# RD Sharma Class 12 Ex 11.8 Solutions Chapter 11 Differentiation

Here we provide RD Sharma Class 12 Ex 11.8 Solutions Chapter 11 Differentiation for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 11.8 Solutions Chapter 11 Differentiation book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 11.8 Solutions Chapter 11 Differentiation

### 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 + x)]

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 – x) / (1 + x))

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)

### Question 11: Differentiate sin-1(2x√(1-x2)) with respect to tan-1(x/√(1-x2)) if -1/√2<x <1/√2.

Solutions:

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 2θ) —–(i) [sin 2θ = 2 sin θ cos θ]

Let, v = tan-1(x/√(1-x2))

v = tan-1(sin θ/ √(1 – sin2θ))

v = tan-1(sin θ/cosθ) [sin2θ + cos2θ = 1]

v = tan-1(tan θ) —–(ii) [tan θ = sin θ/cos θ]

Here, -1/√2 < x <1/√2

⇒ -1/√2 < sin θ <1/√2

⇒ – π/4 < θ < π/4

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

u = 2 sin-1x

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

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

v = sin-1x

dv/dx = 1/ √(1 – x2) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 2 (Ans)

### Question 12: Differentiate tan-1(2x/(1-x2)) with respect to cos-1((1-x2)/(1+x2)) if 0<x<1

Solutions:

Let u = tan-1(2x/(1 – x2))

Substitute x = tan θ, ⇒ θ = tan-1x

u = tan-1(2 tan θ/(1 – tan2θ))

u = tan-1(tan 2θ) —–(i) [tan 2θ = 2 tan θ/(1 – tan2θ)]

Now, let v = cos-1((1 – tan2θ)/(1 + tan2θ))

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

Here, 0 < x <1

⇒ 0 < tan θ < 1

⇒ 0 < θ < π/4

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

u = 2 tan-1x

du/dx = 2/(1 + x2) —–(iii) [d(tan-1x)/dx = 1/(1 + x2)]

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

v = 2 tan-1x

dv/dx = 2/(1 + x2) —–(iv) [d(tan-1x)/dx = 1/(1 + x2)]

Dividing equation (iii) by (iv)

du/dv = 1 (Ans)

### Question 13: Differentiate tan-1((x-1)/(x+1)) with respect to sin-1(3x-4x3) if -1/2<x<1/2.

Solutions:

Let u = tan-1((x – 1)/(x + 1))

Substitute x = tan θ ⇒ θ = tan-1x

u = tan-1((tan θ – 1)/(tan θ + 1))

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

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

Here, -1/2 < x < 1/2

⇒ -1/2 < tan θ < 1/2

⇒ -tan-1(1/2) < θ < tan-1(1/2)

So, from equation (i), u = θ – π/4 [tan-1(tan θ) = θ, θ ∈ [- π/2, π/2]]

u = tan-1x – π/4

du/dx = 1/(1 + x2) —–(ii) [d(tan-1x)/dx = 1/(1 + x2)]

Now, let v = sin-1(3x – 4x3)

Substitute x = sin θ

v = sin-1(3 sin θ – 4 sin3θ)

v = sin-1(sin 3θ) —–(iii) [sin 3θ = 3 sin θ – 4 sin3θ]

Here -1/2<x<1/2

⇒ -1/2 < sin θ < 1/2

⇒ – π/6 < θ < π/6

From equation (ii), v = 3θ [sin-1(sin θ) = θ, θ ∈ (-π/2 , π/2)]

v = 3 sin-1x

dv/dx = 3/ √(1 – x2) —–(iv) [d(sin-1x)/dx = 1/√(1 – x2)]

Dividing equation (ii) by (iv)

du/dv = (1/3) * √(1 – x2)/(1 + x2) (Ans)

### Question 14: Differentiate tan-1(cosx/(1+sinx)) with respect to sec-1x.

Solutions:

Let u = tan-1(cos x/(1 + sin x))

u = tan-1((cos2(x/2) – sin2(x/2))/(cos2(x/2) + sin2(x/2) + 2sin(x/2)cos(x/2)) [cos 2x = cos2x – sin2x , cos2x + sin2x = 1, sin 2x = 2sin x cos x]

Using [a2 – b2 =(a-b)(a+b) ,and (a+b)2 = a2 + b2 + 2ab],

u = tan-1(((cos(x/2) – sin(x/2)) * (cos(x/2) + sin(x/2)))/(cos(x/2) + sin(x/2))2

u = tan-1((cos(x/2) – sin(x/2))/(cos(x/2 + sin(x/2)))

Dividing numerator and denominator by cos(x/2)

u = tan-1((1 – tan(x/2))/(1 + tan(x/2)))

u = tan-1((tan π/4 – tan(x/2))/(1 + tan(x/2)*tan π/4))

u = tan-1(tan(π/4 – x/2) [tan(A – B) = (tan A – tan B)/(1 + tan A * tan B]

u = π/4 – x/2

du/dx = -1/2 —–(i)

Now, let v = sec-1x

dv/dx = 1/(x√(x2-1)) —–(ii)

Dividing equation (i) by (ii)

du/dv = ½ * (- x√(x2-1)) (Ans.)

### Question 15: Differentiate sin-1(2x/(1+x2)) with respect to tan-1(2x/(1-x2)) if -1<x<1.

Solution:

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

Substitute x = tan θ ⇒ θ = tan-1x

u = sin-1((2 tanθ/ (1 + tan2θ))

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

Now, let v = tan-1(2x/(1 – x2))

v = tan-1(2 tan θ/(1 – tan2θ))

v = tan-1(tan 2θ) —–(ii) [tan 2θ = 2 tan θ/(1 – tan2θ)]

Here, -1<x<1

⇒ -1<tan θ< 1

⇒ – π/4 < θ < π/4

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

u = 2tan-1x

du/dx = 2/(1 + x2) —–(iii)

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

v = 2tan-1x

dv/dx = 2/(1+x2) ——(iv)

Dividing equation (iii) by (iv)

du/dv = 1 (Ans)

### Question 16: Differentiate cos-1(4x3-3x) with respect to tan-1(√(1-x2)/x) if 1/2<x<1.

Solution:

Let u = cos-1(4x3-3x)

Substitute x = cos θ ⇒ θ = cos-1x

u = cos-1(4 cos3θ – 3cos θ)

u = cos-1(cos3θ) —–(i) [cos 3θ = 4 cos3θ – 3cos θ]

Now, let v = tan-1(√(1 – x2)/x)

v = tan-1(√(1 – cos2θ)/cos θ)

v = tan-1(sin θ/cos θ)

v = tan-1(tan θ) —–(ii)

Here, 1/2 < x < 1

⇒ 1/2 < cos θ < 1

⇒ 0 < θ < π/3

From equation (i), u = 3θ [cos-1(cos θ) = θ, θ ∈ [0 , π]

u = 3cos-1x

du/dx = -3/√(1-x2) —–(iii)

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

v = cos-1x

dv/dx = -1/√(1-x2) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 3 (Ans)

### Question 17: Differentiate tan-1(x/√(1-x2)) with respect to sin-1(2x√(1-x2)) if -1/√2<x<1/√2.

Solution:

Let u = tan-1(x/√(1-x2))

Substitute x = sin θ ⇒ θ = sin-1x

u = tan-1(sinθ/√(1 – sin2θ))

u = tan-1(sin θ/cos θ)

u = tan-1(tan θ) —–(i)

And, v = sin-1(2x√(1-x2))

v = sin-1(2 sin θ √(1-sin2θ))

v = sin-1(2 sin θ cos θ)

v = sin-1(sin 2θ) —–(ii)

Here, -1/√2<x<1/√2

⇒ -1/√2 < sin θ < 1/√2

⇒ – π/4 < θ < π/4

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

u = sin-1x

du/dx = 1/ √(1-x2) —–(iii)

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

v = 2sin-1x

dv/dx = 2/√(1-x2) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 1/2 (Ans)

### Question 18: Differentiate sin-1√(1-x2) with respect to cot-1(x/√(1-x2)) if 0<x<1.

Solution:

Let u = sin-1√(1 – x)

Substitute x = cos θ ⇒ θ = cos-1x

u = sin-1√(1 – cos2θ)

u = sin-1(sin θ) —–(i)

And, v = cot-1(x/√(1-x2))

v = cot-1(cos θ/√(1 – cos2θ))

v = cot-1(cos θ/sin θ)

v = cot -1(cot θ)

v = tan-1(tan θ) —–(ii) [tan-1(θ) = cot-1(1/θ)]

Here, 0 < x <1

0 < cos θ <1

0 < θ < π/2

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

u = cos-1x

du/dx = -1/ √(1-x2) —–(iii)

And, from equation (ii), v = θ [tan-1(tan θ) = θ, θ ∈ [- π/2, π/2]]

v = cos-1x

dv/dx = -1/ √(1-x2) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 1 (Ans)

### Question 19: Differentiate sin-1(2ax√(1-a2x2)) with respect to √(1-a2x2) if -1/√2<ax<1/√2

Solution:

Let u = sin-1(2ax√(1-a2x2))

Substitute ax = sin θ ⇒ θ = sin-1(ax)

u = sin-1(2sin θ √(1 – sin2θ))

u = sin-1(2 sin θ cos θ)

u = sin-1(sin 2θ) —–(i)

And, let v = √(1-a2x2)

dv/dx = -a2x/ √(1 – a2x2) —–(ii)

Here, -1/√2<ax<1/√2

⇒ -1/√2 < sin θ <1/√2

⇒ – π/4 < θ < π/4

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

u = 2sin-1(ax)

du/dx = 2a/ √(1 – a2x2) —–(iii)

Dividing equation (iii) by (ii)

du/dv = -2/ax (Ans)

### Question 20: Differentiate tan-1((1-x)/(1+x)) with respect to √(1-x2) if -1<x<1 .

Solution:

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

Substitute x = tan θ ⇒ θ = tan-1x

u = tan-1((1 – tan θ)/(1 + tan θ))

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

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

Here, -1 < x < 1

⇒ -1 < tan θ < 1

⇒ – π/4 < θ < π/4

So, from equation (i), u = π/4 – θ [tan-1(tan θ) = θ, θ ∈ [- π/2, π/2]]

u = π/4 – tan-1x

du/dx = -1/(1 + x2) —–(ii)

And, v = √(1 – x2)

dv/dx = -2x/(2 * √(1 – x2)) = – x/√(1 – x2) —–(iii)

Dividing equation (ii) by (iii)

du/dv = √(1 – x2)/(x * (1 + x2)) (Ans)

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