Class 12 RD Sharma Solutions Chapter 11 Differentiation Exercise 11.8 | Set 1 (original) (raw)

Last Updated : 23 Jul, 2025

Differentiation is a branch of calculus that involves finding the rate of change of one variable with respect to another variable. We differentiate a function in order to find its derivative which is a measure of how that function (the output) changes as its input changes.

Differentiation Notation

when a variable y is expressed as a function of another variable x such that y = f(x), we can differentiate y with respect to x. Mathematically, it can be expressed as:

dy/dx or d/dx (f(x)) or f'(x) or y'

**Question 1: Differentiate x 2 with respect to x 3 .

**Solution:

_Let u = x 2 , and let v = x 3

_Differentiating u with respect to x,

_du/dx = 2x -----(i)

_Differentiating v with respect to x,

_dv/dx = 3x 2 ------(ii)

_Dividing equation (i) by (ii)

_(du/dx) / (dv/dx) = 2x/3x 2

****(du/dv) = 2/3x (Ans)**

**Question 2: Differentiate log (1+x 2 ) with respect to tan -1 x.

**Solution:

Let u = log(1 + x2)

Differentiating it with respect to x, using chain rule

du/dx = 1/(1+x2)* 2x = 2x/(1+x2) -----(i)

Now, let v = tan-1x

Differentiating it with respect to x

dv/dx = 1/(1+x2) -----(ii) [ d(tan-1x)/dx = 1/(1 + x2)]

Dividing equation (i) by (ii)

(du/dx) / (dv/dx) = {2x/(1+x2)} / {1/(1+x2)}

**du/dv = 2x (Ans)

**Question 3: Differentiate (log x) x with respect to log x.

**Solution:

Let u = (log x)x

Taking log on both the sides

log u = x log(log x) [log ab = b log a]

Differentiating above equation with respect to x

(1/u) * (du/dx) = 1* log (log x) + x*(1/log x)*(1/x) [d(log x)/dx = 1/x]

du/dx = u*(log (log x) + 1/log x)

du/dx = (log x)x * ((log x * log(log x) + 1) / log x)

du/dx = (log x)x-1*(1 + log x * log(log x)) -----(i)

Now, let v = log x

dv/dx = 1/x -----(ii)

Dividing equations (i) by (ii)

**du/dv = x(log x) x-1 *(1 + log x * log(log x)) (Ans)

**Question 4: Differentiate sin -1 √(1-x 2 ) with respect to cos -1 x, if

****(i) x ∈ (0, 1)**

****(ii) x ∈ (-1, 0)**

**Solution:

****(i)** Let u = sin-1√(1-x2)

Substitute x = cos θ, in above equation ⇒ θ = cos-1x

u = sin-1√(1-cos2θ)

u = sin-1(sin θ) -----(i) [ sin2θ + cos2θ = 1 ]

And, v = cos-1x -----(ii)

Now, x ∈ (0,1)

⇒ cos θ ∈ (0,1)

⇒ θ ∈ (0,π/2)

So, from equation (i),

u = θ [ sin-1(sin θ) = θ, θ ∈ (0,π/2) ]

u = cos-1x [ d(cos-1x)/dx = -1/√(1-x2) ]

Differentiating above equation with respect to x

du/dx = -1/√(1-x2) -----(iii)

Differentiating equation (ii) with respect to x

dv/dx = -1/√(1-x2) -----(iv)

Dividing equation (iii) by (iv)

**du/dv = 1 (Ans)

****(ii)** Let u = sin-1√(1-x2)

Substitute x = cos θ, in above equation ⇒ θ = cos-1x

u = sin-1√(1-cos2θ)

u = sin-1(sin θ) -----(i)

And, v = cos-1x -----(ii)

Now, x ∈ (-1,0)

⇒ cos θ ∈ (-1,0)

⇒ θ ∈ (π/2, π)

So, from equation (i),

u = π - θ [ sin-1(sin θ) = π - θ, θ ∈ (π/2, 3π/2) ]

u = π - cos-1x [ d(cos-1x)/dx = -1/√(1-x2) ]

Differentiating above equation with respect to x

du/dx = +1/√(1-x2) -----(iii)

Differentiating equation (ii) with respect to x

dv/dx = -1/√(1-x2) -----(iv)

Dividing equation (iii) by (iv)

**du/dv = -1 (Ans)

**Question 5: Differentiate sin -1 (4x√(1-4x 2 )) with respect to √(1-4x 2 )

****(i) x ∈ (-1/2, -1/2√2)**

****(ii) x ∈ (1/2√2, 1/2)**

**Solution:

****(i)** Let u = sin-1(4x√(1-4x2))

Substitute 2x = cos θ ⇒ θ = cos-1(2x)

u = sin-1(2 cos θ * √(1 - cos2θ)) [ sin2θ + cos2θ = 1 ]

u = sin-1(2 cos θ sin θ)

u = sin-1(sin 2θ) -----(i) [sin 2θ = 2 sin θ cos θ]

Let, v = √(1-4x2)

dv/dx = 1/(2 * √(1-4x2)) * (-8x) = -4x/√(1-4x2) -----(ii)

