NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Maths
Chapter : 13
Chapter Name : Limits and Derivatives

Exercise 13.1

Q1 Evaluate the Given limit: lim x→3 x + 3

Answer. lim x→3 x + 3 = 3 + 3 = 6

Page : 301 , Block Name : Exercise 13.1

Q2 Evaluate the Given limit: lim
22
x→π (x − )
7

Answer. lim x→π (x −
22

7
) = (π −
22

7
)

Page : 301 , Block Name : Exercise 13.1

Q3 Evaluate the Given limit: lim r→1 πr
2

Answer. lim r→1 πr
2
= π(1)
2
= π

Page : 301 , Block Name : Exercise 13.1

Q4 Evaluate the Given limit: lim
4x+3
x→4
x−2

4(4)+3
Answer. lim
4x+3 16+3 19
x→4 = = =
x−2 4−2 2 2

Page : 301 , Block Name : Exercise 13.1

10 5

Q5 Evaluate the Given limit: lim
x +x +1
x→−1
x−1

10 5 10 5
(−1) +(−1) +1
Answer. lim
x +x +1 1−1+1 1
x→−1 = = = −
x−1 −1−1 −2 2

Page : 301 , Block Name : Exercise 13.1

5
(x+1) −1
Q6 Evaluate the Given limit: lim x→0 x

5
(x+1) −1
Answer. lim x→0
x

Put x + 1 = y so that y → 1 as x → 0.
5 5
(x + 1) − 1 y − 1
Accordingly, lim = lim
x y→1 y − 1

5 5
y − 1
= lim
y→1 y − 1
n n
5−1 x −a n−1
= 5.1 [limx→a = na ]
x−a

=5
5
(x+5) −1
∴ limx→0 = 5
x

Page : 301 , Block Name : Exercise 13.1

2

Q7 Evaluate the Given limit: lim
3x −x−10
x→2 2
x −4

Answer. At x = 2, the value of the given rational function takes the form
0

0
2
3x − x − 10 (x − 2)(3x + 5)
∴ lim = lim
2
x→2 x − 4 x→2 (x − 2)(x + 2)

3x + 5
= lim
x→2 x + 2

3(2) + 5
=
2 + 2

11
=
4

Page : 301 , Block Name : Exercise 13.1

4

Q8 Evaluate the Given limit: lim
x −81
x→3 2
2x −5x−3

Answer. At x = 2, the value of the given rational function takes the form 0

0
2
4 (x − 3)(x + 3) (x + 9)
x − 81
∴ lim = lim
2
x→3 2x − 5x − 3 x→3 (x − 3)(2x + 1)

2
(x + 3) (x + 9)
= lim
x→3 2x + 1
2
(3 + 3) (3 + 9)
=
2(3) + 1

6 × 18
=
7

108
=
7

Page : 301 , Block Name : Exercise 13.1

Q9 Evaluate the Given limit: lim
ax+b
x→0
cx+1

a(0)+b
Answer. lim
ax+b

x→0 = = b
cx+1 c(0)+1

Page : 301 , Block Name : Exercise 13.1

1

Q10 Evaluate the Given limit: lim
z 3 −1
z→1 1

z 6 −1

1

Answer. lim
z 3 −1
z→1 1

z 6 −1

At z = 1, the value of the given function takes the form
0

0
1

Put z 6
= x so that z →1 as x → 1.
1
2
z3 1 x − 1
Accordingly, lim = lim
1
z → 1 x→1 x − 1
z6 − 1
2 2
x − 1
= lim
x→1 x − 1
n n
x −a n−1
[limx→a = na ]
2−1 x−a
= 2.1

= 2

Page : 301 , Block Name : Exercise 13.1

2

Q11 Evaluate the Given limit: lim
ax +bx+c
x→1 ,a + b + c ≠ 0
2
cx +bx+a

Answer.
2 2
ax + bx + c a(1) + b(1) + c
lim =
2 2
x→1 cx + bx + a c(1) + b(1) + a

a + b + c
=
a + b + c

= 1 [a + b + c ≠ 0]

Page : 301 , Block Name : Exercise 13.1

1 1
+

Q12 Evaluate the Given limit: lim
x 2

x→−2
x+2

1 1
+

Answer. lim
x 2

x→−2
x+2

At x = –2, the value of the given function takes the form
0

0

2+x
( )
1 1 2x

Now, lim = lim
x→−2 x → −2 x + 2 x→−2 x + 2

1
= lim
x→−2 2x

1 −1
= =
2(−2) 4

Page : 301 , Block Name : Exercise 13.1

Q13 Evaluate the Given limit: lim
sin ax
x→0
bx

Answer. lim
sin ax
x→0
bx

At x = 0, the value of the given function takes the form
0

0

sin ax sin ax ax
Now, lim = lim ×
x→0 bx x→0 ax bx

sin ax a
= lim ( ) × ( )
x→0 ax b

a sin ax
= lim ( ) [x → 0 ⇒ ax → 0]
b ax→0 ax

a
= × 1
b
a
=
b

Page : 301 , Block Name : Exercise 13.1

Q14 Evaluate the Given limit: lim
sin ax
x→0 , a, b ≠ 0
sin bx

Answer. lim
sin ax
x→0 , a, b ≠ 0
sin bx
sin ax
( )×ax

Now, lim
ax
sin ax
x→0 = limx→0
sin bx sin bx
( )×bx
bx

sin ax
lima→0 ( ) x → 0 ⇒ ax → 0
ax
a
= ( ) × [ ]
b sin bx
limbx→0 ( ) and x → 0 ⇒ bx → 0
bx

a 1 sin y
= ( ) × [limy→0 = 1]
b 1 y

a
=
b

Page : 301 , Block Name : Exercise 13.1

sin(π−x)
Q15 Evaluate the Given limit: lim x→π
π(π−x)

sin(π−x)
Answer. lim x→π
π(π−x)

It is seen that x → π ⇒ (π – x) → 0

sin(π − x) 1 sin(π − x)
∴ lim = lim
x→π π(π − x) π (π−x)→0 (π − x)

1 sin y
= × 1 [lim = 1]
π y→0 y

1
=
π

Page : 302 , Block Name : Exercise 13.1

Q16 Evaluate the Given limit: lim
cos x
x→0 π−x

Answer. lim
cos x cos 0 1
x→0 = =
π−x π−0 π

Page : 302 , Block Name : Exercise 13.1

Q17 Evaluate the Given limit: lim
cos 2x−1
x→0
cos x−1

Answer. lim
cos 2x−1
x→0
cos x−1

At x = 0, the value of the given function takes the form .
0

0

Now,
2
cos 2x−1 1−2 sin x−1 2 x
limx→0 = limx→0 x
[cos x = 1 − 2 sin ]
cos x−1 2
1−2 sin −1 2
2
2
sin x 2
2
( )×x
sin x x2
= limx→0 x
= limx→0
2
sin 2 x
⎛ sin ⎞
2 2 x2
×
2 4
⎝ ( x ⎠
)
2
2
sin x
limx→0 ( )
2
x
= 4
2 x
⎛ sin ⎞
2
limx→0
2
⎝ (x) ⎠
2
2
sin x
(limx→0 )
x
x
= 4 [x → 0 ⇒ → 0]
2
x 2
sin
2
(limx→0 x
)

2

2 sin y
1
= 4 [limy→0 = 1]
2 y
1

= 4

Page : 302 , Block Name : Exercise 13.1

Q18 Evaluate the Given limit: lim
ax+x cos x
x→0
b sin x

Answer. lim
ax+x cos x
x→0
b sin x

At x = 0, the value of the given function takes the form .
0

0

Now,

ax + x cos x 1 x(a + cos x)
lim = lim
x→0 b sin x b x→0 sin x

1 x
= lim ( ) × lim (a + cos x)
b x→0 sin x x→0

1 1
= × × lim (a + cos x)
b sin x x→0
(limx→0 )
x

1 sin x
= × (a + cos 0) [limx→0 = 1]
b x

a+1
=
b

Page : 302 , Block Name : Exercise 13.1

Q19 Evaluate the Given limit: lim x→0 x sec x

Answer. lim
x 0 0
x→0 x sec x = limx→0
cos x
= = = 0
cos 0 1

Page : 302 , Block Name : Exercise 13.1

Q20 Evaluate the Given limit: lim
sin ax+bx
x→0 a, b, a + b ≠ 0
ax+sin bx

Answer. At x = 0, the value of the given function takes the form
0

0

Now,
sin ax+bx
limx→0
ax+sin bx

sin ax
( )ax+bx
ax

= limx→0
sin bx
ax+bx( )
bx

sin ax
(lima→0 )×limx→0 (ax)+limx→0 bx
ax

= [ As x → 0 ⇒ ax → 0 and bx → 0]
sin bx
limx→0 ax+limx→0 bx(limhx→0 )
bx

limx→0 (ax)+lim bx sin x
= [limx→0 = 1]
limx→0 ax+lim bx x

limx→0 (ax+bx)
=
limx→0 (ax+bx)

= limr→0 (1)

= 1

Page : 302 , Block Name : Exercise 13.1

Q21 Evaluate the Given limit: lim x→0 (csc x − cot x)

