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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) √
Differentiating w.r.t x,
(
)
2. Solution :
(i)
Differentiating w.r.t x,
Class XI www.vedantu.com RS Aggarwal Solutions
(
)
(ii)
√
Differentiating w.r.t x,
(
)
(iii)
√
Differentiating w.r.t x,
3. Solution :
(i) 3x-5
Differentiating with respect to x,
( ) (
)
(ii) ⁄
Class XI www.vedantu.com RS Aggarwal Solutions
Differentiating with respect to x,
(
)
(iii) 6. √
Differentiating with respect to x,
4. Solution :
(i) 6x5 + 4x3 – 3x2 + 2x – 7
Differentiating with respect to x,
(6x5 + 4x3 – 3x2 + 2x – 7) = 30x5-1 + 12x3-1 – 6x2-1 + 2x1-1 + 0 =
30x4 + 12x2 – 6x1 + 2x
(ii) ⁄
√ √
Differentiating with respect to x,
( ⁄
√ √
)
= 5×
=
Class XI www.vedantu.com RS Aggarwal Solutions
(iii) ax3 + bx2 + cx + d, where a, b, c, d are constants
Differentiating with respect to x,
(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
Differentiating with respect to x,
= 4.3x3-1 + 3.logn(2).2x + 6×
+ 5 ×- cosec2x
= 12x2 + 3.logn(2).2x -3
- 5 cosec2x
(ii)
√
√ ⁄
Class XI www.vedantu.com RS Aggarwal Solutions
⁄
Differentiating with respect to x,
(
⁄
)
(
)
( )
(
)
( )
6. Solution :
(i) 4 cot -
cos x +
-
+
+ 9
= 4 cot -
cos x + – + + 9
Differentiating with respect to x,
Class XI www.vedantu.com RS Aggarwal Solutions
(
)
(
( ) (
( )
)
( )
(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
Differentiating with respect to x,
(-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)
Class XI www.vedantu.com RS Aggarwal Solutions
( )( )
( )
( ) ( )
( )
= (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
Class XI www.vedantu.com RS Aggarwal Solutions
(iv) .
/
.
/ = 2.
/ .
/
= 2(x +
-
+
) = 2(x +
)
(v) .
/
.
/ = 3.
/ (2x -
)
= 3(2x3 -
-
+
)
= 3(2x3 -
+
)
(vi) (2x2 + 5x – 1) (x – 3)
(2x2 + 5x – 1) (x – 3)
Class XI www.vedantu.com RS Aggarwal Solutions
= (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) ( )( )
( )( )
( )( ) ( )( )
Class XI www.vedantu.com RS Aggarwal Solutions
2( )
( ) ( )
( )( )3 ( )( )
*( ) ( ) + ( )( )
*( + ( )( )
( )
(iii)
√
√
√
( ) ( )
√
√ ( )
√ ( )
√
√
Class XI www.vedantu.com RS Aggarwal Solutions
(iv) ( )√
√
( )√
√
√ ( )√ ( )√
√
√ 2( )
√
√ ( )3 ( )√
√ {( )
√ )} ( )√
√ {
√ }
4
5
√
(
)
√
(V)
√
Class XI www.vedantu.com RS Aggarwal Solutions
√ √
√
√
√
√ ( )
( )
(vi)
( ) ( )
( )
9. Solution :
(i) Given that y = 6x5 – 4x4 – 2x2 + 5x – 9
Differentiating with respect to x,
(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
Class XI www.vedantu.com RS Aggarwal Solutions
(ii) Given y = (sin x + tan x)
Differentiating with respect to x,
(sinx + tanx) = cos x + sec2 x
Putting x=
.
= cos
+ sec2
=
+ 4 =
(iii) Given y= (
= 2cosec x-3cot x
Differentiating with respect to 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. =
=
√
.√
√ / (
=
.√
√ /
√
)
Class XI www.vedantu.com RS Aggarwal Solutions
= √
√ √
√
= √ = RHS
11. Solution :
To prove: (2xy)
.
/.
