# Class 12 RD Sharma Solutions- Chapter 11 Differentiation – Exercise 11.8 | Set 2

**Question 11: Differentiate sin**^{-1}(2xâˆš(1-x^{2})) with respect to tan^{-1}(x/âˆš(1-x^{2})) if -1/âˆš2<x <1/âˆš2.

^{-1}(2xâˆš(1-x

^{2})) with respect to tan

^{-1}(x/âˆš(1-x

^{2})) if -1/âˆš2<x <1/âˆš2.

**Solutions:**

Let u = sin

^{-1}(2xâˆš(1-x^{2}))Substitute x = sin Î¸ â‡’ Î¸ = sin

^{-1}xu = sin

^{-1}(2 sin Î¸ âˆš(1 – sin^{2}Î¸))u = sin

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

^{-1}(sin 2Î¸) —–(i) [sin 2Î¸ = 2 sin Î¸ cos Î¸]Let, v = tan

^{-1}(x/âˆš(1-x^{2}))v = tan

^{-1}(sin Î¸/ âˆš(1 – sin^{2}Î¸))v = tan

^{-1}(sin Î¸/cosÎ¸) [sin^{2}Î¸ + cos^{2}Î¸ = 1]v = tan

^{-1}(tan Î¸) —–(ii) [tan Î¸ = sin Î¸/cos Î¸]Here, -1/âˆš2 < x <1/âˆš2

â‡’ -1/âˆš2 < sin Î¸ <1/âˆš2

â‡’ – Ï€/4 < Î¸ < Ï€/4

So, from equation (i), u = 2 Î¸ [sin

^{-1}(sin Î¸) = Î¸, Î¸ âˆˆ (0,Ï€/2)]u = 2 sin

^{-1}xdu/dx = 2/ âˆš(1 – x

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

^{-1}(tan Î¸) = Î¸, Î¸ âˆˆ [- Ï€/2, Ï€/2]]v = sin

^{-1}xdv/dx = 1/ âˆš(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 Î¸, â‡’ Î¸ = tan

^{-1}xu = tan

^{-1}(2 tan Î¸/(1 – tan^{2}Î¸))u = tan

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

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

^{-1}(cos 2Î¸) —–(ii) [cos 2Î¸ = (1 – tan^{2}Î¸)/(1 + tan^{2}Î¸)]Here, 0 < x <1

â‡’ 0 < tan Î¸ < 1

â‡’ 0 < Î¸ < Ï€/4

So, from equation (i), u = 2Î¸ [tan

^{-1}(tan Î¸) = Î¸, Î¸ âˆˆ [- Ï€/2, Ï€/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 = 2Î¸ [cos

^{-1}(cos Î¸) = Î¸, Î¸ âˆˆ [0 , Ï€]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 Î¸ â‡’ Î¸ = tan

^{-1}xu = tan

^{-1}((tan Î¸ – 1)/(tan Î¸ + 1))u = tan

^{-1}((tan Î¸ – tan Ï€/4)/(1 + tan Î¸ tan Ï€/4))u = tan

^{-1}(tan(Î¸ – Ï€/4) —–(i) [tan(A – B) = (tan A – tan B)/(1 + tan A * tan B]Here, -1/2 < x < 1/2

â‡’ -1/2 < tan Î¸ < 1/2

â‡’ -tan

^{-1}(1/2) < Î¸ < tan^{-1}(1/2)So, from equation (i), u = Î¸ – Ï€/4 [tan

^{-1}(tan Î¸) = Î¸, Î¸ âˆˆ [- Ï€/2, Ï€/2]]u = tan

^{-1}x – Ï€/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 Î¸

v = sin

^{-1}(3 sin Î¸ – 4 sin^{3}Î¸)v = sin

^{-1}(sin 3Î¸) —–(iii) [sin 3Î¸ = 3 sin Î¸ – 4 sin^{3}Î¸]Here -1/2<x<1/2

â‡’ -1/2 < sin Î¸ < 1/2

â‡’ – Ï€/6 < Î¸ < Ï€/6

From equation (ii), v = 3Î¸ [sin

^{-1}(sin Î¸) = Î¸, Î¸ âˆˆ (-Ï€/2 , Ï€/2)]v = 3 sin

^{-1}xdv/dx = 3/ âˆš(1 – x

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

du/dv = (1/3) * âˆš(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 Ï€/4 – tan(x/2))/(1 + tan(x/2)*tan Ï€/4))u = tan

