + x-1 +

x-1 +

x-5/2 +

x1/2 – x-1 -

x-1 + x-5/2 =

x1/2 +

x-5/2


(iv) .

/

.

/ = 2.

/ .

/

= 2(x +

-

+

) = 2(x +

)

(v) .

/

.

/ = 3.

/ (2x -

)

= 3(2x3 -

-

+

)

= 3(2x3 -

+

)

(vi) (2x2 + 5x – 1) (x – 3)

(2x2 + 5x – 1) (x – 3)


= (2x2 + 5x – 1)

(x-3) + (x – 3)

(2x2 + 5x – 1)

= (2x2 + 5x – 1)×1 + (x – 3)(4x + 5) = 2x2 + 5x – 1 + 4x2 -7x -15 =

6x2 -2x -16

8. Solution :

(i)

( ) ( )

( )

(ii) ( )( )

( )( )

( )( ) ( )( )


2( )

( ) ( )

( )( )3 ( )( )

*( ) ( ) + ( )( )

*( + ( )( )

( )

(iii)

√

√

√

( ) ( )

√

√ ( )

√ ( )

√

√


(iv) ( )√

√

( )√

√

√ ( )√ ( )√

√

√ 2( )

√

√ ( )3 ( )√

√ {( )

√ )} ( )√

√ {

√ }

4

5

√

(

)

√

(V)

√


√ √

√

√

√

√ ( )

( )

(vi)

( ) ( )

( )

9. Solution :

(i) Given that y = 6x5 – 4x4 – 2x2 + 5x – 9


(6x5 - 4x4 – 2x2 + 5x – 9) = 30x4 -16x3 – 4x + 5

Putting x = -1

= 30(-1)4 -16(-1)3 – 4(-1) + 5 = 30+16+4+5 = 55


(ii) Given y = (sin x + tan x)


(sinx + tanx) = cos x + sec2 x

Putting x=

.

= cos

+ sec2

=

+ 4 =

(iii) Given y= (

= 2cosec x-3cot x


(2cosec x-3cot x) = 2(- cosecx cotx) – 3(- cosec2x)

Putting x=

= 2(- cosec

cot

) – 3(- cosec2

) = - 2×√ + 3×2 = 6 - 2×√

10. Solution :

To show:

√

L.H.S. =

=

√

.√

√ / (

=

.√

√ /

√

)


= √

√ √

√

= √ = RHS

11. Solution :

To prove: (2xy)

.

/.

LHS = (2xy)

LHS = (2xy) (√

√

) (

√

√

) (

(√

√

)

√

√

)

LHS = (√

√

) (√

√

)

LHS = (√

√

)

12. Solution :

√


√

= cotx

Differentiating y with respect to x

( ) =

13. Solution :

cos x = ( )⁄

( )⁄

y = cos x

Differentiating y with respect to x

= - sin x

EXERCISE 28 B

Solution :

Let f(x) = ax + b

( ) h 0lim

( ) ( )

...(i)

f(x)=ax + b

f(x+h)=a(x + h) + b


=ax + ah + b

Putting values in (i), we get

f’(x) = h 0lim

( )

= h 0lim

( )

= h 0lim

= a

f’(x) = a

2. Solution :

Let f(x)=a

F’(x) =h 0lim

( ) ( )

...(i)

f(x)=a

f(x+h)=a(x+h)2+

( )


F’(x) =h 0lim

0 ( )

( )1 0

1


=h 0lim

0 ( )

( )1 0

1

=h 0lim

( )

( )

=h 0lim

,( ) - 0

( )

1

=h 0lim

, - 0

( )

( )1

=h 0lim

[ ] 0

( )1

=h 0lim

[ ] 0

( )1

=h 0lim

0 ( )

( )1

Putting h = 0, we get


=a[(0)+2x]-

( ) =2ax-

3. Solution :

Let f(x) = 3x2 + 2x – 5

F’(x) =h 0lim

( ) ( )

...(i)

f(x) = 3x2 + 2x – 5

f(x + h) = 3(x + h)2 + 2(x + h) – 5

= 3(x2 + h2 + 2xh) + 2x + 2h – 5

= 3x2 + 3h2 + 6xh + 2x + 2h – 5


f’(x) =h 0lim

( )


= h 0lim

= h 0lim

)

=

h 0lim


f’(x) = 3(0) + 6x + 2 = 6x + 2

4.. Solution :

Let f(x) = x3 – 2x2 + x + 3

f’(x) =h 0lim

( ) ( ))

…(i)

f(x) = x3 – 2x2 + x + 3

f(x + h) = (x + h)3 – 2(x + h)2 + (x + h) + 3



( ) h 0lim

( ) ( ) ( ) , -

=h 0lim

( ) ( ) ( )