Here, x ∈ (-1/2,-1/2√2)

⇒ 2x ∈ (-1, -1/√2)

⇒ θ ∈ (¾ π, π)

So, from equation (i), u = π - 2θ [ sin-1(sin θ) = π - θ, θ ∈ (π/2, 3π/2) ]

u = π - 2cos-1(2x)

du/dx = 0 - 2* (-1/√(1-4x2)) * 2 = 4/√(1-4x2) -----(iii) [ d(cos-1x)/dx = -1/√(1-x2) ]

Dividing equation (iii) by (ii)

**du/dv = -(1/x) (Ans.)

****(ii)** Let u = sin-1(4x√(1-4x2))

Substitute 2x = cos θ ⇒ θ = cos-1(2x)

u = sin-1(2 cos θ * √(1 - cos2θ)) [ sin2θ + cos2θ = 1 ]

u = sin-1(2 cos θ sin θ)

u = sin-1(sin 2θ) -----(i) [sin 2θ = 2 sin θ cos θ]

Let, v = √(1-4x2)

dv/dx = 1/(2 * √(1-4x2)) * (-8x) = -4x/√(1-4x2) -----(ii)

Here, x ∈ (1/2√2, 1/2)

⇒ 2x ∈ (1/√2, 1)

⇒ θ ∈ (0, π/4)

So, from equation (i), u = 2θ [ sin-1(sin θ) = θ, θ ∈ (-π/2, π/2) ]

u = 2cos-1(2x)

du/dx = 2* (-1/√(1-4x2)) * 2 = -4/√(1-4x2) -----(iii) [ d(cos-1x)/dx = -1/√(1-x2) ]

Dividing equation (iii) by (ii)

**du/dv = (1/x) (Ans.)

**Question 6: Differentiate tan -1 ((√(1+x 2 )-1)/x) with respect to sin- 1 (2x/(1+x 2 )) if -1<x<1.

**Solution:

Let u = tan-1((√(1+x2)-1) / x)

Substitute x = tan θ ⇒ θ = tan-1x

u = tan-1((√(1+tan2θ)-1) / tan θ)

u = tan-1((sec θ -1) / tan θ) [sec2θ = 1 + tan2θ]

u = tan-1((1 - cos θ) / sin θ) [sec θ = 1/cos θ]

u = tan-1((2sin2(θ/2) / 2 sin(θ/2) cos(θ/2)) [1- cos2θ = 2 sin2θ, and sin2θ = 2sinθcosθ]

u = tan-1((sin(θ/2) / cos(θ/2))

u = tan-1(tan(θ/2)) -----(i) [tanθ = sinθ/cosθ]

Now, let v = sin-1(2x/1+x2)

v = sin-1(2 tanθ / (1 + tan2θ))

v = sin-1(sin 2θ) -----(ii) [sin2θ = 2 tanθ / (1 + tan2θ)]

Here, -1<x<1

⇒ -1<tan θ <1

⇒ - π/4 < θ < π/4

Therefore, from (i), u = θ/2 [ tan-1(tan θ) = θ, θ ∈ [ - π/2, π/2] ]

u = 1/2 * tan-1x

Differentiating it with respect to x,

du/dx = ½*(1 + x2)) -----(iii) [ d(tan-1x)/dx = 1/(1 + x2 )]

From equation (ii), v = 2θ [ sin-1(sin θ) = θ, θ ∈ [ - π/2, π/2] ]

v = 2tan-1x

Differentiating it with respect to x,

dv/dx = 2/(1 + x2) -----(iv)

Dividing equation (iii) by (iv)

**du/dv = 1/4 (Ans)

**Question 7: Differentiate sin -1 (2x√(1-x 2 )) with respect to sec -1 (1/√(1-x 2 )) if

****(i) x ∈ (0,1/√2)**

****(ii) x ∈ (1/√2,1)**

**Solution:

****(i)** Let u = sin-1(2x √(1-x2))

Substitute x = sin θ ⇒ θ = sin-1x

u = sin-1(2 sin θ √(1 - sin2θ))

u = sin-1(2 sin θ cos θ) [sin2θ + cos2θ = 1]

u = sin-1(sin 2 θ) -----(i)

And, let v = sec-1(1/√(1-x2))

v = sec-1(1/√(1-sin2θ))

v = sec-1(1/ cos θ) [sin2θ + cos2θ = 1]

v = sec-1(sec θ) [sec θ = 1/ cos θ]

v = cos-1(1/sec θ)

v = cos-1(cos θ) ----(ii) [sec-1x=cos-1(1/x)]

Here, x ∈ (0,1/√2)

sin θ ∈ (0,1/√2)

θ ∈ (0, π / 4)

From equation (i), u = 2θ [ sin-1(sin θ) = θ, θ ∈ [ - π / 2, π / 2] ]

u = 2sin-1x

du/dx = 2/√(1-x2) -----(iii) [d(sin-1x)/dx = 1/√(1 - x2)]

And, from equation (ii), v = θ [ cos-1(cos θ) = θ, θ ∈ [0, π]

v = sin-1x

dv/dx = 1/√(1-x2) ----(iv)