LHS = (2xy)
LHS = (2xy) (√
√
) (
√
√
) (
(√
√
)
√
√
)
LHS = (√
√
) (√
√
)
LHS = (√
√
)
12. Solution :
√
Class XI www.vedantu.com RS Aggarwal Solutions
√
= 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
Class XI www.vedantu.com RS Aggarwal Solutions
=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+
( )
Putting values in (i), we get
F’(x) =h 0lim
0 ( )
( )1 0
1
Class XI www.vedantu.com RS Aggarwal Solutions
=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
Class XI www.vedantu.com RS Aggarwal Solutions
=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
Putting values in (i), we get
f’(x) =h 0lim
( )
Class XI www.vedantu.com RS Aggarwal Solutions
= h 0lim
= h 0lim
)
=
h 0lim
Putting h = 0, we get
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
Putting values in (i), we get
Class XI www.vedantu.com RS Aggarwal Solutions
( ) h 0lim
( ) ( ) ( ) , -
=h 0lim
( ) ( ) ( )
=h 0lim
,( ) ] ,( ) - ,
= h 0lim
[ ] [ ]
= h 0lim
[ ] , -
=h 0lim
Putting h = 0, we get
Class XI www.vedantu.com RS Aggarwal Solutions
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
Putting values in (i), we get
( ) h 0lim
( )
=h 0lim
( )
( )
= 8x8-1 = 8x7
6…Solution :
f(x)=
( ) h 0lim
( ) ( )
….(i)
Class XI www.vedantu.com RS Aggarwal Solutions
f(x)=
and f(x+h)=
( )
Putting values in (i), we get
( ) h 0lim
( )
=h 0lim
( )
( ) = (-3) x3-1 = -3x-4 = -
7…Solution :
f(x) =
and ( ) lh 0lim
( ) ( )
….(i)
f(x) =
and f(x+h) =
( )
Putting values in (i), we get
( ) lh 0lim
( )
=
lh 0lim
( )
( )
Class XI www.vedantu.com RS Aggarwal Solutions
= (-5) x5-1 = -5x-6 =-
8….Solution :
f(x) = √
and ( ) lh 0lim
( ) ( )
….(i)
f(x) = √ and f(x+h)=√ ( )
=√
Putting values in (i), we get
( ) lh 0lim
√ √
Class XI www.vedantu.com RS Aggarwal Solutions
=lh 0lim
√ √
√ √
√ √
=lh 0lim
(√ ) (√ )
(√ √ )
=lh 0lim
(√ √ )
=lh 0lim
(√ √ )
=lh 0lim
√ √
Putting h = 0, we get
√ ( ) √
Class XI www.vedantu.com RS Aggarwal Solutions
=
√ √ =
√
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
(√ √
Class XI www.vedantu.com RS Aggarwal Solutions
=lh 0lim
(√ √
Putting h = 0, we get
=
√ ( ) √ =
√
10. ….Solution :
f(x) =
√
and ( ) lh 0lim
( ) ( )
…..(i)
f (x) =
√ f(x+h) =
√
Putting values in (i), we get ( ) lh 0lim
√
√
( ) lh 0lim
√ √
(√ ) (√ )
=lh 0lim
√ √
(√ )(√ ) √ √
√ √
Class XI www.vedantu.com RS Aggarwal Solutions
=lh 0lim
(√ ) (√ ) (√ ) √ )
=lh 0lim
(√ ) (√ ) (√ ) √ )
=lh 0lim
(√ ) (√ ) (√ ) √ )
Putting h = 0, we get
=
(√ )(√ )(√ √ ) =
(√ ) ( √ )
=
(√ )
11….Solution :
f(x) =
√ ( )
lh 0lim
( ) ( )
…..(i)
f (x) =
√ f(x+h) =
√
Putting values in (i), we get ( ) lh 0lim
√
√
Class XI www.vedantu.com RS Aggarwal Solutions
=lh 0lim
√ √
(√ ) (√ )
=lh 0lim
√ √
(√ ) (√ )
√ √
(√ ) (√ )
=lh 0lim
(√ )
(√ )
(√ ) (√ ) (√ ) √ )
=lh 0lim
(√ ) (√ ) (√ ) √ )
=lh 0lim
(√ ) (√ ) (√ ) √ )
Putting h = 0, we get
=
(√ ) ( √ )
=
(√ ) =
(√ )
12…Solution :
Class XI www.vedantu.com RS Aggarwal Solutions
f(x) =
√
and ( ) lh 0lim
( ) ( )
…..(i)
f (x+h) =
√
Putting values in (i), we get ( ) lh 0lim
√
√
= lh 0lim
√ √
(√ )(√ )
= lh 0lim
√ √
(√ )(√ ) √ √
√ √
=
2 2
h 0
6x 5 6x 6h 5
h 6x 6h 5 6x 5 6x 6h 5lim
=lh 0lim
(√ ) (√ ) (√ ) √ )
Class XI www.vedantu.com RS Aggarwal Solutions
=lh 0lim
(√ ) (√ ) (√ ) √ )
=lh 0lim
(√ ) (√ ) (√ ) √ )
Putting h = 0, we get
=
(√ ( ) ) (√ ) (√ ) √ )
(√ ) ( √
=
(√ )
13…..Solution :
f(x) =
√ ( )
lh 0lim
( ) ( )
…..(i)
Class XI www.vedantu.com RS Aggarwal Solutions
f(x+h) =
√ ( )
√
Putting values in (i), we get ( ) lh 0lim
√
√
= lh 0lim
√ √
(√ )(√ )
=lh 0lim
√ √
(√ )(√ ) √ √
√ √
=lh 0lim
(√ )
(√ )
(√ ) (√ ) (√ ) √ )
=lh 0lim
(√ ) (√ ) (√ ) √ )
=lh 0lim
(√ ) (√ ) (√ ) √ )
=lh 0lim
(√ ) (√ ) (√ ) √ )
Putting h = 0, we get
Class XI www.vedantu.com RS Aggarwal Solutions
=
(√ ( )) (√ ) (√ ) √ ( )) =
(√ )
14… Solution :
f(x) =
( )
lh 0lim
( ) ( )
…..(i)
f(x+h)= ( )
( )
Putting values in (i), we get ( ) lh 0lim
=lh 0lim
( )( ) ( )( )
( )( )
=lh 0lim
, -
(( )( ))
=lh 0lim
(( )( ))
=lh 0lim
(( )( ))
Class XI www.vedantu.com RS Aggarwal Solutions
Putting h = 0, we get
=
(( ( ) )( )) =
( )
15….Solution :
f(x) =
( )
lh 0lim
( ) ( )
…..(i)
f(x+h)= ( )
( )
Putting values in (i), we get ( ) lh 0lim
=lh 0lim
( )( ) ( )( )
( )( )
=lh 0lim
h , h -
( )( )
=lh 0lim
h h
( )( )
=lh 0lim
( )( )
Class XI www.vedantu.com RS Aggarwal Solutions
Putting h = 0, we get
=
( )
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
Putting values in (i), we get
’(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
Class XI www.vedantu.com RS Aggarwal Solutions
=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
Putting h = 0, we get
=
2
2
x 1
x
17…Solution :
f(x) = √
( ) lh 0lim
( ) ( )
…..(i)
f(x+h) = √ ( )
Class XI www.vedantu.com RS Aggarwal Solutions
=√ ( )
Putting values in (i), we get
( ) lh 0lim
√ ( ) √
=lh 0lim
√ ( ) √
√ ( ) √
√ ( ) √
=lh 0lim
(√ ( ))
(√ )
(√ ( ) (√ )
=lh 0lim
( )
√ ( ) √ )
lh 0lim
3x 3h 3x 3x 3h 3x2sin sin
2 2
h cos 3x 3h cos3x
Class XI www.vedantu.com RS Aggarwal Solutions
=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
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
=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
Class XI www.vedantu.com RS Aggarwal Solutions
= 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
Class XI www.vedantu.com RS Aggarwal Solutions
=
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:
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
= 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
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
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:
(
)
=
Class XI www.vedantu.com RS Aggarwal Solutions
3 Solution :
.
( )/=
( )
( )
=
( )
4
.
( )/=
( )
( )
= ( )
( )
= ( )
( )
5.Solution:
0( )
( )1= ( ) ( )
( )
=
( )
=
( )
6 Solution:
.
/ =
( ) ( ) ( )
( )
=
( )
Class XI www.vedantu.com RS Aggarwal Solutions
=
( )
7 Solution:
0( )
( )1 =
( ) ( ) ( )
( )
=
( )
=
( )
8 Solution:
.
/=( ) ( ) ( ) ( )
( )
=
( )
=
( ) =
( )
( ) =
( )
( )
9 Solution:
0
( )1 =
( ) ( ) ( )
( )
=
( ) =
( )
10. Solution
0
1= ( ) ( ) ( )
( )
Class XI www.vedantu.com RS Aggarwal Solutions
= , ( ) ( )-
( )
11 Solution
[(√ √ )
(√ √ )] =
√ (√ √ ) (√ √ )
√
(√ √ )
=
√
√
√
√
(√ √ ) =
√
√ (√ √ )
12. Solution
0
1 =
( ) ( ) .
/
( )
= ( ) ( )
( )
13 Solution:
0
√ 1 =
(√ ) ( ) .