=h 0lim

,( ) ] ,( ) - ,

= h 0lim

[ ] [ ]

= h 0lim

[ ] , -

=h 0lim



f’(x) = (0)2 + 2x(0) + 3x2 – 2(0) – 4x + 1 = 3x2 – 4x + 1

5…Solution :

Let f(x) = x8

( ) h 0lim

( ) ( )

…(i)

f(x) = x8

f(x + h) = (x + h)8


( ) h 0lim

( )

=h 0lim

( )

( )

= 8x8-1 = 8x7

6…Solution :

f(x)=

( ) h 0lim

( ) ( )

….(i)


f(x)=

and f(x+h)=

( )


( ) h 0lim

( )

=h 0lim

( )

( ) = (-3) x3-1 = -3x-4 = -

7…Solution :

f(x) =

and ( ) lh 0lim

( ) ( )

….(i)

f(x) =

and f(x+h) =

( )


( ) lh 0lim

( )

=

lh 0lim

( )

( )


= (-5) x5-1 = -5x-6 =-

8….Solution :

f(x) = √

and ( ) lh 0lim

( ) ( )

….(i)

f(x) = √ and f(x+h)=√ ( )

=√


( ) lh 0lim

√ √


=lh 0lim

√ √

√ √

√ √

=lh 0lim

(√ ) (√ )

(√ √ )

=lh 0lim

(√ √ )

=lh 0lim

(√ √ )

=lh 0lim

√ √


√ ( ) √


=

√ √ =

√

9.. Solution :

f(x)= √

and ( ) lh 0lim

( ) ( )

….(i)

f(x) = √

f(x+h) = √ ( ) = √

Putting values in (i), we get ( ) lh 0lim

√ √

=lh 0lim

√ √

√ √

√ √

=lh 0lim

(√ √


=lh 0lim

(√ √


=

√ ( ) √ =

√

10. ….Solution :

f(x) =

√

and ( ) lh 0lim

( ) ( )

…..(i)

f (x) =

√ f(x+h) =

√


√

√

( ) lh 0lim

√ √

(√ ) (√ )

=lh 0lim

√ √

(√ )(√ ) √ √

√ √


=lh 0lim

(√ ) (√ ) (√ ) √ )

=lh 0lim

(√ ) (√ ) (√ ) √ )

=lh 0lim

(√ ) (√ ) (√ ) √ )


=

(√ )(√ )(√ √ ) =

(√ ) ( √ )

=

(√ )

11….Solution :

f(x) =

√ ( )

lh 0lim

( ) ( )

…..(i)

f (x) =

√ f(x+h) =

√


√

√


=lh 0lim

√ √

(√ ) (√ )

=lh 0lim

√ √

(√ ) (√ )

√ √

(√ ) (√ )

=lh 0lim

(√ )

(√ )

(√ ) (√ ) (√ ) √ )

=lh 0lim

(√ ) (√ ) (√ ) √ )

=lh 0lim

(√ ) (√ ) (√ ) √ )


=

(√ ) ( √ )

=

(√ ) =

(√ )

12…Solution :


f(x) =

√

and ( ) lh 0lim

( ) ( )

…..(i)

f (x+h) =

√


√

√

= lh 0lim

√ √

(√ )(√ )

= lh 0lim

√ √

(√ )(√ ) √ √

√ √

=

2 2

h 0

6x 5 6x 6h 5

h 6x 6h 5 6x 5 6x 6h 5lim

=lh 0lim

(√ ) (√ ) (√ ) √ )


=lh 0lim

(√ ) (√ ) (√ ) √ )

=lh 0lim

(√ ) (√ ) (√ ) √ )


=

(√ ( ) ) (√ ) (√ ) √ )

(√ ) ( √

=

(√ )

13…..Solution :

f(x) =

√ ( )

lh 0lim

( ) ( )

…..(i)


f(x+h) =

√ ( )

√


√

√

= lh 0lim

√ √

(√ )(√ )

=lh 0lim

√ √

(√ )(√ ) √ √

√ √

=lh 0lim

(√ )

(√ )

(√ ) (√ ) (√ ) √ )

=lh 0lim

(√ ) (√ ) (√ ) √ )

=lh 0lim

(√ ) (√ ) (√ ) √ )

=lh 0lim

(√ ) (√ ) (√ ) √ )



=

(√ ( )) (√ ) (√ ) √ ( )) =

(√ )

14… Solution :

f(x) =

( )

lh 0lim

( ) ( )

…..(i)

f(x+h)= ( )

( )