^{-1}(tan(Ï€/4 – x/2) [tan(A – B) = (tan A – tan B)/(1 + tan A * tan B]u = Ï€/4 – x/2

du/dx = -1/2 —–(i)

Now, let v = sec

^{-1}xdv/dx = 1/(xâˆš(x

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

du/dv = Â½ * (- xâˆš(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 Î¸ â‡’ Î¸ = tan

^{-1}xu = sin

^{-1}((2 tanÎ¸/ (1 + tan^{2}Î¸))u = sin

^{-1}(sin 2Î¸) —–(i) [sin2Î¸ = 2 tanÎ¸/(1 + tan^{2}Î¸)]Now, let v = tan

^{-1}(2x/(1 – x^{2}))v = tan

^{-1}(2 tan Î¸/(1 – tan^{2}Î¸))v = tan

^{-1}(tan 2Î¸) —–(ii) [tan 2Î¸ = 2 tan Î¸/(1 – tan^{2}Î¸)]Here, -1<x<1

â‡’ -1<tan Î¸< 1

â‡’ – Ï€/4 < Î¸ < Ï€/4

So, from equation (i), u = 2Î¸ [sin

^{-1}(sin Î¸) = Î¸, Î¸ âˆˆ (-Ï€/2 , Ï€/2)]u = 2tan

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

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

^{-1}(tan Î¸) = Î¸, Î¸ âˆˆ [- Ï€/2, Ï€/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}(âˆš(1-x^{2})/x) if 1/2<x<1.

**Solution:**

Let u = cos

^{-1}(4x^{3}-3x)Substitute x = cos Î¸ â‡’ Î¸ = cos

^{-1}xu = cos

^{-1}(4 cos^{3}Î¸ – 3cos Î¸)u = cos

^{-1}(cos3Î¸) —–(i) [cos 3Î¸ = 4 cos^{3}Î¸ – 3cos Î¸]Now, let v = tan

^{-1}(âˆš(1 – x^{2})/x)v = tan

^{-1}(âˆš(1 – cos^{2}Î¸)/cos Î¸)v = tan

^{-1}(sin Î¸/cos Î¸)v = tan

^{-1}(tan Î¸) —–(ii)Here, 1/2 < x < 1

â‡’ 1/2 < cos Î¸ < 1

â‡’ 0 < Î¸ < Ï€/3

From equation (i), u = 3Î¸ [cos

^{-1}(cos Î¸) = Î¸, Î¸ âˆˆ [0 , Ï€]u = 3cos

^{-1}xdu/dx = -3/âˆš(1-x

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

^{-1}(tan Î¸) = Î¸, Î¸ âˆˆ [- Ï€/2, Ï€/2]]v = cos

^{-1}xdv/dx = -1/âˆš(1-x2) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 3 (Ans)

**Question 17: **Differentiate tan^{-1}(x/âˆš(1-x^{2})) with respect to sin^{-1}(2xâˆš(1-x^{2})) if -1/âˆš2<x<1/âˆš2.

**Solution:**

Let u = tan

^{-1}(x/âˆš(1-x^{2}))Substitute x = sin Î¸ â‡’ Î¸ = sin

^{-1}xu = tan

^{-1}(sinÎ¸/âˆš(1 – sin^{2}Î¸))u = tan

^{-1}(sin Î¸/cos Î¸)u = tan

^{-1}(tan Î¸) —–(i)And, v = sin

^{-1}(2xâˆš(1-x^{2}))v = sin

^{-1}(2 sin Î¸ âˆš(1-sin^{2}Î¸))v = sin

^{-1}(2 sin Î¸ cos Î¸)v = sin

^{-1}(sin 2Î¸) —–(ii)Here, -1/âˆš2<x<1/âˆš2

â‡’ -1/âˆš2 < sin Î¸ < 1/âˆš2

â‡’ – Ï€/4 < Î¸ < Ï€/4

From equation (i), u = Î¸ [tan

^{-1}(tan Î¸) = Î¸, Î¸ âˆˆ [- Ï€/2, Ï€/2]]u = sin

^{-1}xdu/dx = 1/ âˆš(1-x

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

^{-1}(sin Î¸) = Î¸, Î¸ âˆˆ (-Ï€/2, Ï€/2)]v = 2sin

^{-1}xdv/dx = 2/âˆš(1-x

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

du/dv = 1/2 (Ans)

