# Class 12 RD Sharma Solutions- Chapter 11 Differentiation – Exercise 11.8 | Set 1

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

^{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.

^{2}) with respect to tan

^{-1}x.

**Solution:**

Let u = log(1 + x

^{2})Differentiating it with respect to x, using chain rule

du/dx = 1/(1+x

^{2})* 2x = 2x/(1+x^{2}) —–(i)Now, let v = tan

^{-1}xDifferentiating it with respect to x

dv/dx = 1/(1+x

^{2}) —–(ii) [ d(tan^{-1}x)/dx = 1/(1 + x^{2})]Dividing equation (i) by (ii)

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

^{2})} / {1/(1+x^{2})}

du/dv = 2x (Ans)

**Question 3: Differentiate (log x)**^{x} with respect to 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 a

^{b}= 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

^{-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-x^{2})Substitute x = cos θ, in above equation ⇒ θ = cos

^{-1}xu = sin

^{-1}√(1-cos^{2}θ)u = sin

^{-1}(sin θ) —–(i) [ sin^{2}θ + cos^{2}θ = 1 ]And, v = cos

^{-1}x —–(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

^{-1}x [ d(cos^{-1}x)/dx = -1/√(1-x^{2}) ]Differentiating above equation with respect to x

du/dx = -1/√(1-x

^{2}) —–(iii)Differentiating equation (ii) with respect to x

dv/dx = -1/√(1-x

^{2}) —–(iv)Dividing equation (iii) by (iv)

du/dv = 1 (Ans)

(ii)Let u = sin^{-1}√(1-x^{2})Substitute x = cos θ, in above equation ⇒ θ = cos

^{-1}xu = sin

^{-1}√(1-cos^{2}θ)u = sin

^{-1}(sin θ) —–(i)And, v = cos

^{-1}x —–(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

^{-1}x [ d(cos^{-1}x)/dx = -1/√(1-x^{2}) ]Differentiating above equation with respect to x

du/dx = +1/√(1-x

^{2}) —–(iii)Differentiating equation (ii) with respect to x

dv/dx = -1/√(1-x

^{2}) —–(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})

^{-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√(1-4x^{2}))Substitute 2x = cos θ ⇒ θ = cos

^{-1}(2x)u = sin

^{-1}(2 cos θ * √(1 – cos^{2}θ)) [ sin^{2}θ + cos^{2}θ = 1 ]u = sin

^{-1}(2 cos θ sin θ)u = sin

^{-1}(sin 2θ) —–(i) [sin 2θ = 2 sin θ cos θ]Let, v = √(1-4x

^{2})dv/dx = 1/(2 * √(1-4x

^{2})) * (-8x) = -4x/√(1-4x^{2}) —–(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/√(1-4x

^{2})) * 2 = 4/√(1-4x^{2}) —–(iii) [ d(cos^{-1}x)/dx = -1/√(1-x^{2}) ]Dividing equation (iii) by (ii)

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

(ii)Let u = sin^{-1}(4x√(1-4x^{2}))Substitute 2x = cos θ ⇒ θ = cos

^{-1}(2x)u = sin

^{-1}(2 cos θ * √(1 – cos^{2}θ)) [ sin^{2}θ + cos^{2}θ = 1 ]u = sin

^{-1}(2 cos θ sin θ)u = sin

^{-1}(sin 2θ) —–(i) [sin 2θ = 2 sin θ cos θ]Let, v = √(1-4x

^{2})dv/dx = 1/(2 * √(1-4x

^{2})) * (-8x) = -4x/√(1-4x^{2}) —–(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/√(1-4x

^{2})) * 2 = -4/√(1-4x^{2}) —–(iii) [ d(cos^{-1}x)/dx = -1/√(1-x^{2}) ]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.

^{-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+x^{2})-1) / x)Substitute x = tan θ ⇒ θ = tan

^{-1}xu = tan

^{-1}((√(1+tan^{2}θ)-1) / tan θ)u = tan

^{-1}((sec θ -1) / tan θ) [sec^{2}θ = 1 + tan^{2}θ]u = tan

^{-1}((1 – cos θ) / sin θ) [sec θ = 1/cos θ]u = tan

^{-1}((2sin^{2}(θ/2) / 2 sin(θ/2) cos(θ/2)) [1- cos2θ = 2 sin^{2}θ, 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+x^{2})v = sin

^{-1}(2 tanθ / (1 + tan^{2}θ))v = sin

^{-1}(sin 2θ) —–(ii) [sin2θ = 2 tanθ / (1 + tan^{2}θ)]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