=lh 0lim

( )( ) ( )( )

( )( )

=lh 0lim

, -

(( )( ))

=lh 0lim

(( )( ))

=lh 0lim

(( )( ))



=

(( ( ) )( )) =

( )

15….Solution :

f(x) =

( )

lh 0lim

( ) ( )

…..(i)

f(x+h)= ( )

( )


=lh 0lim

( )( ) ( )( )

( )( )

=lh 0lim

h , h -

( )( )

=lh 0lim

h h

( )( )

=lh 0lim

( )( )



=

( )

16….Solution :

f(x) =

’(x ) =lh 0lim

f x h f x

h

…..(i)

f(x+h)=

2 2 2x h 1 x h 2xh 1

x h x h


’(x ) =lh 0lim

2 2 2x h 2xh 1 x 1

x h xh

=lh 0lim

2 2 2x h 2xh 1 x x 1 x h

x h x

h


=lh 0lim

3 2 3 2x xh x x x h x h

h x h x

=lh 0lim

2xh x 1

x h x


=

2

2

x 1

x

17…Solution :

f(x) = √

( ) lh 0lim

( ) ( )

…..(i)

f(x+h) = √ ( )


=√ ( )


( ) lh 0lim

√ ( ) √

=lh 0lim

√ ( ) √

√ ( ) √

√ ( ) √

=lh 0lim

(√ ( ))

(√ )

(√ ( ) (√ )

=lh 0lim

( )

√ ( ) √ )

lh 0lim

3x 3h 3x 3x 3h 3x2sin sin

2 2

h cos 3x 3h cos3x


=lh 0lim

6x 3h 3h2sin sin

2 2

h cos 3x 3h cos3x

h 0 h 0

h 0

3hsin

3 6x 3h22 sin3h2 2

2

1

cos 3x 3h cos3x

lim lim

lim

=

h 0 h 0

6x 3h 13 1

2 cos 3x 3h cos3xlim lim

Putting h=0 , we get

=

6x 3 0 13 sin

2 cos 3x 3 0 cos3x

1

2

3sin3xf ' x

2 cos3x

18 Solution:

Let f(x)= secx


f’(x)=

h 0

f x h f x........(i)

hlim

f x h sec x h

Putting values in (i) , we get

h 0

sec x h secxf ' x

hlim

22

h 0

sec x h secx

h sec x h secxlim

=

h 0

sec x h sec x

h sec x h secxlim

=

h 0

1 1

cos x h cos x

h sec x h secxlim

=

h 0

cos x cos x h

cos x h cos x

h sec x h sec xlim


h 0

cos x cox x h

h cos x h cos x sec x h sec xlim

=

h 0

x x h x h2sin sin x

2 2

h cos x h cos x sec x h secxlim

=

h 0

2x n n2sin sin

2 2

h cos x h cos x sec x h secxlim

=

h 0 h 0 h 0

hsin

1 h 122 sin xh2 2 cos x h cos x sec x h secx2

lim lim lim

=

h 0 h 0

h 11 sin x

2 cos x h cos x sex x h secxlim lim

h 0

sin x1

xlim


Putting h=0 ,we get

=

0 1sin x

2 cos x 0 cos x sec x 0 secx

=

1sin x

cos x cos x sec x sec x

= xx

x

sec2cos

sin2

=1

tan x secx2

19. Solution:

Let f(x)=2tan x

h 0

f x h f xf ' x .....(i)

hlim

2f x h tan (x h)

Putting values in (i) , we get

2 2

h 0

tan x h tan xf ' x

hlim

=

h 0

tan x h tan x tan x h tan x

hlim


h 0

sin x h sin x hsin x sin x

cos x h cos x cos x h cos x

hlim

h 0

sin x h cos x sin x cos(x h) sin x h cos x sin x cos(x h)

cos x h cos x cos x h cos x

hlim

2 2

h 0

sin[ x h x] sin x h x]

h cos x h cos xlim

2 2h 0

sinh sin 2x h

h cos x h cos xlim

2 2

h 0 h 0 h 0

1 sinh 1sin 2x h

hcos x cos x hlim lim lim

2 2

h 0 h 0

1 11 sin 2x h

cos x cos x hlim lim

Putting h=0 , we get

2 2

1 1sin 2x 0

cos x cos x 0

= 2sin x 12 sec x secx

cosx cosx


=2tanxsec2 x

20. Solution:

Letf(x)=sin(2x+3)

F’(x)=

h 0

f x h f x.......(i)

hlim

F(x+h)=sin[2(x+h)+3]

