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Class 12 RD Sharma Solutions- Chapter 11 Differentiation – Exercise 11.8 | Set 1
• Difficulty Level : Hard
• Last Updated : 13 Jan, 2021

### Question 1: Differentiate x2 with respect to x3.

Solution:

Let u = x2, and let v = x3

Differentiating u with respect to x,

du/dx = 2x —–(i)

Differentiating v with respect to x,

dv/dx = 3x2 ——(ii)

Dividing equation (i) by (ii)

(du/dx) / (dv/dx) = 2x/3x2

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

### Question 2: Differentiate log (1+x2) with respect to tan-1x.

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-x2) 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√(1-4x2)) with respect to √(1-4x2)

(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+x2)-1)/x) with respect to sin-1(2x/(1+x2)) 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-x2)) with respect to sec-1(1/√(1-x2)) 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+x2)) with respect to cos-1((1-x2) / (1+x2)), 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 + a2x2).

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 + a2x2)) (Ans)

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