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

**Solution:**

Let u = sin

^{-1}âˆš(1 – x^{2 })Substitute x = cos Î¸ â‡’ Î¸ = cos

^{-1}xu = sin

^{-1}âˆš(1 – cos^{2}Î¸)u = sin

^{-1}(sin Î¸) —–(i)And, v = cot

^{-1}(x/âˆš(1-x^{2}))v = cot

^{-1}(cos Î¸/âˆš(1 – cos^{2}Î¸))v = cot

^{-1}(cos Î¸/sin Î¸)v = cot

^{-1}(cot Î¸)v = tan

^{-1}(tan Î¸) —–(ii) [tan^{-1}(Î¸) = cot^{-1}(1/Î¸)]Here, 0 < x <1

0 < cos Î¸ <1

0 < Î¸ < Ï€/2

So, from equation (i), u = Î¸ [sin

^{-1}(sin Î¸) = Î¸, Î¸ âˆˆ (-Ï€/2 , Ï€/2)]u = cos

^{-1}xdu/dx = -1/ âˆš(1-x

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

^{-1}(tan Î¸) = Î¸, Î¸ âˆˆ [- Ï€/2, Ï€/2]]v = cos

^{-1}xdv/dx = -1/ âˆš(1-x

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

du/dv = 1 (Ans)

**Question 19: **Differentiate sin^{-1}(2axâˆš(1-a^{2}x^{2})) with respect to âˆš(1-a^{2}x^{2}) if -1/âˆš2<ax<1/âˆš2

**Solution:**

Let u = sin

^{-1}(2axâˆš(1-a^{2}x^{2}))Substitute ax = sin Î¸ â‡’ Î¸ = sin

^{-1}(ax)u = sin

^{-1}(2sin Î¸ âˆš(1 – sin^{2}Î¸))u = sin

^{-1}(2 sin Î¸ cos Î¸)u = sin

^{-1}(sin 2Î¸) —–(i)And, let v = âˆš(1-a

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

^{2}x/ âˆš(1 – a^{2}x^{2}) —–(ii)Here, -1/âˆš2<ax<1/âˆš2

â‡’ -1/âˆš2 < sin Î¸ <1/âˆš2

â‡’ – Ï€/4 < Î¸ < Ï€/4

So, from equation (i), u = 2Î¸ [sin

^{-1}(sin Î¸) = Î¸, Î¸ âˆˆ (-Ï€/2 , Ï€/2)]u = 2sin

^{-1}(ax)du/dx = 2a/ âˆš(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 âˆš(1-x^{2}) if -1<x<1 .

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

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

**Solution:**

Let u = tan

^{-1}((1-x)/(1+x))Substitute x = tan Î¸ â‡’ Î¸ = tan

^{-1}xu = tan

^{-1}((1 – tan Î¸)/(1 + tan Î¸))u = tan

^{-1}((tan Ï€/4 – tan Î¸)/(1 + tan Î¸ tan Ï€/4))u = tan

^{-1}(tan(Ï€/4 – Î¸) —–(i) [tan(A – B) = (tan A – tan B)/(1 + tan A * tan B]Here, -1 < x < 1

â‡’ -1 < tan Î¸ < 1

â‡’ – Ï€/4 < Î¸ < Ï€/4

So, from equation (i), u = Ï€/4 – Î¸ [tan

^{-1}(tan Î¸) = Î¸, Î¸ âˆˆ [- Ï€/2, Ï€/2]]u = Ï€/4 – tan

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

^{2}) —–(ii)And, v = âˆš(1 – x

^{2})dv/dx = -2x/(2 * âˆš(1 – x

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

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

## Please

Loginto comment...