Putting value in (i) ,we get

F’(x)=

h 0

sin 2 x h 3 sin 2x 3

hlim

h 0

2(x h) 3 2x 3 2 x h 3 2x 32sin cos

2 2hlim

h 0

2x 2h 3 2x 3 2x 2h 6 2x2sin cos

2 2hlim

h 0

2sin(h)cos(2x h 3)

hlim

=h 0 h 0

sinhcos(2x h 3)

h2lim lim


= h 0

2(1) cos 2x h 3lim

Putting h=0 ,we get

=2cos(2x+0+3) =2cos(2x+3)

21. Solution

Let f(x)=tan(3x+1)

h 0

f x h f xf ' x ....(i)

hlim

f(x)=tan(3x+1)

f(x+h)=tan[3(x+h)+1]

Putting value in (i) ,we get

h 0

tan 3 x h 1 tan 3x 1f '(x)

hlim

=

h 0

sin 3(x h) 1 3x 1

cos 3 x h 1 cos 3x 1

hlim


=

h 0

sin 3x 3h 1 3x 1

cos 3 x h 1 cos 3x 1

hlim

h 0

sin3h

h cos 3 x h 1 cos 3x 1lim

h 0 h 0

sin3h 1

3h cos 3 x h 1 cos 3x 1lim lim

h 0

13(1)

cos 3 x h 1 cos 3x 1lim

Putting h=0,we get

=

13

cos 3 x 0 1 cos 3x 1

= 23sec 3x 1

Hence ,f’(x)= 23sec 3x 1

Exercise 28 C

1. Solution:


It is given that 2x sin x

2 2dx sin x 2x sin x x cos x

dx

=2xsinx+ 2x cos x

2. Solution:

It is given that x(e cos x)

x x xd(e cos x) e cos x e sin x

dx

x xe cos x e sin x

xe cosx sin x

3…Solution:i

It is given that xe cot x

x x x 2de cot x ' e cot x e cosec x

dx =

x x 2e cot x e cosec x =

x 2e cot x cosec x

4. Solution:

It is given that nx cot x


n n 1 n 2dx cot x nx x cosec x

dx

= n 1 n 2nx cot x x cosec x = n 1 2x nx cot x cosec x

5. Solution:

It is given that 3x secx

3 2 3dx secx 3x secx x secx tan x

dx

2 33x secx x secx tan x 2x secx 3 x tan x

6: Solution:

It is given that 2x 3x 1 sin x

2 2dx 3x 1 sin x 2x 3 sin x x 3x 1 cos x

dx

=

2sin x 2x 3 cos x x 3x 1

7. Solution:

It is given that 4x tan x

4 3 4 2dx tan x ' 4x tan x x sec x

dx


= 3 4 24x tan x x sec x 3 2x 4tan x xsec x

8. Solution:

It is given that 2 x3x 5 4x 3 e

2 x 2 x xd3x 5 4x 3 e 3 4x 3 e 3x 5 8x e

dx

=2 x 2 x x12x 9 3e 24x 3xe 40x 5e =

2 x x36x x 3e 40 9 2e

9….Solution:

It is given that 2 3x 4x 5 x 2

2 3 3 2 2dx 4x 5 x 2 2x 4 x 2 x 4x 5 3x

dx

=4 3 4 3 22x 4x 4x 8 3x 12x 15x

=4 3 25x 16x 15x 4x 8

10. Solution:

It is given that 2 2x 2x 3 x 7x 5


2 2dx 2x 3 x 7x 5

dx

= 2 22x 2 x 7x 5 x 2x 3 2x 7

= 3 2 2 3 2 22x 14x 10x 2x 14x 10 2x 7x 4x 14x 6x 21

3 24x 27x 32x 11

11. Solution :

It is given that tan x secx cot x cosecx

d

tan x secx cot x cosecxdx

=

secx secx tan x cot x cosecx tan x secx cosecx cosecx cot x ’

= secx tan x secx cot x cosecx cosecx cosecx cot x

= secx tan x secx cosecx cot x cosecx

12. Solution :

Let F(x) = 3 xx cos x 2 tan x


3 2 3dx cos x 3x cos x x sin x

dx

= 2 33x cos x x sin x = 2x 3cosx xsin x

x x x 2d2 tan x 2 log 2 tan x 2 sec x

dx

=x 22 (log 2 tan x sec x)

Therefore, F’(x)= 2 x 2x 3cos x xsin x 2 log2tan x sec x

Exercise 28 D

1. Solution :

.

/=

( ) (using Quotient rule)

= ( )

2. Solution:

(

)

=


3 Solution :

.

( )/=

( )

( )

=

( )

4

.

( )/=

( )

( )

= ( )

( )

= ( )

( )

5.Solution:

0( )

( )1= ( ) ( )

( )

=

( )

=

( )

6 Solution:

.