Dividing equation (iii) by (iv)

**du/dv = 2 (Ans)

****(ii)** Let u = sin-1(2x √(1-x2))

Substitute x = sin θ ⇒ θ = sin-1x

u = sin-1(2 sin θ √(1 - sin2θ))

u = sin-1(2 sin θ cos θ) [sin2θ + cos2θ = 1]

u = sin-1(sin 2 θ) -----(i)

And, let v = sec-1(1/√(1-x2))

v = sec-1(1/√(1-sin2θ))

v = sec-1(1/ cos θ) [sin2θ + cos2θ = 1]

v = sec-1(sec θ) [sec θ = 1/ cos θ]

v = cos-1(1/sec θ)

v = cos-1(cos θ) ----(ii) [sec-1x=cos-1(1/x)]

Here, x ∈ (1/√2, 1)

sin θ ∈ (1/√2, 1)

θ ∈ (π / 4, π/ 2)

From equation (i), u = 2θ [ sin-1(sin θ) = θ, θ ∈ [ - π / 2, π / 2] ]

u = 2sin-1x

du/dx = 2/√(1-x2) -----(iii) [d(sin-1x)/dx = 1/√(1 - x2)]

And, from equation (ii), v = θ [ cos-1(cos θ) = θ, θ ∈ [ 0 , π ]

v = sin-1x

dv/dx = 1/√(1-x2) ----(iv)

Dividing equation (iii) by (iv)

**du/dv = 2 (Ans)

**Question 8: Differentiate (cos x) sin x with respect to (sin x) cos x .

**Solution:

Let u = (cos x)sin x

Taking log on both sides,

log u = log(cos x)sin x

log u = sin x * log(cos x)

Differentiating above equation with respect to x, using product and chain rule,

1/u * du/dx = sin x * (1/cos x) * (- sin x) + cos x * log(cos x)

1/u * du/dx = (- sin x)* tan x + cos x * log (cos x)

du/dx = u * [(- sin x)* tan x + cos x * log (cos x)]

du/dx = (cos x)sin x * [cos x * log (cos x) - sin x * tan x ] -----(i)

And, let v = (sin x)cos x

Similarly, take log and differentiating the above equation, we get

dv/dx = (sin x)cos x * [cot x * cos x - sin x * log(sin x)] -----(ii)

Dividing equation (i) by (ii)

**du/dv = {(cos x)sin x * [cos x * log (cos x) - sin x * tan x ]} / {(sin x)cos x * [cot x * cos x - sin x * log(sin x)]} (Ans)

**Question 9: Differentiate sin -1 (2x / (1+x 2 )) with respect to cos -1 ((1-x 2 ) / (1+x 2 )), if 0<x<1.

**Solution:

Let u = sin-1(2x / (1+x2))

Substitute x = tan θ ⇒ θ = tan-1x

u = sin-1(2 tan θ / (1 + tan2θ))

u = sin-1(sin 2θ) -----(i) [ sin2θ = 2 tanθ/(1 + tan2θ) ]

Let v = cos-1((1 - x2 ) / (1 + x2 ))

v = cos-1((1 - tan2θ) / (1 + tan2θ))

v = cos-1(cos 2θ) -----(ii) [ cos2θ = (1 - tan2θ) / (1 + tan2θ) ]

Here, 0 < x <1

⇒ 0 < tan θ < 1

⇒ 0 < θ < π/4

From equation (i), u = 2θ [ sin-1(sin θ) = θ , θ ∈ [ -π/2 , π/2 ] ]

u = 2 tan-1x

Differentiating above equation with respect to x,

du/dx = 2/(1 + x2) -----(iii)

From equation (ii), v = 2θ [ cos-1(cos θ) = θ, θ ∈ [0, π]

v = 2 tan-1x

Differentiating above equation with respect to x,

dv/dx = 2/(1 + x2) -----(iv)

Dividing equation (iii) by (iv)

**du/dv = 1 (Ans)

**Question 10: Differentiate tan -1 ((1 + ax) / (1 - ax)) with respect to √(1 + a 2 x 2 ).

**Solution:

Let u = tan-1((1 + ax) / (1 - ax))

Substitute ax = tan θ ⇒ θ = tan-1(ax)

u = tan-1((1 + tan θ) / (1 - tan θ))

u = tan-1((tan π/4 + tan θ) / (1 - tan π/4 * tan θ))

u = tan-1(tan (π/4 + θ) [ tan(A + B) = (tan A + tan B) / (1 - tan A * tan B) ]

u = π/4 + θ

u = π/4 + tan-1(ax)

Differentiating above equation with respect to x,

du/dx = 0 + 1 / (1 + (ax)2) * a = a/(1 + a2x2) -----(i)

Now, let v = √(1 + a2x2)

dv/dx = 1/(2*√(1 + a2x2)) * a2 * 2x = a2x / √(1 + a2x2) -----(ii)

Dividing equation (i) by (ii)

**du/dv = 1/(ax*√(1 + a 2 x 2 )) (Ans)

Summary

Exercise 11.8 in RD Sharma's Class 12 mathematics textbook likely covers advanced applications of differentiation. This may include: