**Question 11: Differentiate sin**^{-1}(2x√(1-x^{2})) with respect to tan^{-1}(x/√(1-x^{2})) if -1/√2<x <1/√2.

^{-1}(2x√(1-x

^{2})) with respect to tan

^{-1}(x/√(1-x

^{2})) if -1/√2<x <1/√2.

**Solutions:**

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 2θ) —–(i) [sin 2θ = 2 sin θ cos θ]Let, v = tan

^{-1}(x/√(1-x^{2}))v = tan

^{-1}(sin θ/ √(1 – sin^{2}θ))v = tan

^{-1}(sin θ/cosθ) [sin^{2}θ + cos^{2}θ = 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

^{-1}xdu/dx = 2/ √(1 – x

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

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

^{-1}xdv/dx = 1/ √(1 – x

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

du/dv = 2 (Ans)

**Question 12: **Differentiate tan^{-1}(2x/(1-x^{2})) with respect to cos^{-1}((1-x^{2})/(1+x^{2})) if 0<x<1

**Solutions:**

Let u = tan

^{-1}(2x/(1 – x^{2}))

Substitute x = tan θ, ⇒ θ = tan

^{-1}xu = tan

^{-1}(2 tan θ/(1 – tan^{2}θ))u = tan

^{-1}(tan 2θ) —–(i) [tan 2θ = 2 tan θ/(1 – tan^{2}θ)]Now, let v = cos

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

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

^{-1}xdu/dx = 2/(1 + x

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

^{-1}(cos θ) = θ, θ ∈ [0 , π]v = 2 tan

^{-1}xdv/dx = 2/(1 + x

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

du/dv = 1 (Ans)

**Question 13: **Differentiate tan^{-1}((x-1)/(x+1)) with respect to sin^{-1}(3x-4x^{3}) if -1/2<x<1/2.

**Solutions:**

Let u = tan

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

^{-1}xu = 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

^{-1}x – π/4du/dx = 1/(1 + x

^{2}) —–(ii) [d(tan^{-1}x)/dx = 1/(1 + x^{2})]Now, let v = sin

^{-1}(3x – 4x^{3})Substitute x = sin θ

v = sin

^{-1}(3 sin θ – 4 sin^{3}θ)v = sin

^{-1}(sin 3θ) —–(iii) [sin 3θ = 3 sin θ – 4 sin^{3}θ]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

^{-1}xdv/dx = 3/ √(1 – x

^{2}) —–(iv) [d(sin^{-1}x)/dx = 1/√(1 – x^{2})]Dividing equation (ii) by (iv)

du/dv = (1/3) * √(1 – x^{2})/(1 + x^{2}) (Ans)

**Question 14: **Differentiate tan^{-1}(cosx/(1+sinx)) with respect to sec^{-1}x.

**Solutions:**

Let u = tan

^{-1}(cos x/(1 + sin x))u = tan

^{-1}((cos^{2}(x/2) – sin^{2}(x/2))/(cos^{2}(x/2) + sin^{2}(x/2) + 2sin(x/2)cos(x/2)) [cos 2x = cos^{2}x – sin^{2}x , cos^{2}x + sin^{2}x = 1, sin 2x = 2sin x cos x]Using [a

^{2}– b^{2}=(a-b)(a+b) ,and (a+b)^{2}= a^{2}+ b^{2}+ 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

^{-1}xdv/dx = 1/(x√(x

^{2}-1)) —–(ii)Dividing equation (i) by (ii)

du/dv = ½ * (- x√(x^{2}-1)) (Ans.)

**Question 15: **Differentiate sin^{-1}(2x/(1+x^{2})) with respect to tan^{-1}(2x/(1-x^{2})) if -1<x<1.

**Solution:**

Let u = sin

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

^{-1}xu = sin

^{-1}((2 tanθ/ (1 + tan^{2}θ))u = sin

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

^{-1}(2x/(1 – x^{2}))v = tan

^{-1}(2 tan θ/(1 – tan^{2}θ))v = tan

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

^{-1}xdu/dx = 2/(1 + x

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

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

^{-1}xdv/dx = 2/(1+x

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

du/dv = 1 (Ans)

**Question 16: **Differentiate cos^{-1}(4x^{3}-3x) with respect to tan^{-1}(√(1-x^{2})/x) if 1/2<x<1.

**Solution:**

Let u = cos

^{-1}(4x^{3}-3x)Substitute x = cos θ ⇒ θ = cos

^{-1}xu = cos

^{-1}(4 cos^{3}θ – 3cos θ)u = cos

^{-1}(cos3θ) —–(i) [cos 3θ = 4 cos^{3}θ – 3cos θ]Now, let v = tan

^{-1}(√(1 – x^{2})/x)v = tan

^{-1}(√(1 – cos^{2}θ)/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

^{-1}xdu/dx = -3/√(1-x

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

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

^{-1}xdv/dx = -1/√(1-x2) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 3 (Ans)

**Question 17: **Differentiate tan^{-1}(x/√(1-x^{2})) with respect to sin^{-1}(2x√(1-x^{2})) if -1/√2<x<1/√2.

**Solution:**

Let u = tan

^{-1}(x/√(1-x^{2}))Substitute x = sin θ ⇒ θ = sin

^{-1}xu = tan

^{-1}(sinθ/√(1 – sin^{2}θ))u = tan

^{-1}(sin θ/cos θ)u = tan

^{-1}(tan θ) —–(i)And, v = sin

^{-1}(2x√(1-x^{2}))v = sin

^{-1}(2 sin θ √(1-sin^{2}θ))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

^{-1}xdu/dx = 1/ √(1-x

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

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

^{-1}xdv/dx = 2/√(1-x

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

du/dv = 1/2 (Ans)

**Question 18: **Differentiate sin^{-1}√(1-x^{2}) with respect to cot^{-1}(x/√(1-x^{2})) if 0<x<1.

**Solution:**

Let u = sin

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

^{-1}xu = sin

^{-1}√(1 – cos^{2}θ)u = sin

^{-1}(sin θ) —–(i)And, v = cot

^{-1}(x/√(1-x^{2}))v = cot

^{-1}(cos θ/√(1 – cos^{2}θ))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

^{-1}xdu/dx = -1/ √(1-x

^{2}) —–(iii)And, from equation (ii), v = θ [tan

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

^{-1}xdv/dx = -1/ √(1-x

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

du/dv = 1 (Ans)

**Question 19: **Differentiate sin^{-1}(2ax√(1-a^{2}x^{2})) with respect to √(1-a^{2}x^{2}) if -1/√2<ax<1/√2

**Solution:**

Let u = sin

^{-1}(2ax√(1-a^{2}x^{2}))Substitute ax = sin θ ⇒ θ = sin

^{-1}(ax)u = sin

^{-1}(2sin θ √(1 – sin^{2}θ))u = sin

^{-1}(2 sin θ cos θ)u = sin

^{-1}(sin 2θ) —–(i)And, let v = √(1-a

^{2}x^{2})dv/dx = -a

^{2}x/ √(1 – a^{2}x^{2}) —–(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 – a

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

du/dv = -2/ax (Ans)

**Question 20: Differentiate tan**^{-1}((1-x)/(1+x)) with respect to √(1-x^{2}) if -1<x<1 .

^{-1}((1-x)/(1+x)) with respect to √(1-x

^{2}) if -1<x<1 .

**Solution:**

Let u = tan

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

^{-1}xu = 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

^{-1}xdu/dx = -1/(1 + x

^{2}) —–(ii)And, v = √(1 – x

^{2})dv/dx = -2x/(2 * √(1 – x

^{2})) = – x/√(1 – x^{2}) —–(iii)Dividing equation (ii) by (iii)

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