/ =

( ) ( ) ( )

( )

=

( )


=

( )

7 Solution:

0( )

( )1 =

( ) ( ) ( )

( )

=

( )

=

( )

8 Solution:

.

/=( ) ( ) ( ) ( )

( )

=

( )

=

( ) =

( )

( ) =

( )

( )

9 Solution:

0

( )1 =

( ) ( ) ( )

( )

=

( ) =

( )

10. Solution

0

1= ( ) ( ) ( )

( )


= , ( ) ( )-

( )

11 Solution

[(√ √ )

(√ √ )] =

√ (√ √ ) (√ √ )

√

(√ √ )

=

√

√

√

√

(√ √ ) =

√

√ (√ √ )

12. Solution

0

1 =

( ) ( ) .

/

( )

= ( ) ( )

( )

13 Solution:

0

√ 1 =

(√ ) ( ) .

√ /

(√ )

=

√ (√ )

14.Solution

[

( )]

( ) ( ) ( )

( )


( )

( )

15. Solution :

(

)

( ) ( ) ( )

( )

)

( )

( )

16. Solution:

[

]

( ) ( ) ( )

( )

( )

( )


17.. Solution :

[

]

( ) ( ) ( ) ( )

( )

( )

( )

( )

18.Solution :

[

]

( )( ) ( )( )

( )


( )( ) ( )( )

( )

( ) ,( ) ( )( )-

( )

( ) ,( ) ( )-

( )

( ) , -

( )

, -

( )

19.. Solution :

.(

/

( ) ( ) ( )

( )


( )( ) ( )

( )

20.. Solution :

(

)

( ) ( ) ( ) ( )

( )

( ) ( ) ( )

( )

,( ) ( ) ( )-

,( ) ( ) ( )-

,( ) ( )-

( )

( )


21. Solution :

(

√ )

{ , - √ } {( ) (

√ )}

(√ )

{ , - √ } {( ) (

√ )}

{ , - √ } {( ) (

√ )}

* , -+ *( )+

√

* , -+ *( )+

22.. Solution :


( ( )

( ))

( )( ) , ( )-( )

( )

( )

( )

( )

( )

23.. Solution :

(

( ))

( )( ) , -, ( ) -

( )

( ),( ) ( )( )-

( )

( )


( )

( )

24.. Solution :

6

7

( )( ) , -, -

( )

, -

( )

( ) ( )

( )

25.. Solution :


[( )

( )]

( )( ) ( )( )

( )

( ) ( )

( )

( )

( )

( )

( )

26. Solution :

.

/

( )( ) ( )( )

( )


( )

( )

= - x

(ii)

.

/

( )( ) ( )( )

( )

= secx tanx = - x

EXERCISE 28 E

1.Solution

It is given that sin4x

d

dx(sin4x) = cos(4x)

d(4x) 4cos4x

dx

d d(cosnu) sin(nu) (nu)

dx dx

2. Solution :

It is given that cos5x.


d d(cos5x) sin(5x) (5x) 5sin5x

dx dx

d d(cosnu) sin(nu) (nu)

dx dx

3. Solution:

It is given that tan3x.

2 2d dtan3x sec 3x (3x) 3sec 3x

dx dx

2d d(tan nu) sec (nu) (nu)

dx dx

4. Solution :

It is given that cosx3

3 3 3 2 3d dcos x sin x (x ) 3x sin(x )

dx dx

nn 1d d dx

(cosnu) sin nu (nu) and nxdx dx dx

5. Solution :

It is given that 2cot x

2 2d dcot x dxcot x 2cot x cot x(cosec x)

dx dx dx

nn 1d d dx

(cosnu) sin nu (nu) and nxdx dx dx


6. Solution :

It is given that 3tan x

3 2 2 2d d(tan x) dxtan x 3tan x 3tan x (sec x)

dx dx dx

na a 1 n 1d d(tan nu) d(nu) dx

(tan nu) a tan nu and nxdx dx dx dx

7. Solution :

It is given that tan x

2

2d d d 1 sec ( x )tan x sec x x (x)

dx dx dx 2 x

2d d d(tan nu) sec nu nu and (nu)

dx dx dx

8. Solution :

It is given that 2xe

2 2x x2 2 xd de e (x ) 2xe

dx dx

n

at at n 1d d dx(e ) e at and nx

dx dx dx

9. Solution :

It is given that cot xe


cot x cot x cot xd de e (cot x) e cosecx

dx dx

10. Solution :

It is given that sin x

d 1 d d 1sin x sin x x cosx

dx dx dx2 sin x 2 sin x

11. Solution

It is given that 6

5 7x

6 5 5 5d d

5 7x 5 7x (5 7x) 6 5 7x 7 42 5 7xdx dx

12. Solution :

It is given that 5

3 4x

5 5 5 5d d

3 4x 4 3 4x (3 4x) 4 3 4x ( 4) 16 3 4xdx dx

n n 1d dy(y ) ny

dx dx

13. Solution :

It is given that 4

23x x 1


4 3 3 3

2 2 2 2 2d d3x x 1 4 3x x 1 (3x x 1) 4 3x x 1 (3 6x 1) 4 3x x 1 6x 1

dx dx

14.Solution:

It is given that 2ax bx c

2dax bx c 2ax b

dx

15.Solution:

It is given that

32

32

1x x 3

x x 3

3 4

2 2

42

d 1x x 3 3 x x 3 (2x 1) 3 2x 1

dx x x 3

16. Solution ;

It is given that 2sin 2x 3

2d d dsin 2x 3 2sin 2x 3 sin(2x 3) 2x 3 4sin 2x 3 cos 2x 3

dx dx dx

17. Solution :

It is given that 2 3cos x


2 3 3 3 2 2 3 3dcos x 2cos sin x 3x 6x cos x sin x

dx

a a 1d d dcos nu a cos nu cosnu nu

dx dx dx

18. Solution:

It is given that y = 3sin X

3 3 3

3

2 3

3 2

3 3

d 1 d dsin x sin x x

dx dx dx2 sin x

3x cos x1cos x 3x

2 sin x 2 sin x

19.Solution:

It is given that xsin x

d 1 dxsin x xsin x

dx dx2 xsin x

sin x x cos x1sin x x cos x

2 xsin x 2 xsin x

20. Solution:

It is given that cot x


2

2

d 1 d dcot x cot x x

dx dx dx2 cot x

1 1sec x

2 x2 cot x

sec x

4 x cot x

21. Solution :

It is given that cos3xsin5x

dcos3xsin5x

dx

d cos3x d sin5xsin5x cos3x sin5x 3sin3x cos3x

dx dx

5cos5x 5cos 3x cos 5x 3sin 5x 3sin 3x

22. Solution :

It is given that sin xsin 2x

d sin 2x d sin xdsin xsin 2x sin x sin 2x

dx dx dx

sin x 2cos2x sin 2x sin x

2sin(x)

