RD Sharma Class 12 Ex 11.8 Solutions Chapter 11 Differentiation

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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 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-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)

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 – x² )

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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