√ /
(√ )
=
√ (√ )
14.Solution
[
( )]
( ) ( ) ( )
( )
Class XI www.vedantu.com RS Aggarwal Solutions
( )
( )
15. Solution :
(
)
( ) ( ) ( )
( )
)
( )
( )
16. Solution:
[
]
( ) ( ) ( )
( )
( )
( )
Class XI www.vedantu.com RS Aggarwal Solutions
17.. Solution :
[
]
( ) ( ) ( ) ( )
( )
( )
( )
( )
18.Solution :
[
]
( )( ) ( )( )
( )
Class XI www.vedantu.com RS Aggarwal Solutions
( )( ) ( )( )
( )
( ) ,( ) ( )( )-
( )
( ) ,( ) ( )-
( )
( ) , -
( )
, -
( )
19.. Solution :
.(
/
( ) ( ) ( )
( )
Class XI www.vedantu.com RS Aggarwal Solutions
( )( ) ( )
( )
20.. Solution :
(
)
( ) ( ) ( ) ( )
( )
( ) ( ) ( )
( )
,( ) ( ) ( )-
,( ) ( ) ( )-
,( ) ( )-
( )
( )
Class XI www.vedantu.com RS Aggarwal Solutions
21. Solution :
(
√ )
{ , - √ } {( ) (
√ )}
(√ )
{ , - √ } {( ) (
√ )}
{ , - √ } {( ) (
√ )}
* , -+ *( )+
√
* , -+ *( )+
22.. Solution :
Class XI www.vedantu.com RS Aggarwal Solutions
( ( )
( ))
( )( ) , ( )-( )
( )
( )
( )
( )
( )
23.. Solution :
(
( ))
( )( ) , -, ( ) -
( )
( ),( ) ( )( )-
( )
( )
Class XI www.vedantu.com RS Aggarwal Solutions
( )
( )
24.. Solution :
6
7
( )( ) , -, -
( )
, -
( )
( ) ( )
( )
25.. Solution :
Class XI www.vedantu.com RS Aggarwal Solutions
[( )
( )]
( )( ) ( )( )
( )
( ) ( )
( )
( )
( )
( )
( )
26. Solution :
.
/
( )( ) ( )( )
( )
Class XI www.vedantu.com RS Aggarwal Solutions
( )
( )
= - 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.
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
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 :
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
=
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
According to the quotient rule of differentiation
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
Class XI www.vedantu.com RS Aggarwal Solutions
=
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
Class XI www.vedantu.com RS Aggarwal Solutions
=
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
Class XI www.vedantu.com RS Aggarwal Solutions
=
3 3 2 3
22
2xa 2x 2xa 2x
1 x
=
2
22
4xa
1 x
According to the chain rule of differentiation
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
According to the chain rule of differentiation
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
According to the chain rule of differentiation
dy dy dt
dx dt dx
Class XI www.vedantu.com RS Aggarwal Solutions
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
Class XI www.vedantu.com RS Aggarwal Solutions
=
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
Class XI www.vedantu.com RS Aggarwal Solutions
= 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
According to the quotient rule of differentiation
If t = m
n
dm dndyn m
dx dx dx
Class XI www.vedantu.com RS Aggarwal Solutions
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
According to the chain rule of differentiation
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:
Class XI www.vedantu.com RS Aggarwal Solutions
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
According to the chain rule of differentiation
dy dy dt
dx dt dx
Class XI www.vedantu.com RS Aggarwal Solutions
=
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
cos x sin x cos x sin x
cos x sin x cos x sin x
=
21 cos x sin xy y
1 cos x sin x
= 2dyy 1 0
dx
THEREFORE, IT IS PROVED.
42. Solution:
Class XI www.vedantu.com RS Aggarwal Solutions
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 cos x sin x
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
Class XI www.vedantu.com RS Aggarwal Solutions
=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
THEREFORE, IT IS PROVED.
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
Class XI www.vedantu.com RS Aggarwal Solutions
=
1 1
21
1 x 1 1 x 1
1 x
=
1
21
1 x 1 x
1 x
=
2
2
1 x
According to the chain rule of differentiation
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)
Class XI www.vedantu.com RS Aggarwal Solutions
=
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
THEREFORE, IT IS PROVED.
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
Class XI www.vedantu.com RS Aggarwal Solutions
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
According to the chain rule of differentiation
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
Class XI www.vedantu.com RS Aggarwal Solutions
=
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
Class XI www.vedantu.com RS Aggarwal Solutions
THEREFORE, IT IS PROVED.