Answer. At x = 0, the value of the given function takes the form 0

0

Now,

lim ( cosecx − cot x)
x→0

1 cos x
= lim ( − )
x→0 sin x sin x

1 − cos x
= lim ( )
x→0 sin x

1−cos x
( )
x

= lim
x→0 sin x
( )
x

1−cos x
limx→0
x
=
sin x
limx→0
x

0 1−cos x sin x
= [limx→0 = 0 and limx→0 = 1]
1 x x

= 0

Page : 302 , Block Name : Exercise 13.1

Q22 Evaluate the Given limit: lim
tan 2x
π
x→ π
2 x−
2

Answer. lim
tan 2x
π
x→ π
2 x−
2

At x = , the value of the given function takes the form
π 0

2 0

Now, put x − π
= y
so that
x →
π
,y → 0
2 2
π
tan 2(y+ )
tan 2x 2

∴ limx→ π π
= limy→0
2 x− y
2

tan(π+2y)
= limy→0
y

tan 2y
= limy→0 [tan(π + 2y) = tan 2y]
y

sin 2y
limy→0
y cos 2y

sin 2y 2
= limy→0 ( × )
2y cos 2y

sin 2y 2
= ( lim ) × lim ( ) [y → 0 ⇒ 2y → 0]
2y→0 2y y→0 cos 2y sin x
[limx→0 = 1]
x
2
= 1 ×
cos 0
2
= 1 ×
1

= 2

Page : 302 , Block Name : Exercise 13.1

2x + 3, x ≤ 0
Q23 Find lim x→0 f (x) and limx→1 f (x), where f (x) = {
3(x + 1), x > 0

Answer. The given function is

2x + 3, x ≤ 0
f (x) = {
3(x + 1), x > 0

limx→0 f (x) = limx→0 [2x + 3] = 2(0) + 3 = 3

limx→0+ f (x) = limx→0 3(x + 1) = 3(0 + 1) = 3

∴ limx→0+ f (x) = limx→0+ f (x) = limx→0 f (x) = 3

limx→1 f (x) = limx→1 3(x + 1) = 3(1 + 1) = 6

limx→1+ f (x) = limx→1 3(x + 1) = 3(1 + 1) = 6

∴ limx→1− f (x) = limx→1+ f (x) = limx→1 f (x) = 6

Page : 302 , Block Name : Exercise 13.1

2
x − 1, x ≤ 1
Q24 Find lim x→1 f (x), where f (x) = {
2
−x − 1, x > 1

Answer. The given function is
2
x − 1, x ≤ 1
f (x) = {
2
−x − 1, x > 1

2 2
lim + f (x) = limx→1 [x − 1] = 1 − 1 = 1 − 1 = 0
x→1

2 2
lim + f (x) = limx→1 [−x − 1] = −1 − 1 = −1 − 1 = −2
x→1

It is observed that limx→1+ f (x) ≠ limx→1+ f (x)

Hence, limx→1 f (x) does not exist.

Page : 302 , Block Name : Exercise 13.1

|x|
, x ≠ 0
Q25 Evaluate lim x→0 f (x), where f (x) = {
x

0, x = 0

Answer. The given function is
|x|
, x ≠ 0
x
f (x) = {
0, x = 0

|x|
lim f (x) = lim [ ]
x→0
+
x→0 x

−x
= lim ( ) [ When x is negaitve |x| = −x]
x→0 x

= lim (−1)
x→0

= −1

|x|
lim f (x) = lim [ ]
x→0
+
x→0
+
x
x
= lim [ ] [ When x is positive, |x| = x]
x→0
+ x

= lim (1)
x→0

= lim (1)
x→0

= 1

It is observed that lim 0 f (x) ≠ lim + f (x)
x→0 x→0

Hence, limx→0 f (x) does not exist.

Page : 302 , Block Name : Exercise 13.1

x
, x ≠ 0
Q26 Find lim
|x|
x→0 f (x), where f (x) = {

0, x = 0

Answer. The given function is
x
, x ≠ 0
|x|
f (x) = {
0, x = 0

x
lim f (x) = lim [ ]
x→0
−
x→0
0 |x|

x
= lim [ ] [ When x < 0, |x| = −x]
x→0 −x

= lim (−1)
x→0

= −1

x
lim f (x) = lim [ ]
x→0
+
x→0
+
|x|

x
= lim [ ] [ When x > 0, |x| = x]
x→0
+ x

= lim (1)
x→0

= 1

It is observed that limx→0+ f (x) ≠ limx→0+ f (x)

Hence, limx→0 f (x) does not exist.

Page : 302 , Block Name : Exercise 13.1

Q27 Find lim x→5 f (x), where f (x) = |x| − 5

Answer. The given function is f(x) = lim x→5 f (x), where f (x) = |x| − 5

lim f (x) = lim [|x| − 5]
x→5 x→5
[ When x > 0, |x| = x]
= lim (x − 5)
x→5

=5-5

=0
lim f (x) = lim (|x| − 5)
+
x→−5 x→5
[ When x > 0, |x| = x]
= lim (x − 5)
x→4

= 5 − 5

= 0

∴ limx→5 f (x) = lim + f (x) = 0
x→5

Hence, limx→5 f (x) = 0

Page : 302 , Block Name : Exercise 13.1

a + bx, x < 1
⎧
⎪

Q28 Suppose f (x) = ⎨ 4, x = 1
⎩
⎪
b − ax, x > 1

limx→1 f (x) = f (1) what are possible values of a and b?

Answer. The given function is
⎧ a + bx, x < 1
⎪
f (x) = ⎨ 4, x = 1
⎩
⎪
b − ax x > 1

lim + f (x) = limx→1 (a + bx) = a + b
x→1

lim + f (x) = limx→1 (b − ax) = b − a
x→1

f (1) = 4

It is given that limx→1 f (x) = f (1)

∴ lim − f (x) = lim + f (x) = limx→1 f (x) = f (1)
x→1 x→1

⇒ a + b = 4 and b − a = 4

On solving these two equations, we obtain a = 0 and b = 4

Thus, the respective possible values of a and b are 0 and 4.

Page : 302 , Block Name : Exercise 13.1

Q29
Let a1 , a2 , … , an be fixed real numbers and define a function

f (x) = (x − a1 ) (x − a2 ) … (x − an )

What is lim x→a1 f (x)? For some a ≠ a11 a2 , … , an , compute limx→a f (x)

Answer. The given function is f (x) = (x − a ) (x − a ) … (x − a )
1 2 n

lim f (x) = lim [(x − a1 ) (x − a2 ) … (x − an )]
x→a1 x→a1

= [ lim (x − a1 )] [ lim (x − a2 )] … [ lim (x − an )]
x→a1 x→a1 x→a1

= (a1 − a1 ) (a1 − a2 ) … (a1 − an ) = 0

∴ lim f (x) = 0
x→a1

Now, lim f (x) = lim [(x − a1 ) (x − a2 ) … (x − an )]
x→a x→a

= [lim (x − a1 )] [lim (x − a2 )] … [lim (x − an )]
x→a x→a x→a

= (a − a1 ) (a − a2 ) … (a − an )

∴ lim f (x) = (a − a1 ) (a − a2 ) … (a − an )
x→a

Page : 303 , Block Name : Exercise 13.1

|x| + 1, x < 0
⎧
⎪

Q30 If f (x) = ⎨ 0, x = 0
⎩
⎪
|x| − 1, x > 0

For what value (s) of a does lim x→a f (x) exists?

Answer. The given function is
⎧ |x| + 1, x < 0
⎪

f (x) = ⎨ 0, x = 0
⎩
⎪
|x| − 1, x > 0

When a = 0

lim f (x) = lim (|x| + 1) [ If x < 0, |x| = −x]
+
x→0 x→0

= lim (−x + 1)
x→0

= −0 + 1

= 1

lim f (x) = lim (|x| − 1) [ If x > 0, |x| = x]
+ +
x→0 x→0

= lim (x − 1)
+
x→0

= 0 − 1

= −1

Here, it is observed that lim − f (x) ≠ lim + f (x)
x→0 x→0

∴ limx→0 f (x) does not exist.

When a < 0,
lim f (x) = lim (|x| + 1)
x→σ x→∞

= lim (−x + 1) [x < a < 0 ⇒ |x| = −x]
x→a

= −a + 1

lim f (x) = lim (|x| + 1) [a < x < 0 ⇒ |x| = −x]
− ∗
x→a x→a

= lim (−x + 1)
x→∞

= −a + 1

∴ limx→a f (x) = limx→a′ f (x) = −a + 1

Thus, limit of f (x) exists at x = a, where a < 0

When a > 0
limx→a− f (x) = limx→a− (|x| − 1)

= limx→a (x − 1) [0 < x < a ⇒ |x| = x]

= a-1
limx→a+ f (x) = limx→a+ (|x| − 1)

= limx→a (x − 1) [0 < a < x ⇒ |x| = x]

=a-1
∴ limx→a− f (x) = limx→a∗ f (x) = a − 1

Thus, limit of f (x) exists at x = a, where a > 0

Thus, lim x→a f (x) exists for all a ≠ 0.