cos 2x sin 2x sin x

23. Solution :


It is given that cos sin ax b

dcos sin ax b

dx =

1

21

sin sin ax b cos ax b ax b a2

= 1

2a

sin sin ax b cos ax b ax b2

24. Solution :

It is given that 2xe sin3x

2xde sin3x

dx= 2x 2x[sin3x 2 e ] e 3cos3x

= 2xe 2sin3x 3cos3x

25. Solution :

It is given that 3xde cos2x

dx

3xde cos2x

dx= 3x 3x[cos2x 3 e ] e 2sin 2x

= 3xe 3cos2x 2sin 2x

26. Solution:

It is given that 5xe cot 4x


5xde cot 4x

dx

= 5x 5x 2[cot 4x 5e e ] e 4cosec 4x

= 5x 2e 5cot 4x 4cosec 4x

27. Solution:

It is given that 3 xcos x e

3 xdcos x e

dx= 3 x x 2cos [x e ] e 3x

= x 3 2cos e x 3x

= 3 x x 3 2sin x e e x 3x

28. Solution :

Let y= xsin x cosx

e , t xsin x cosx,p x and q sinx

xsin x cos xt tdyy e e e

dt

According to product of differentiation

d cos xdp dqdt dx q p

dx dx dx

= [sin x 1 ] x cosx sin x

=xcosx

According to the chain rule of differentiation


dy dtdy dx

dt dx

= xsin x cosxe xcosx

29. Solution :

Let y = x x

x x x x

x x

e e,m e e ,n e e

e e

According to the quotient rule of differentiation

If y=m

n

2

dm dnn m

dx dxdy dxn

x x x x x x x x

2x x

e e e e e e e e

e e

=

2 2x x x x

2x x

e e e e

e e

=

x x x x x x x x

2x x

e e e e e e e e

e e


=

x x

2x x

2e 2e

e e

=

2

x x

4

e e

30. Solution :

Let y = 2x 2x

2x 2x 2x 2x

2x 2x

e e,m e e ,n e e

e e


If y=m

n

2

dm dnn m

dx dxdy dxn

2x 2x 2x 2x x 2x 2x 2x

22x 2x

e e 2e 2e e2 e 2e 2e

e e

=

2 22x 2x 2x 2x

22x 2x

2 e e 2 e e

e e

=

x 2x 2x 2x 2x 2x x 2x

22x 2x

2 e2 e e e e e e2 e

e e


=

2x 2x

22x 2x

2 2e 2e

e e

=

2

2x 2x

8

e e

31. Solution :

Let y = 2 2

2 2

2 2

1 x 1 x,m 1 x ,n 1 x , t

1 x 1 x

If t = m

n

2

dm dnn m

dx dxdt dxn

3 3

22

2x 2x 2x 2x

1 x

=

2

2

4x

1 x

According to chain of differentiation

dy dtdy dx

dt dx

=

11

2 2

2 22

1 1 x 4x

2 1 x 1 x


=

11

2 2

2 22

2x 1 x 4x

1 1 x 1 x

12 2

12

2 2

2x 1 x 1

1 11 x

= 1 3

2 22 22x 1 x 1 x

32. Solution:

Let y= 2 2 2 2

2 2 2 2

2 2 2 2

a x a x,m a x ,n a x , t

a x a x

If t = m

n

2

dm dnn m

dt dx dx

dx n

=

2 2 2 2

22 2

a x 2x a x 2x

a x


=

3 3 2 3

22

2xa 2x 2xa 2x

1 x

=

2

22

4xa

1 x


dy dy dt

dx dt dx

11

2 2 22

2 2 22 2

1 a x 4xa

2 a x a x

12 2 2 2

12

2 2 2

2xa a x 1

1 1a x

1 3

2 2 2 2 22 22xa a x a x

33. Solution:

Let y= 1 sin x 1 sin x

,m 1 sin x,n 1 sin x, t1 sin x 1 sin x


If t = m

n

2

dm dnn m

dt dx dx

dx n

=

2

1 sin x cos x 1 sin x cos x

1 sin x

=

2

cos x sin x cos x cos x sin x cos x

1 sin x

=

2

2cos x

1 sin x


dy dy dt

dx dt dx

11

2

2

1 1 sin x 2cos x

2 1 sin x 1 sin x

1

2

12

2

cos x 1 sin x 1

1 11 sin x


1 3

2 2cos x 1 sin x 1 sin x

34. Solution :

Let y= x x

x x

x x

1 e 1 e,m 1 e ,n 1 e , t

1 e 1 e

If t = m

n

2

dm dnn m

dt dx dx

dx n

=

x x x x

2x

1 e e 1 e e

1 e

=

x 2x x 2x

2x

e e e e

1 e

=

x

2x

2e

1 e


dy dy dt

dx dt dx


11

x x2

x 2x

1 1 e 2e

2 1 e 1 e

1x x 2

12

x 2

e 1 e 1

1 11 e

1 3

x x x2 2e 1 e 1 e

35. Solution :

Let y = 2x 3

2x 3e x,m e x ,n cosec2x

cosec2x

If t = m

n

2

dm dnn m

dt dx dx

dx n

=

2x 2 2x 3

2

cosec2x 2e 3x e x 2cosec2xcot 2x

cosec2x


=

2x 2 2x 3

2

2e cosec2x 3x cosec2x 2e cosec2x cot 2x 2x cosec2x cot 2x

cosec2x

=

2x 2

2

2e cosec2x 1 cot 2x 3x cosec2x 1 cot 2x

cosec2x

x 2

2

1 cot 2x 2e cosec2x 3x cosec2x

cosec2x

x 2

2

1 cot 2x 2e 3x cosec2x

cosec2x

x 2

1

1 cot 2x 2e 3x

cosec2x

x 21 cot 2x 2e 3x sin 2x

36. Solution :

Let y = sin sin x cosx t sin x cosx

dy dy dt

dx dt dx


= 1

12

1cos sin sin x cosx sin x cosx cosx sin x

2

= 1

21

cos sin sin x cosx sin x cosx cosx sin x2

37. Solution:

Let y= x xe log sin 2x ,m e and n log sin2x

According to the product rule of differentiation

= x x 1log sin 2x e e 2cos2x

sin 2x

= x 2cos2xe log sin 2x

sin 2x

= xe log sin 2x 2cot 2x

38. Solution:

Let y=

2 22 2

22

1 x 1 xcos ,m 1 x ,n 1 x , t

1 x1 x


If t = m

n

dm dndyn m

dx dx dx


2

dm dnn m

dt dx dx

dx n

=

2 2

22

1 x 2x 1 x 2x

1 x

=

3 3

22

2x 2x 2x 2x

1 x

=

2

2

4x

1 x


dy dy dt

dx dt dx

=

2

2 2

1 x 4xsin

1 x 1 x 2

=

2

2 2

1 x 4xsin

1 x 1 x 2

39. Solution:


Let y=

2 22 2

22

1 x 1 xsin ,m 1 x ,n 1 x , t

1 x1 x

If t = m

n

2

dm dnn m

dt dx dx

dx n

=

2 2

22

1 x 2x 1 x 2x

1 x

=

3 3

22

2x 2x 2x 2x

1 x

=

2

2

4x

1 x


dy dy dt

dx dt dx


=

2

2 2

1 x 4xcos

1 x 1 x 2

40. Solution :

Let y= 2

2sin x x,m sin x x ,n cot 2x

cot 2x

If y = m

n

2

dm dnn m

dy dx dx

dx n

=

2 2 2

2

cot 2x cos x 2x cot 2x 2cosec 2xsin x 2x cosec 2x

cosec2x

=

2 2

2

cot 2x cos x 2x 2cosec 2x sin x x

cosec2x

=

2 2

2 2

2cosec 2x sin x x cot 2x cos x 2x

cosec2x cosec2x

=

2

2

2 sin x x cos2x cos x 2x

11sin 2x

sin 2x

= 22 sin x x cos2xsin 2x cos x 2x


41. Solution:

Let y= cos x sin x

,m cos x sin x, y cos x sin xcos x sin x

If y = m

n

2

dm dnn m

dy dx dx

dx n

=

2

cot x sin x sin x cos x cos x sin x sin x cos x

cos x sin x

=

2 2

2

cos x sin x cos x sin x

cos x sin x

=

2 2

2 2



=

21 cos x sin xy y

1 cos x sin x

= 2dyy 1 0

dx

THEREFORE, IT IS PROVED.

42. Solution:


Let y= cos x sin x

,m cos x sin x,n cos x sin xcos x sin x

If y = m

n

2

dm dnn m

dy dx dx

dx n

=

2

cot x sin x sin x cos x cos x sin x sin x cos x

cos x sin x

=

2 2

2


cos x sin x

=

2 2 2 2

2

cos x sin x 2cos xsin x cos x sin x 2cos xsin x

cos x sin x

=

2 2

2

2 cos x sin x

cos x sin x

=

2 2

2

(1)cos x sin x 1

cos x sin x2


=2

1

cos x sin x

2 2

=2

0 0

1

cos x cos45 sin xsin 45

1 1

= 2

1cosa cosb sin asin b cos a b

cos x4

= 2sec x4


43. Solution:

Let y= 1 1

1 1

1 1

1 x 1 x,m 1 x ,n 1 x , t

1 x 1 x

If t = m

n

2

dm dnn m

dt dx dx

dx n


=

1 1

21

1 x 1 1 x 1

1 x

=

1

21

1 x 1 x

1 x

=

2

2

1 x


dy dtdydx dt dx

=

11

1 2

1 21

1 1 x 2

2 1 x 1 x

=

11 2

1 21

1 1 x 1

1 1 x 1 x

=

11 2

1 111

1 2

1 x 1 1 x1

1 x1 x1 x

(Multiplying and dividing by 1-x)


=

11

1 2

11 2

1 x 11

1 x 1 x1 x

=

11 2

1 21 2

1 x 1 y1

1 x 1 x 1 x1 x

Therefore 2 dy1 x y

dx

2 dy1 x y 0

dx


44. Solution:

secx tan xy

secx tan x

1 sin x1 sin xcos x cos xy

1 sin x 1 sin x

cos x cos x

1 sin xm 1 sin x,n 1 sin x, t

1 sin x


If t = m

n

2

dm dnn m

dt dx dx

dx n

=

2

1 sin x cos x 1 sin x cos x

1 sin x

=

2

cos x sin xcos x cos x sin xcos x

1 sin x

=

2

2cos x

1 sin x


dy dtdydx dt dx

=

11

12

2

1 1 sin x 2cos x

2 1 sin x 1 sin x

=

1

2

12

2

cos x 1 sin x 1

1 11 sin x


=

31 3

22 2

1 sin xcos x 1 sin x 1 sin x

1 sin x

(Multiplying and dividing by 3

21 sin x

=

33 1 3

22 2 2

1cos x 1 sin x 1 sin x

1 sin x

= 3 1 3 3

2 2 2 2cos x 1 sin x 1 sin x 1 sin x

= 3

1 2 2cos x 1 sin x 1 sin x

= 1 3

cosx 1 sin x cosx

= 1 3 1

1 sin x cosx

=2

1 sin x

cos x

=1 1

1 1 sin x

cos x cos x

1 sin xsec

cosx cosx

= secx secx tan x

chapter-28- differentiation

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