^{-1}xDifferentiating it with respect to x,

du/dx = ½*(1 + x

^{2})) —–(iii) [ d(tan^{-1}x)/dx = 1/(1 + x^{2 })]From equation (ii), v = 2θ [ sin

^{-1}(sin θ) = θ, θ ∈ [ – π/2, π/2] ]v = 2tan

^{-1}xDifferentiating it with respect to x,

dv/dx = 2/(1 + x

^{2}) —–(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

^{-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-x^{2}))Substitute x = sin θ ⇒ θ = sin

^{-1}xu = sin

^{-1}(2 sin θ √(1 – sin^{2}θ))u = sin

^{-1}(2 sin θ cos θ) [sin^{2}θ + cos^{2}θ = 1]u = sin

^{-1}(sin 2 θ) —–(i)And, let v = sec

^{-1}(1/√(1-x^{2}))v = sec

^{-1}(1/√(1-sin^{2}θ))v = sec

^{-1}(1/ cos θ) [sin^{2}θ + cos^{2}θ = 1]v = sec

^{-1}(sec θ) [sec θ = 1/ cos θ]v = cos

^{-1}(1/sec θ)v = cos

^{-1}(cos θ) —-(ii) [sec^{-1}x=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

^{-1}xdu/dx = 2/√(1-x

^{2}) —–(iii) [d(sin^{-1}x)/dx = 1/√(1 – x^{2})]And, from equation (ii), v = θ [ cos

^{-1}(cos θ) = θ, θ ∈ [0, π]v = sin

^{-1}xdv/dx = 1/√(1-x

^{2}) —-(iv)Dividing equation (iii) by (iv)

du/dv = 2 (Ans)

(ii)Let u = sin^{-1}(2x √(1-x^{2}))Substitute x = sin θ ⇒ θ = sin

^{-1}xu = sin

^{-1}(2 sin θ √(1 – sin^{2}θ))u = sin

^{-1}(2 sin θ cos θ) [sin^{2}θ + cos^{2}θ = 1]u = sin

^{-1}(sin 2 θ) —–(i)And, let v = sec

^{-1}(1/√(1-x^{2}))v = sec

^{-1}(1/√(1-sin^{2}θ))v = sec

^{-1}(1/ cos θ) [sin^{2}θ + cos^{2}θ = 1]v = sec

^{-1}(sec θ) [sec θ = 1/ cos θ]v = cos-

^{1}(1/sec θ)v = cos

^{-1}(cos θ) —-(ii) [sec^{-1}x=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

^{-1}xdu/dx = 2/√(1-x

^{2}) —–(iii) [d(sin^{-1}x)/dx = 1/√(1 – x^{2})]And, from equation (ii), v = θ [ cos

^{-1}(cos θ) = θ, θ ∈ [ 0 , π ]v = sin

^{-1}xdv/dx = 1/√(1-x

^{2}) —-(iv)Dividing equation (iii) by (iv)

du/dv = 2 (Ans)

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

^{-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+x^{2}))Substitute x = tan θ ⇒ θ = tan

^{-1}xu = sin

^{-1}(2 tan θ / (1 + tan^{2}θ))u = sin

^{-1}(sin 2θ) —–(i) [ sin2θ = 2 tanθ/(1 + tan^{2}θ) ]Let v = cos

^{-1}((1 – x^{2 }) / (1 + x^{2 }))v = cos

^{-1}((1 – tan^{2}θ) / (1 + tan^{2}θ))v = cos

^{-1}(cos 2θ) —–(ii) [ cos2θ = (1 – tan^{2}θ) / (1 + tan^{2}θ) ]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

^{-1}xDifferentiating above equation with respect to x,

du/dx = 2/(1 + x

^{2}) —–(iii)From equation (ii), v = 2θ [ cos

^{-1}(cos θ) = θ, θ ∈ [0, π]v = 2 tan

^{-1}xDifferentiating above equation with respect to x,

dv/dx = 2/(1 + x

^{2}) —–(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}).

^{-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 + a^{2}x^{2}) —–(i)Now, let v = √(1 + a

^{2}x^{2})dv/dx = 1/(2*√(1 + a

^{2}x^{2})) * a^{2}* 2x = a^{2}x / √(1 + a^{2}x^{2}) —–(ii)Dividing equation (i) by (ii)

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

Attention reader! Don’t stop learning now. Participate in the **Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students**.