Page : 303 , Block Name : Exercise 13.1

f (x)−2
Q31 If the function f (x) satisfies lim x→1 = π, evaluate limx→1 f (x)
x2 −1

Answer.
f (x)−2
limx→1 2
= π
x −1

lim(f (x)−2)
⇒ 2
= π
lim(x −1)

2
⇒ limx→1 (f (x) − 2) = π limx→1 (x − 1)

2
⇒ limx→1 (f (x) − 2) = π (1 − 1)

⇒ limx→1 (f (x) − 2) = 0

⇒ limx→1 f (x) − limx→1 2 = 0

⇒ limx→1 f (x) − 2 = 0

∴ limx→1 f (x) = 2

Page : 303 , Block Name : Exercise 13.1

2
⎧ mx + n, x < 0
⎪

Q32 If f (x) = ⎨ nx + m, 0 ≤ x ≤ 1 . For what integers m and n does both limx→0 f (x)
⎩
⎪ 3
nx + m, x > 1

and lim x→1 f (x) exist?

Answer. The given function is
2
⎧ mx + n, x < 0
⎪

f (x) = ⎨ nx + m, 0 ≤ x ≤ 1
⎩
⎪ 3
nx + m, x > 1
2
lim f (x) = lim (mx + n)
−
x→0 x→0

2
= m(0) + n

= n

lim f (x) = lim (nx + m)
+
x→0 x→0

= n(0) + m

= m

lim f (x) exists if m = n
x→0

Thus, lim f (x) = lim (nx + m)
−
x→1 x→1

= n(1) + m

=m+n
3
lim f (x) = lim (nx + m)
+
x→1 x→1

3
= n(1) + m

= m + n

∴ limx→1− f (x) = limx→1+ f (x) = limx→1 f (x)

Thus, lim x→1 f (x) exists for any integral value of m and n.

Page : 303 , Block Name : Exercise 13.1

Exercise 13.2

Q1 Find the derivative of x – 2 at x = 10. 2

Answer.
2
Let f (x) = x − 2. Accordingly

f (10 + h) − f (10)
′
f (10) = lim
h→0 h
2 2
[(10 + h) − 2] − (10 − 2)
= lim
h→0 h
2 2 2
10 +2⋅10⋅h+h −2−10 +2
= limh→0
h

2
20h+h
= limh→0
h

= limh→0 (20 + h) = (20 + 0) = 20

Thus, the derivative of x – 2 at x = 10 is 20. 2

Page : 312 , Block Name : Exercise 13.2

Q2 Find the derivative of x at x = 1.

Answer. Let f(x) = x. Accordingly,

f (1 + h) − f (1)
′
f (1) = lim
h→0 h

(1 + h) − 1
= lim
h→0 h

h
= lim
h→0 h

= lim (1)
h→0

= 1

Thus, the derivative of x at x = 1 is 1.

Page : 312 , Block Name : Exercise 13.2

Q3 Find the derivative of 99x at x = l00.

Answer. Let f(x) = 99x. Accordingly,
f (100 + h) − f (100)
′
f (100) = lim
h→0 h

99(100 + h) − 99(100)
= lim
h→0 h
99×100+99h−99×100
= limh→0
h

99h
= limh→0
h

= limh→0 (99) = 99

Thus, the derivative of 99x at x = 100 is 99.

Page : 312 , Block Name : Exercise 13.2

Q4 Find the derivative of the following functions from rst principle.
3
(i) x − 27 (ii) (x − 1)(x − 2)

1 x+1
(iii) 2
( iv )
x x−1

Answer. (i) Let f (x) = x 3
− 27. . Accordingly, from the rst principle,
f (x + h) − f (x)
′
f (x) = lim
h→0 h
3 3
[(x + h) − 27] − (x − 27)
= lim
h→0 h
3 3 2 2 3
x +h +3x h+3xh −x
= limh→0
h

3 2 2
h +3x h+3xh
= limh→0
h

2 2
= limh→0 (h + 3x + 3xh)

2 2
= 0 + 3x + 0 = 3x

(ii) Let f(x) = (x – 1) (x – 2). Accordingly, from the rst principle,

f (x + h) − f (x)
′
f (x) = lim
h→0 h

(x + h − 1)(x + h − 2) − (x − 1)(x − 2)
= lim
h→0 h
2 2 2
(x + hx − 2x + hx + h − 2h − x − h + 2) − (x − 2x − x + 2)
= lim
h→0 h
2
(hx+hx+h −2h−h)

= limh→0
h

2
2hx+h −3h
= limh→0
h

= limh→0 (2x + h − 3)

= (2x + 0 − 3)

= 2x − 3

(iii) Let f (x) = 1
2
. Accordingly, from the rst principle,
x

f (x + h) − f (x)
′
f (x) = lim
h→0 h
1 1
−
2 2
(x+h) x
= lim
h→0 h

2 2
1 x − (x + h)
= lim [ ]
2 2
h→0 h x (x + h)

2 2 2
1 x −x −h −2hx
= limh→0 [ 2 2
]
h x (x+h)

2
1 −h −2hx
= limh→0 [ 2 2
]
h x (x+h)

−h−2x
= limh→0 [ 2 2
]
x (x+h)

0−2x −2
= =
2 2 3
x (x+0) x

(iv) Let f (x) = x+1

x−1
. Accordingly, from the rst principle,
f (x + h) − f (x)
′
f (x) = lim
h→0 h

x+h+1 x+1
( − )
x+h−1 x−1

= lim
h→0 h

1 (x − 1)(x + h + 1) − (x + 1)(x + h − 1)
= lim [ ]
h→0 h (x − 1)(x + h − 1)
2 2
(x +hx+x−x−h−1)−(x +hx−x+x+h−1)
1
= limh→0 [ ]
h (x−1)(x+h−1)

1 −2h
= limh→0 [ ]
h (x−1)(x+h−1)

−2
= limh→0 [ ]
(x−1)(x+h−1)

−2 −2
= = 2
(x−1)(x−1) (x−1)

Page : 312 , Block Name : Exercise 13.2

Q5 For the function:
100 99 2
x x x
f (x) = + + … + + x + 1
100 99 2

Prove That f (1) = 100f (0) ′ ′

Answer. The given function is
100 99 2
x x x
f (x) = + + … + + x + 1
100 99 2

100 99 2
d d x x x
f (x) = [ + + … + + x + 1]
dx dx 100 99 2

100 99 2
d d x d x d x d d
f (x) = ( ) + ( ) + … + ( ) + (x) + (1)
dx dx 100 dx 99 dx 2 dx dx

d n n−1
On using theorem (x ) = nx , we obtain
dx

99 98
d 100x 99x 2x
f (x) = + + … + + 1 + 0
dx 100 99 2

′ yp 98
∴ f (x) = x + x + … + x + 1

At x = 0

′
f (0) = 1

At x = 1
′ 99 98
f (1) = 1 + 1 + … + 1 + 1 = [1 + 1 + … + 1 + 1]100 terms

= 1x 100 = 100
Thus, f (1) = 100 × f (0)
′ 1

Page : 312 , Block Name : Exercise 13.2

Q6 Find the derivative of x n
+ ax
n−1
+ a x
2 n−2
+ … + a
n−1
x + a
n
for some xed real number a.

Answer.
n n−1 2 n−2 n−1 n
Let f (x) = x + ax + a x + … + a x + a

′ d n n−1 2 n−2 n−1 n
∴ f (x) = (x + ax + a x + … + a x + a )
dx
d n d n−1 2 d n−2 n−1 d n d
= (x ) + a (x ) + a (x ) + … + a (x) + a (1)
dx dx dx dx dx
d n n−1
On using theorem x = nx , we obtain
dx

′ n−1 n−2 2 n−3 n−1 n
f (x) = nx + a(n − 1)x + a (n − 2)x + … + a + a (0)

n−1 n−2 2 n−3 n−1
= nx + a(n − 1)x + a (n − 2)x + … + a

Page : 313 , Block Name : Exercise 13.2

Q7 For some constants a and b, nd the derivative of:
2
(i) (x-a) (x-b) (ii) 2 x−a
(ax + b) (iii)
x−b

Answer. (i) Let f (x) = (x – a) (x – b)

2
⇒ f (x) = x − (a + b)x + ab

d
′ 2
∴ f (x) = (x − (a + b)x + ab)
dx
d d d
2
= (x ) − (a + b) (x) + (ab)
dx dx dx
d n n−1
On using theorem (x ) = nx , we obtain
dx

′
f (x) = 2x − (a + b) + 0 = 2x − a − b

2
(ii) Let f (x) = (ax 2
+ b)
2 4 2 2
⇒ f (x) = a x + 2abx + b

′ d 2 4 2 2 2 d 4 d 2 d 2
∴ f (x) = (a x + 2abx + b ) = a (x ) + 2ab (x ) + (b )
dx dx dx dx

d
n n−1
On using theorem x = nx , we obtain
dx
′ 2 3 2
f (x) = a (4x ) + 2ab(2x) + b (0)

2 3
= 4a x + 4abx
2
= 4ax (ax + b)

(x−a)
Let f (x) =
(x−b)

(iii)
′ d x−a
⇒ f (x) = ( )
dx x−b

By quotient rule,
d d
(x − b) (x − a) − (x − a) (x − b)
′ dx dx
f (x) =
2
(x − b)

(x − b)(1) − (x − a)(1)
=
2
(x − b)

x − b − x + a
=
2
(x − b)

a − b
=
2
(x − b)

Page : 313 , Block Name : Exercise 13.2

n n

Q8 Find the derivative of for some constant a.
x −a

x−a

Answer.
n n
x −a
Let f (x) =
x−a

n n
′ d x −a
⇒ f (x) = ( )
dx x−a

By quotient rule,

d n n n n d
(x − a) (x − a ) − (x − a ) (x − a)
′ dx dx
f (x) =
2
(x − a)

n−1 n n
(x − a) (nx − 0) − (x − a )
=
2
(x − a)

n n−1 n n
nx − anx − x + a
=
2
(x − a)

Page : 313 , Block Name : Exercise 13.2

Q9 Find the derivative of
(i) 2x −
3 3
(ii) (5x + 3x − 1) (x − 1)
4

(iii) x
−3
(5 + 3x) (iv) x
5
(3 − 6x
−9
)
2

(iv) −4 −5 2 x
x (3 − 4x ) (vi) −
x+1 3x−1

Answer. (i) Let
3
f (x) = 2x −
4

d 3
′
f (x) = (2x − )
dx 4

d d 3
= 2 (x) − ( )
dx dx 4

= 2 − 0

= 2

3
(ii) Let f (x) = (5x + 3x − 1) (x − 1)

By Leibnitz product rule,

d d
′ 3 3
f (x) = (5x + 3x − 1) (x − 1) + (x − 1) (5x + 3x − 1)
dx dx
3 2
= (5x + 3x − 1) (1) + (x − 1) (5.3x + 3 − 0)

3 2
= (5x + 3x − 1) + (x − 1) (15x + 3)
3 3 2
= 5x + 3x − 1 + 15x + 3x − 15x − 3

3 2
= 20x − 15x + 6x − 4

−3
(iii) Let f (x) = x (5 + 3x)

By Leibnitz product rule,

d d
′ −3 −3
f (x) = x (5 + 3x) + (5 + 3x) (x )
dx dx
−3 −3−1
= x (0 + 3) + (5 + 3x) (−3x )

−3 −4
= x (3) + (5 + 3x) (−3x )

−3 −4 −3
= 3x − 15x − 9x
−3 −4
= −6x − 15x

−3 5
= −3x (2 + )
x

−3
−3x
= (2x + 5)
x

−3
= (5 + 2x)
x4

5 −9
(iv) Let f (x) = x (3 − 6x )

By Leibnitz product rule,

d d
′ 5 −9 −9 5
f (x) = x (3 − 6x ) + (3 − 6x ) (x )
dx dx
5 −9−1 −9 4
= x {0 − 6(−9)x } + (3 − 6x ) (5x )

5 −10 4 −5
= x (54x ) + 15x − 30x

−5 4 −5
= 54x + 15x − 30x
−5 4
= 24x + 15x

24
4
= 15x +
5
x

−4 −5
(v) Let f (x) = x (3 − 4x )

By Leibnitz product rule,

d d
′ −4 −5 −5 −4
f (x) = x (3 − 4x ) + (3 − 4x ) (x )
dx dx
−4 −5−1 −5 −4−1
= x {0 − 4(−5)x } + (3 − 4x ) (−4)x

−4 −6 −5 −5
= x (20x ) + (3 − 4x ) (−4x )

−10 −5 −10
= 20x − 12x + 16x
−10 −5
= 36x − 12x

12 36
= − +
5 10
x x

(vi) Let f(x) =
2
2 x
−
x+1 3x−1

2
′ d 2 d x
f (x) = ( ) − ( )
dx x+1 dx 3x−1

By quotient rule,
d d d 2 2 d
(x + 1) (2) − 2 (x + 1) (3x − 1) (x ) − x (3x − 1)
′ dx dx dx dx
f (x) = [ ] − [ ]
dx dx

2
(x + 1)(0) − 2(1) (3x − 1)(2x) − (x ) (3)
= [ ] − [ ]
2 2
(x + 1) (3x − 1)

2 2
−2 6x − 2x − 3x
= − [ ]
2 2
(x + 1) (3x − 1)

2 2
−2 3x − 2x
= − [ ]
2 2
(x + 1) (3x − 1)

−2 x(3x − 2)
= −
2 2
(x + 1) (3x − 1)

Page : 313 , Block Name : Exercise 13.2

Q10 Find the derivative of cos x from rst principle.

Answer. Let f (x) = cos x. Accordingly, from the rst principle,
f (x + h) − f (x)
′
f (x) = lim
h→0 h

cos(x + h) − cos x
= lim
h→0 h
cos x cos h−sin x sin h−cos x
= limh→0 [ ]
h

− cos x(1−cosh h)−sin x sin h
= limh→0 [ ]
h

− cos x(1−cos h) sin x sin h
= limh→0 [ − ]
h h

1−cos h sin h
= − cos x (limh→0 ) − sin x limh→0 ( )
h h

1−cos h sin h
= − cos x(0) − sin x(1) [limh→0 = 0 and limh→0 = 1]
h h

′
∴ f (x) = − sin x

Page : 313 , Block Name : Exercise 13.2

Q11 Find the derivative of the following functions:
(i) sin x cos x (ii) sec x (iii) 5sec x + 4cos x
(iv) cosec x (v) 3cot + 5cosec x
(vi) 5sin - 6cos x + 7 (vii) 2tan x - 7sec x

Answer. (i) Let f (x) = sin x cos x. Accordingly, from the rst principle,
f (x + h) − f (x)
′
f (x) = lim
h→0 h

sin(x + h) cos(x + h) − sin x cos x
= lim
h→0 h

1
= lim [2 sin(x + h) cos(x + h) − 2 sin x cos x]
h→0 2h
1
= limh→0 [sin 2(x + h) − sin 2x]
2h

1 2x+2h+2x 2x+2h−2x
= limh→0 [2 cos ⋅ sin ]
2h 2 2

1 4x+2h 2h
= limh→0 [cos sin ]
h 2 2

1
= limh→0 [cos(2x + h) sin h]
h

sin h
= limh→0 cos(2x + h) ⋅ limh→0
h

= cos(2x + 0).1

= cos 2x

(ii) Let f (x) = sec x. Accordingly, from the rst principle,
f (x + h) − f (x)
′
f (x) = lim
h→0 h

sec(x + h) − sec x
= lim
h→0 h

1 1 1
= lim [ − ]
h→0 h cos(x + h) cos x

1 cos x − cos(x + h)
= lim [ ]
h→0 h cos x cos(x + h)
x+x+h x−x−h
−2 sin( ) sin( )
1 1 2 2

= limh→0 [ ]
cos x h cos(x+h)

2x+h h
−2 sin( ) sin(− )
1 1 2 2

= ⋅ limh→0 [ ]
cos x h cos(x+h)

h
sin( )
2x+h 2
[sin( ) ]
2 h
( )
1 2
= limh→0
cos x cos(x+h)

h 2x+h
sin( ) sin( )
2 2
1
= ⋅ lim h ⋅ limh→0
cos x →0 h cos(x+h)
2 ( )
2

1 sin x
= ⋅ 1 ⋅
cos x cos x

= sec x tan x

(iii) Let f (x) = 5 sec x + 4 cos x. Accordingly, from the rst principle,
f (x+h)−f (x)
′
f (x) = limh→0
h

5 sec(x+h)+4 cos(x+h)−[5 sec x+4 cos x]
= limh→0
h

[sec(x+h)−sec x] [cos(x+h)−cos x]
= 5 limh→0 + 4 limh→0
h h

1 1 1 1
= 5 limh→0 [ − ] + 4 limh→0 [cos(x + h) − cos x]
h cos(x+h) cos x h

cos x−cos(x+h)
1 1
= 5 limh→0 [ ] + 4 limh→0 [cos x cos h − sin x sin h − cos x]
h cos x cos(x+h) h

x+x+h x−x−h
−2 sin( ) sin( )
5 2 2
1 1
= limh→0 [ ] + 4 limh→0 [− cos x(1 − cos h) − sin x sin h]
cos x h cos(x+h) h

2x+h h
−2 sin( ) sin(− )
2 2 (1−cos h) sin h
5 1
= ⋅ limh→0 [ ] + 4 [− cos x limh→0 − sin x limh→0 ]
cos x h cos(x+h) h h

2x+h h
sin( ) sin( )
5 2 2

= ⋅ [limh→0 ⋅ limh→0 ] − 4 sin x
cos x cos(x+h) h

2

5 sin x
= ⋅ ⋅ 1 − 4 sin x
cos x cos x

= 5 sec x tan x ⋅ −4 sin x

(iv) Let f (x) = cosec x. Accordingly, from the rst principle,
f (x+h)−f (x)
′
f (x) = limh→0
h

′ 1
f (x) = limh→0 [csc(x + h) − cosecx]
h

1 1 1
= limh→0 [ − ]
h sin(x+h) sin x

1 sin x−sin(x+h)
= limh→0 [ ]
h sin(x+h) sin x

x+x+h x−x−h
2 cos( )⋅sin( )
1 2 2

= limh→0 [ ]
h sin(x+h) sin x

2x+h h
2 cos( ) sin(− )
1 2 2

= limh→0 [ ]
h sin(x+h) sin x

h
sin( )
2x+h 2
− cos( )⋅
2 h
( )
2
= limh→0
sin(x+h) sin x

2x+h h
− cos( ) sin( )
2 2

= limh→0 ( ) ⋅ lim h
sin(x+h) sin x →0 h
2 ( )
2

− cos x
= ( ) .1
sin x sin x

= − csc x cot x

(v) Let f (x) = 3cot x + 5cosec x. Accordingly, from the rst principle,
f (x + h) − f (x)
′
f (x) = lim
h→0 h

3 cot(x + h) + 5 cosec(x + h) − 3 cot x − 5 cosecx
= lim .......(1)
h→0 h

1 1
= 3 lim [cot(x + h) − cot x] + 5 lim [ cosec(x + h) − cosecx]
h→0 h h→0 h
1
Now, lim [cot(x + h) − cot x]
h→0 h

1 cos(x + h) cos x
= lim [ − ]
h→0 h sin(x + h) sin x

1 cos(x + h) sin x − cos x sin(x + h)
= lim [ ]
h→0 h sin x sin(x + h)

1 sin(x − x − h)
= lim [ ]
h→0 h sin x sin(x + h)

1 sin(−h)
= lim [ ]
h→0 h sin x sin(x + h)

sin h 1
= − (lim ) ⋅ (lim )
h→0 h h→0 sin x ⋅ sin(x + h)
1 −1 2
= −1 ⋅ = 2
= − cosec x. . . (2)
sin x⋅sin(x+0) sin x

1
limh→0 [ cosec(x + h) − cosecx]
h

1 1 1
= limh→0 [ − ]
h sin(x+h) sin x

sin x−sin(x+h)
1
= limh→0 [ ]
h sin(x+h) sin x

x+x+h x−x−h
2 cos( )⋅sin( )
2 2
1
= limh→0 [ ]
h sin(x+h) sin x

2x+h h
2 cos( ) sin(− )
2 2
1
= limh→0 [ ]
h sin(x+h) sin x

h
sin( )
2x+h 2
− cos( )⋅
2 h
( )
2
= limh→0
sin(x+h) sin x

2x+h h
− cos( ) sin( )
2 2

= limh→0 ( ) ⋅ limh→0
sin(x+h) sin x h
( )
2

− cos x
= ( ) ⋅ 1
sin x sin x

= − cosec cot x. . . . . . (3)

From (1), (2), and (3), we obtain
′ 2
f (x) = −3 cosec x − 5 cosecx cot x

(vi) Let f (x) = 5sin x – 6cos x + 7. Accordingly, from the rst principle,
f (x+h)−f (x)
′
f (x) = limh→0
h

1
= limh→0 [5 sin(x + h) − 6 cos(x + h) + 7 − 5 sin x + 6 cos x − 7]
h

1
= limh→0 [5{sin(x + h) − sin x} − 6{cos(x + h) − cos x}]
h

1 x+h+x x+h−x cos x cos h−sin x sin h−cos x
= 5 limh→0 [2 cos( ) sin( )] − 6 limh→0
h 2 2 h

1 2x+h h − cos x(1−cos h)−sin x sin h
= 5 limh→0 [2 cos( ) sin ] − 6 limh→0 [ ]
h 2 2 h

h
sin
2x+h 2 1−cos h sin h
= 5 [limh→0 cos( )] [limh ] − 6 [(− cos x) (limh→0 ) − sin x limh→0 ( )]
2 h
h h
2

= 5 cos x ⋅ 1 − 6[(− cos x) ⋅ (0) − sin x.1]

= 5 cos x + 6 sin x

(vii) Let f (x) = 2 tan x – 7 sec x. Accordingly, from the rst principle,

f (x + h) − f (x)
′
f (x) = lim
h→0 h

1
= lim [2 tan(x + h) − 7 sec(x + h) − 2 tan x + 7 sec x]
h→0 h

1
= lim [2{tan(x + h) − tan x} − 7{sec(x + h) − sec x}]
h→0 h

1 sin(x+h) sin x 1 1 1
= 2 limh→0 [ − ] − 7 limh→0 [ − ]
h cos(x+h) cos x h cos(x+h) cos x

1 sin(x+h) cos x−sin x cos(x+h) 1 cos x−cos(x+h)
= 2 limh→0 [ ] − 7 limh→0 [ ]
h cos x cos(x+h) h cos x cos(x+h)

x+x+h x−x−h
−2 sin( ) sin( )
1 sin(x+h−x) 1 2 2

= 2 limh→0 [ ] − 7 limh→0 [ ]
h cos x cos(x+h) h cos x cos(x+h)

2x+h h
−2 sin( ) sin(− )
sin h 1 1 2 2

= 2 limh→0 [( ) ] − 7 limh→0 [ ]
h cos x cos(x+h) h cos x cos(x+h)

1 sin x
= 2.1 ⋅ − 7.1 ( )
cos x cos x cos x cos x

2
= 2 sec x − 7 sec x tan x

Page : 313 , Block Name : Exercise 13.2

Miscellaneous Exercise

Q1 Find the derivative of the following functions from rst principle:
−1
(i) − x ( ii) (−x) ( iii) sin(x + 1)

π
(iii)cos(x − )
8

Answer. (i) Let f(x) = –x. Accordingly, f(x + h) = -(x + h)
By rst principle,
f (x + h) − f (x)
′
f (x) = lim
h→0 h

−(x + h) − (−x)
= lim
h→0 h

−x − h + x
= lim
h→0 h

−h
= lim
h→0 h

= lim (−1) = −1
h→0

(ii) Let f (x) = (−x) −1
=
1

−x
=
−1

x
. Accordingly, f (x + h) = −1

(x+h)

By rst principle,

f (x + h) − f (x)
′
f (x) = lim
h→0 h

1 −1 −1
= lim [ − ( )]
h→0 h x + h x

1 −x + (x + h) 1
= lim [ + ]
h→0 h x + h x

1 −x + (x + h)
= lim [ ]
h→0 h x(x + h)

1 −x + (x + h)
= lim [ ]
h→0 h x(x + h)

1 h
= lim [ ]
h→0 h x(x + h)

1
= lim
h→0 x(x + h)

1 1
= =
2
x ⋅ x x

(iii) Let f(x) = sin (x + 1). Accordingly,f (x + h) = sin(x + h + 1)
By rst principle,
f (x + h) − f (x)
′
f (x) = lim
h→0 h

1
= lim [sin(x + h + 1) − sin(x + 1)]
h→0 h

1 x + h + 1 + x + 1 x + h + 1 − x − 1
= lim [2 cos( ) sin( )]
h→0 h 2 2

1 2x + h + 2 h
= lim [2 cos( ) sin( )]
h→0 h 2 2
h
sin( )
2x+h+2 2

= limh→0 [cos( ) ⋅ ]
2 h
( )
2

h h
sin( ) sin( )
2x+h+2 2 2 h
= limh→0 cos( ) ⋅ lim h [ As h → 0 ⇒ → 0]
2 →0 b h
2
2 +0 +0
2 2

2x+0+2
= cos( ) ⋅ 1
2 sin x
[ limx→0 = 1]
x
= cos(x + 1)

(iv) Let f (x) = cos(x −
π π
) Accordingly, f (x + h) = cos(x + h − )
8 8

By rst principle,
f (x + h) − f (x)
′
f (x) = lim
h→0 h

1 π π
= lim [cos(x + h − ) − cos(x − )]
h→0 h 8 8

π π
π π
(x+h− +x− ) x+h− −x+
8 8
1 8 8
= limh→0 [−2 sin sin( )]
h 2 2

π
2x+h−
1 h
] ⎤
4
= limh→0 [−2 sin( ) sin
h 2 2
⎥
⎥
h ⎥
2x+h−
π
sin( ) ⎥
4
2
⎥
= limh→0 [− sin( ) ]
2 2
⎦

h
π
2x+h− sin( )
2 h
4
= limh→0 [− sin( )] ⋅ lim h [ As h → 0 ⇒ → 0]
2 →0 h 2
2 ( )
2

π
2x+0−
4
= − sin( ).1
2

π
= − sin(x − )
8

Page : 317 , Block Name : Miscellaneous Exercise

Q2 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are
xed non-zero constants and m and n are integers): (x + a)

Answer. Let f(x) = x + a. Accordingly, f(x + h) = -(x + h)
By rst principle,
f (x + h) − f (x)
′
f (x) = lim
h→0 h

x + h + a − x − a
= lim
h→0 h

h
= lim ( )
h→0 h

= lim (1)
h→0

= 1

Page : 317 , Block Name : Miscellaneous Exercise

Q3 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are
xed non-zero constants and m and n are integers): (px + q) ( + s) r

x

Answer. f (x) = (px + q) (
r
+ s)
x

By Leibnitz product rule,
r r
′ ′
f (x) = (px + q) ( + s) + ( + s) (px + q)
x x
′ r
−1
= (px + q)(rx + s) + ( + s) (p)
x
r
−2
= (px + q) (−rx ) + ( + s) p
x

−r r
= (px + q) ( ) + ( + s) p
2
x x

−pr qr pr
= − + + ps
2
x x x
qr
= ps −
2
x

Page : 317 , Block Name : Miscellaneous Exercise

Q4 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are
xed non-zero constants and m and n are integers):(ax + b)(cx + d) 2

Answer.
2
Let f (x) = (ax + b)(cx + d)

By Leibnitz product rule,

d d
′ 2 2
f (x) = (ax + b) (cx + d) + (cx + d) (ax + b)
dx dx
d d
2 2 2 2
= (ax + b) (c x + 2cdx + d ) + (cx + d) (ax + b)
dx dx

d d d d d
2 2 2 2
= (ax + b) [ (c x ) + (2cdx) + d ] + (cx + d) [ ax + b]
dx dx dx dx dx

2 2
= (ax + b) (2c x + 2cd) + (cx + d ) a

2
= 2c(ax + b)(cx + d) + a(cx + d)

Page : 317 , Block Name : Miscellaneous Exercise

Q5 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are
xed non-zero constants and m and n are integers):
ax+b

cx+d

Answer. Let f (x) =
ax+b

cx+d

By quotient rule,
d d
(cx + d) (ax + b) − (ax + b) (cx + d)
′ dx dx
f (x) = (cx + d)
2
(cx + d)

(cx + d)(a) − (ax + b)(c)
=
2
(cx + d)

acx + ad − acx − bc
=
2
(cx + d)

ad − bc
=
2
(cx + d)

Page : 317 , Block Name : Miscellaneous Exercise

Q6 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are

1
1+
xed non-zero constants and m and n are integers): x

1
1−
x

1 x+1
1+
Answer. f (x) = x x x+1
= = , where x ≠ 0
1 x−1 x−1
1−
x x

By quotient rule,
(x − 1) − (x − 1) − (x + 1) − (x + 1)
′
f (x) = , x ≠ 0, 1
2
(x − 1)

(x − 1)(1) − (x + 1)(1)
= , x ≠ 0, 1
2
(x − 1)

x − 1 − x − 1
= , x ≠ 0, 1
2
(x − 1)

−2
= , x ≠ 0, 1
2
(x − 1)

Page : 317 , Block Name : Miscellaneous Exercise

Q7 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are
xed non-zero constants and m and n are integers):
1
2
ax +bx+c

Answer. Let f (x) = 1

ax2 +bx+c

By quotient rule,
2 d d 2
(ax + bx + c) (1) − (ax + bx + c)
′ dx dx
f (x) =
dx
2
(ax + bx + c) (0) − (2ax + b)
=
2
2
(ax + bx + c)

−(2ax + b)
=
2
2
(ax + bx + c)

Page : 317 , Block Name : Miscellaneous Exercise

Q8 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are
xed non-zero constants and m and n are integers):
ax+b

2
px +qx+r

Answer. Let f (x) =
ax+b

2
px +qx+r

By quotient rule,
2 d d 2
(px + qx + r) (ax + b) − (ax + b) (px + qx + r)
′ dx dx
f (x) =
2
2
(px + qx + r)

2
(px + qx + r) (a) − (ax + b)(2px + q)
=
2
2
(px + qx + r)

2 2
apx + aqx + ar − 2apx − aqx − 2bpx − bq
=
2 2
(px + qx + r)

2
−apx − 2bpx + ar − bq
=
2 2
(px + qx + r)

Page : 317 , Block Name : Miscellaneous Exercise

Q9 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are
2

xed non-zero constants and m and n are integers):
px +qx+r

ax+b

Answer.
2
px +qx+r
Let f (x) =
ax+b

By quotient rule,
d 2 2 d
(ax + b) (px + qx + r) − (px + qx + r) (ax + b)
′ dx dx
f (x) =
2
(ax + b)

2
(ax + b)(2px + q) − (px + qx + r) (a)
=
2
(ax + b)
2 2
2apx + aqx + 2bpx + bq − apx − aqx − ar
=
2
(ax + b)

2
apx + 2bpx + bq − ar
=
2
(ax + b)

Page : 317 , Block Name : Miscellaneous Exercise

Q10 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers):
a b
− + cos 4 2
x x

Answer.
a b
Let f (x) = − + cos x
4 2
x x

d a d b d
′
f (x) = ( ) − ( ) + (cos x)
4 2
dx x dx x dx

d d d
−4 −2
= a (x ) − b (x ) + (cos x)
dx dx dx
−5 −3 d n n−1 d
= a (−4x ) − b (−2x ) + (− sin x) [ (x ) = nx and (cos x) = − sin x]
dx dx

−4a 2b
= 5
+ 3
− sin x
x x

Page : 317 , Block Name : Miscellaneous Exercise

Q11 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers): 4√x − 2

Answer.
Let f (x) = 4√x − 2

d d d
′
f (x) = (4√x − 2) = (4√x) − (2)
dx dx dx

d 1 1 1
−1
= 4 (x 2 ) − 0 = 4 ( x2 )
dx 2

1 2
−
= (2x 2 ) =
√x

Page : 317 , Block Name : Miscellaneous Exercise

Q12 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers): (ax + b) n

Answer. f (x) = (ax + b) . Accordingly, f (x + h) = {a(x + h) + b}
n n
= (ax + ah + b)
n

By rst principle,
f (x + h) − f (x)
′
f (x) = lim
h→0 h
n n
(ax + ah + b) − (ax + b)
= lim
h→0 h
n
n ah n
(ax + b) (1 + ) − (ax + b)
ax+b

= lim
h→0 h
n
ah
(1 + ) − 1
ax+b
n
= (ax + b) lim
h→0 h
2
1 ah n(n−1) ah
n
= (ax + b) limh→0 [{1 + n ( ) + ( ) + …} − 1] (Using binomial theorem)
n ax+b ⌊2 ax+b
–

2 2
1 ah n(n−1)a h
n
= (ax + b) limh→0 [n ( ) + + … (Terms containing higher degrees of h)]
2
h ax+b ⌊2(ax+b)

2
na n(n−1)a h
n
= (ax + b) limb→0 [ + + …]
(ax+b) ⌊2(ax+b)2

n na
= (ax + b) [ + 0]
(ax+b)

n
(ax+b)
= na
(ax+b)

n−1
= na(ax + b)

Page : 317 , Block Name : Miscellaneous Exercise

Q13 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers): (ax + b) (cx + d) n m

Answer.

n m
Let f (x) = (ax + b) (cx + d)

By Leibnitz product rule,

′ n d m m d n
f (x) = (ax + b) (cx + d) + (cx + d) (ax + b)
dx dx
m
Now, let f1 (x) = (cx + d)

m
f1 (x + h) = (cx + ch + d)

f1 (x+h)−f1 (x)
′
f (x) = limh→0
1 h
m m
(cx+ch+d) −(cx+d)
= limh→0
h

m
m 1 ch
= (cx + d) limh→0 [(1 + ) − 1]
h cx+d

2 2
m(m−1) (c h )
m 1 mch
= (cx + d) limh→0 [(1 + + + …) − 1]
2
h (cx+d) 2 (cx+d)

2 2
1 mch m(m−1)c h
m
= (cx + d) limh→0 [ + + … (Terms containing higher degrees of h)]
2
h (cx+d) 2(cx+d)

2
mc m(m−1)c h
m
= (cx + d) limh→0 [ + + …]
(cx+d) 2(cx+d)2

m mc
= (cx + d) [ + 0]
cx+d

m
mc(cx+d)
=
(cx+d)

m−1
= mc(cx + d)
d m m−1
(cx + d) = mc(cx + d) … (2)
dx

d n n−1
Similarly, (ax + b) = na(ax + b) … (3)
dx

Therefore, from (1), (2), and (3), we obtain
′ n m−1 m n−1
f (x) = (ax + b) {mc(cx + d) } + (cx + d) {na(ax + b) }

n−1 m−1
= (ax + b) (cx + d) [mc(ax + b) + na(cx + d)]

Page : 317 , Block Name : Miscellaneous Exercise

Q14 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers): sin (x + a)

Answer.
Let f (x) = sin(x + a)

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

By first principle,

f (x+h)−f (x)
′
f (x) = limh→0
h

sin(x+h+a)−sin(x+a)
= limh→0
h

1 x+h+a+x+a x+h+a−x−a
= limh→0 [2 cos( ) sin( )]
h 2 2

1 2x+2a+h h
= limh→0 [2 cos( ) sin( )]
h 2 2

h
sin( )
2x+2a+h 2

= limh→0 [cos( ){ }]
2 h
( )
2

n
sin( )
2x+2a+h 2 h
= limh→0 cos( ) lim b { } [ As h → 0 ⇒ → 0]
2 →0 h 2
2 ( )
2

2x+2a
= cos( ) × 1
2 sin x
[ limx→0 = 1]
x

= cos(x + a)

Page : 317 , Block Name : Miscellaneous Exercise

Q15 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers): cosec x cot x

Answer.
Let f (x) = csc x cot x

By Leibnitz product rule,

′ ′ ′
f (x) = csc x(cot x) + cot x(csc x)

Let f1 (x) = cot x. Accordingly , f1 (x + h) = cot(x + h)

By first principle,

f1 (x + h) − f1 (x)
′
f (x) = lim
1
h→0 h

cot(x + h) − cot x
= lim
h→0 h

1 cos(x + h) cos x
= lim ( − )
h→0 h sin(x + h) sin x

1 sin x cos(x+h)−cos x sin(x+h)
= limh→0 [ ]
h sin x sin(x+h)

1 sin(x−x−h)
= limh→0 [ ]
h sin x sin(x+h)

1 1 sin(−h)
= limh [ ]
sin x h sin(x+h)

−1 sin h 1
= ⋅ (limh→0 ) (limh→0 )
sin x h sin(x+h)

−1 1
= ⋅ 1 ⋅ ( )
sin x sin(x+0)

−1
=
2
sin x

2
= − cosec x

∴ (cot x)
′
= − csc
2
x ......(2)
Now, let f2 (x) = csc x. Accordingly, f2 (x + h) = cosec(x + h)

By first principle,

f2 (x+h)−f2 (x)
′
f (x) = limh→0
2 h

1
= limh→0 [ cosec(x + h) − cosecx]
h

1 1 1
= limh→0 [ − ]
h sin(x+h) sin x

1 sin x−sin(x+h)
= limh→0 [ ]
h sin x sin(x+h)

x+x+h x−x−h
2 cos( ) sin( )
1 1 2 2

= limh→0 [ ]
sin x h sin(x+h)

2x+h −h
2 cos( ) sin( )
1 1 2 2

= limh→0 = ]
sin x h sin(x+h)

h 2x+h
⎡ − sin( ) cos( ) ⎤
1 2 2

= lim ⎢
⎢ ⋅ ⎥
⎥
sin x h→0 h sin(x + h)
⎣ ( ) ⎦
2

h 2x+h
sin( ) cos( )
−1 2 2

= lim ⋅ lim
sin x h→0 h h→0 sin(x + h)
( )
2

2x+0
cos( )
−1 2

= .1.
sin x sin(x+0)

−1 cos x
= ⋅
sin x sin x

= − cos ec x ⋅ cot x
′
∴ ( cosecx) = − cosecx cot x … (3)

From (1), (2), and (3), we obtain

′ 2
f (x) = cosecx (− cosec x) + cot x(− cosecx cot x)

3 2
= − cosec x − cot x cosecx

Page : 317 , Block Name : Miscellaneous Exercise

Q16 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers): cos x

1+sin x

Answer. Let
cos x
f (x) =
1+sin x

By quotient rule,
d d
(1+sin x) (cos x)−(cos x) (1+sin x)
′ dx dx
f (x) = 2
(1+sin x)

(1+sin x)(− sin x)−(cos x)(cos x)
= 2
(1+sin x)

2 2
− sin x−sin x−cos x
=
(1+sin x)2

2 2
− sin x−(sin x+cos x)
= 2
(1+sin x)

− sin x−1
= 2
(1+sin x)

−(1+sin x)
= 2
(1+sin x)

−1
=
(1+sin x)

Page : 317 , Block Name : Miscellaneous Exercise

Q17 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers):
sin x+cos x

sin x−cos x

Answer. Let f (x) = sin x+cos x

sin x−cos x

By quotient rule,
d d
(sin x−cos x) (sin x+cos x)−(sin x+cos x) (sin x−cos x)
′ dx dx
f (x) =
(sin x−cos x)2

(sin x − cos x)(cos x − sin x) − (sin x + cos x)(cos x + sin x)
=
2
(sin x − cos x)

2 2
−(sin x − cos x) − (sin x + cos x)
=
2
(sin x − cos x)
2 2 2 2
−[sin x+cos x−2 sin x cos x+sin x+cos x+2 sin x cos x]
=
(sin x−cos x)2

−[1 + 1]
=
2
(sin x − cos x)

−2
=
2
(sin x − cos x)

Page : 318 , Block Name : Miscellaneous Exercise

Q18 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers):

Answer. Let
sec x−1
f (x) =
sec x+1

1
−1
cos x 1−cos x
f (x) = =
1 1+cos x
+1
cos x

By quotient rule,
d d
(1+cos x) (1−cos x)−(1−cos x) (1+cos x)
′ dx dx
f (x) =
(1+cos x)2

2
(1+cos x)(sin x)−(1−cos x)
= 2
(1+cos x)

sin x + cos x sin x + sin x − sin x cos x
=
2
(1 + cos x)

2 sin x
=
2
(1 + cos x)

2 sin x 2 sin x
= 2
=
2
1 (sec x+1)
(1+ )
sec x
sec2 x

2
2 sin x sec x
=
2
(sec x + 1)

2 sin x
=
2
(sec x + 1)

2 sec x + 1nx
=
2
(sec x + 1)

Page : 318 , Block Name : Miscellaneous Exercise

Q19 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers):
n
sin x

Answer.
n
Let y = sin x

Accordingly, for n = 1, y = sin x

dy d
∴ = cos x, i.e., sin x = cos x
dx dx

dy d
∴ = (sin x sin x)
dx dx
′ ′
= (sin x) sin x + sin x(sin x) [By leibnitz product rule]

= cos x sin x + sin x cos x

= 2 sin x cos x. . . (1)
3
For n = 3, y = sin x

dy d
2
∴ = (sin x sin x)
dx dx
′
′ 2 2
= (sin x) sin x + sin x(sin x) [ By Leibnitz product rule ]
2
= cos x sin x + sin x(2 sin x cos x)[Using (1)]

2 2
= cos x sin x + 2 sin x cos x

2
= 3 sin x cos x
d n (n−1)
(sin x) = n sin x cos x
dx

Let our assertion be true for n = k .

d k (k−1)
i. e. , (sin x) = k sin x cos x. . . . (2)
dx

Consider

d d
k+1 k
(sin x) = (sin x sin x)
dx dx
′
′ t k
= (sin x) sin x + sin x(sin x) [By leibnitz product rule]

k (k−1)
= cos x sin x + sin x (k sin x cos x)[Using (2)]

k k
= cos x sin x + k sin x cos x

k
= (k + 1) sin x cos x

Thus, our assertion is true for n = k + 1

d n (n−1)
Hence, by mathematical induction, (sin x) = n sin x cos x
dx

Page : 318 , Block Name : Miscellaneous Exercise

Q20 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers):

Answer.
a+b sin x
Let f (x) =
c+d cos x

By quotient rule,
d d
(c+d cos x) (a+b sin x)−(a+b sin x) (c+d cos x)
′ dx dx
f (x) =
2
(c+d cos x)

(c+d cos x)(b cos x)−(a+b cos x)(−d sin x)
= 2
(c+d cos x)

2 2
cb cos x + bd cos x + ad sin x + bd sin x
=
2
(c + d cos x)

2 2
bc cos x + ad sin x + bd (cos x + sin x)
=
2
(c + d cos x)

bc cos x + ad sin x + bd
=
2
(c + d cos x)

Page : 318 , Block Name : Miscellaneous Exercise

Q21 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
sin(x+a)
are xed non-zero constants and m and n are integers): cos x

Answer. Let
sin(x+a)
f (x) =
cos x

By quotient rule,
d d
cos x [sin(x+a)]−sin(x+a) cos x
′ dx dx
f (x) = 2
cos x

d
cos x [sin(x+a)]−sin(x+a)(− sin x)
′ dx
f (x) = 2
. . . . . (i)
cos x

Let g(x) = sin(x + a) ⋅ Accordingly, g(x + h) = sin(x + h + a)

By first principle,

g(x + h) − g(x)
′
g (x) = lim
h→0 h

1
= lim [sin(x + h + a) − sin(x + a)]
h→0 h

1 x + h + a + x + a x + h + a − x − a
= lim [2 cos( ) sin( )]
h→0 h 2 2

1 2x + h + a + h h
= lim [2 cos( ) sin( )]
h→0 h 2 2
h
sin( )
2x+2a+h 2

= limh→0 [cos( ){ }]
2 h
( )
2

h
sin( )
2x+2a+h 2 h
= limh→0 cos( ) lim h { } [ As h → 0 ⇒ → 0]
2 ,0 h 2
2 ( )
2

= cos(x + a) …….(ii)
From (i) and (ii), we obtain

cos x ⋅ cos(x + a) + sin x sin(x + a)
′
f (x) =
2
cos x

cos(x + a − x)
=
2
cos x
cos a
=
2
cos x

Page : 318 , Block Name : Miscellaneous Exercise

Q22 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers): x (5 sin x – 3 cos x) 4

Answer.
4
Let f (x) = x (5 sin x − 3 cos x)

By product rule,

′ 4 d d 4
f (x) = x (5 sin x − 3 cos x) + (5 sin x − 3 cos x) (x )
dx dx

4 d d d 4
= x [5 (sin x) − 3 (cos x)] + (5 sin x − 3 cos x) (x )
dx dx dx

4 3
= x [5 cos x − 3(− sin x)] + (5 sin x − 3 cos x) (4x )

3
= x [5x cos x + 3x sin x + 20 sin x − 12 cos x]

Page : 318 , Block Name : Miscellaneous Exercise

Q23 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers):(x + 1) cos x 2

Answer.

2
Let f (x) = (x + 1) cos x

By product rule,

′ 2 d d 2
f (x) = (x + 1) (cos x) + cos x (x + 1)
dx dx

2
= (x + 1) (− sin x) + cos x(2x)

2
= −x sin x − sin x + 2x cos x

Page : 318 , Block Name : Miscellaneous Exercise

Q24 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers):
2
(ax + sin x) (p + q cos x)

Answer.
2
Let f (x) = (ax + sin x) (p + q cos x)

By product rule,

′ 2 d d 2
f (x) = (ax + sin x) (p + q cos x) + (p + q cos x) (ax + sin x)
dx dx
2
= (ax + sin x) (−q sin x) + (p + q cos x)(2ax + cos x)

2
= −q sin x (ax + sin x) + (p + q cos x)(2ax + cos x)

Page : 318 , Block Name : Miscellaneous Exercise

Q25 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers): (x + cos x)(x − tan x)

Answer.
Let f (x) = (x + cos x)(x − tan x)

By product rule,

′ d d
f (x) = (x + cos x) (x − tan x) + (x − tan x) (x + cos x)
dx dx

d d
= (x + cos x) [ (x) − (tan x)] + (x − tan x)(1 − sin x)
dx dx

d
= (x + cos x) [1 − tan x] + (x − tan x)(1 − sin x)
dx

Let g(x) = tan x . Accordingly, g(x + h) = tan(x + h)

By first principle,

g(x+h)−g(x)
′
g (x) = limh→0
h

tan(x+h)−tan x
= limh→0 ( )
h

1 sin(x+h) sin x
= limh→0 [ − ]
h cos(x+h) cos x

1 sin(x+h) cos x−sin x cos(x+h)
= limh→0 [ ]
h cos(x+h) cos x

1 sin(x+h−x)
= ⋅ lim 1 [ ]
cos x h→ cos(x+h)
h

1 1 sin h
= ⋅ lim [ ]
cos x h→0 h cos(x + h)

1 sin h 1
= ⋅ (lim ) ⋅ ( lim )
cos x h→0 h h→∞ cos(x + h)

1 1
= ⋅ 1 ⋅
cos x cos(x + 0)

1
=
2
cos x
2
= sec x … . . (ii)

Therefore, from ( i ) and (ii), we obtain

′ 2
f (x) = (x + cos x) (1 − sec x) + (x − tan x)(1 − sin x)

2
= (x + cos x) (− tan x) + (x − tan x)(1 − sin x)

2
= − tan x(x + cos x) + (x − tan x)(1 − sin x)

Page : 318 , Block Name : Miscellaneous Exercise

Q26 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers):
4x+5 sin x

3x+7 cos x

Answer.
4x+5 sin x
f (x) =
3x+7 cos x

By quotient rule,
d d
(3x+7 cos x) (4x+5 sin x)−(4x+5 sin x) (3x+7 cos x)
′ dx dx
f (x) = 2
(3x+7 cos x)

d d d d
(3x + 7 cos x) [4 (x) + 5 (sin x)] − (4x + 5 sin x) [3 x + 7 cos x]
dx dx dx dx

=
2
(3x + 7 cos x)

(3x + 7 cos x)(4 + 5 cos x) − (4x + 5 sin x)(3 − 7 sin x)
=
2
(3x + 7 cos x)

2 2
12x + 15x cos x + 28 cos x + 35 cos x − 12x + 28x sin x − 15 sin x + 35 sin x
=
2
(3x + 7 cos x)

2 2
15x cos x + 28 cos x + 28x sin x − 15 sin x + 35 (cos x + sin x)
=
2
(3x + 7 cos x)

35 + 15x cos x + 28 cos x + 28x sin x − 15 sin x
=
2
(3x + 7 cos x)

Page : 318 , Block Name : Miscellaneous Exercise

Q27 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s

are xed non-zero constants and m and n are integers):
2 π
x cos( )
4

sin x

Answer.
2 π
x cos( )
4

f (x) =
sin x

By quotient rule,
d 2 2 d
sin x (x )−x (sin x)
′ π dx dx
f (x) = cos ⋅ [ ]
2
4 sin x

2
π sin x ⋅ 2x − x cos x
=cos ⋅ [ ]
4 2
sin x
π
x cos [2 sin x − x cos x]
4
=
2
sin x

Page : 318 , Block Name : Miscellaneous Exercise

Q28 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers):
x

1+tan x

Answer. Let
x
f (x) =
1+tan x

d d
(1+tan x) (x)−x (1+tan x)
′ dx dx
f (x) = 2
(1+tan x)

d
(1+tan x)−x⋅ (1+tan x)
′ dx
f (x) = 2
. . . . . (i)
(1+tan x)

Let g(x) = 1 + tan x. Accordingly, g(x + h) = 1 + tan(x + h)

By first principle,

g(x+h)−g(x)
′
g (x) = limh→0
h

1+tan(x+h)−1−tan x
= limh→0 [ ]
h

1 sin(x+h) sin x
= limh→0 [ − ]
h cos(x+h) cos x

1 sin(x+h) cos x−sin x cos(x+h)
= limh→0 [ ]
h cos(x+h) cos x

1 sin(x + h − x)
= lim [ ]
h→0 h cos(x + h) cos x

1 sin h
= lim [ ]
h→0 h cos(x + h) cos x

sin h 1
= (lim ) ⋅ (lim )
h→0 h h→0 cos(x + h) cos x

1 2
= 1 × 2
= sec x
cos x

d 2
⇒ (1 + tan x) = sec x. . . . . (ii)
dx

From (i) and (ii), we obtain

2
′ 1+tan x−x sec x
f (x) = 2
(1+tan x)

Page : 318 , Block Name : Miscellaneous Exercise

Q29 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers): (x + sec x) (x – tan x)

Answer.
Let f (x) = (x + sec x)(x − tan x)

By product rule,

′ d d
f (x) = (x + sec x) (x − tan x) + (x − tan x) (x + sec x)
dx dx

d d d d
= (x + sec x) [ (x) − tan x] + (x − tan x) [ (x) + sec x]
dx dx dx dx

.....(i)
d d
= (x + sec x) [1 − tan x] + (x − tan x) [1 + sec x]
dx dx

Let f1 (x) = tan x, f2 (x) = sec x

Accordingly, f1 (x + h) = tan(x + h) and f2 (x + h) = sec(x + h)

f1 (x+h)−f1 (x)
′
f (x) = limh→0 ( )
1 h

tan(x+h)−tan x
= limh→0 ( )
h

tan(x+h)−tan x
= limh→0 [ ]
h

1 sin(x+h) sin x
= limh→0 [ − ]
h cos(x+h) cos x

1 sin(x+h) cos x−sin x cos(x+h)
= limh→0 [ ]
h cos(x+h) cos x

1 sin(x+h−x)
= limh→0 [ ]
h cos(x+h) cos x

1 sin h
= limh→0 [ ]
h cos(x+h) cos x

sin h 1
= (limh→0 ) ⋅ (limh→0 )
h cos(x+h) cos x

1 2
= 1 × = sec x
cos2 x

d 2
⇒ tan x = sec x. . . . (ii)
dx

f2 (x + h) − f2 (x)
′
f (x) = lim ( )
2
h→0 h

sec(x + h) − sec x
= lim ( )
h→0 h

1 1 1
= lim [ − ]
h→0 h cos(x + h) cos x

1 cos x − cos(x + h)
= lim [ ]
h→0 h cos(x + h) cos x

x+x+h x−x−h
−2 sin( )⋅sin( )
1 1 2 2

= ⋅ limh→0 [ ]
cos x h cos(x+h)

2x+h −h
−2 sin( )⋅sin( )
1 1 2 2

= ⋅ limh→0 [ ]
cos x h cos(x+h)

h
sin( )
2x+h 2
⎡ sin( ){ } ⎤
2 h

1
⋅ limh→0 ⎢ ⎥
2
= ⎢ ⎥
cos x ⎢ cos(x+h) ⎥
⎣ ⎦

h
sin( )
2x+h 2 h
{limh→0 sin( )}{lim h }
2 →0 h 2
→0
2
2
= sec x ⋅
limh→0 cos(x+h)

sin x⋅1
= sec x ⋅
cos x

d
⇒ sec x = sec x tan x
dx

From (i), (ii), and (iii), we obtain

′ 2
f (x) = (x + sec x) (1 − sec x) + (x − tan x)(1 + sec x tan x)

Page : 318 , Block Name : Miscellaneous Exercise

Q30 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s
are xed non-zero constants and m and n are integers):
x
n
sin x

Answer. Let
x
f (x) = n
sin x

By quotient rule,

n d d n
sin x x−x sin x
′ dx dx
f (x) = 2n
sin x

It can be easily shown that
d n n−1
sin x = n sin x cos x
dx

Therefore,

n d d n
sin x x − x sin x
′ dx dx
f (x) =
2n
sin x
n n−1
sin x ⋅ 1 − x (n sin x cos x)
=
2n
sin x
n−1
sin x(sin x − nx cos x)
=
2n
sin x
sin x − nx cos x
=
n+1
cin x

Page : 318 , Block Name : Miscellaneous Exercise