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

### Question 11: Differentiate sin-1(2xâˆš(1-x2)) with respect to tan-1(x/âˆš(1-x2)) if -1/âˆš2<x <1/âˆš2.

Solutions:

Let u = sin-1(2xâˆš(1-x2))

Substitute x = sin Î¸ â‡’ Î¸ = sin-1x

u = sin-1(2 sin Î¸ âˆš(1 – sin2Î¸))

u = sin-1(2 sin Î¸ cos Î¸) [sin2Î¸ + cos2Î¸ = 1]

u = sin-1(sin 2Î¸) —–(i) [sin 2Î¸ = 2 sin Î¸ cos Î¸]

Let, v = tan-1(x/âˆš(1-x2))

v = tan-1(sin Î¸/ âˆš(1 – sin2Î¸))

v = tan-1(sin Î¸/cosÎ¸) [sin2Î¸ + cos2Î¸ = 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-1x

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

From equation (ii), v = Î¸ [tan-1(tan Î¸) = Î¸, Î¸ âˆˆ [- Ï€/2, Ï€/2]]

v = sin-1x

dv/dx = 1/ âˆš(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 Î¸, â‡’ Î¸ = tan-1x

u = tan-1(2 tan Î¸/(1 – tan2Î¸))

u = tan-1(tan 2Î¸) —–(i) [tan 2Î¸ = 2 tan Î¸/(1 – tan2Î¸)]

Now, let v = cos-1((1 – tan2Î¸)/(1 + tan2Î¸))

v = cos-1(cos 2Î¸) —–(ii) [cos 2Î¸ = (1 – tan2Î¸)/(1 + tan2Î¸)]

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-1x

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

From equation (ii), v = 2Î¸ [cos-1(cos Î¸) = Î¸, Î¸ âˆˆ [0 , Ï€]

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 Î¸ â‡’ Î¸ = tan-1x

u = 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-1x – Ï€/4

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

Now, let v = sin-1(3x – 4x3)

Substitute x = sin Î¸

v = sin-1(3 sin Î¸ – 4 sin3Î¸)

v = sin-1(sin 3Î¸) —–(iii) [sin 3Î¸ = 3 sin Î¸ – 4 sin3Î¸]

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-1x

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

Dividing equation (ii) by (iv)

du/dv = (1/3) * âˆš(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 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-1x

dv/dx = 1/(xâˆš(x2-1)) —–(ii)

Dividing equation (i) by (ii)

du/dv = Â½ * (- xâˆš(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 Î¸ â‡’ Î¸ = tan-1x

u = sin-1((2 tanÎ¸/ (1 + tan2Î¸))

u = sin-1(sin 2Î¸) —–(i) [sin2Î¸ = 2 tanÎ¸/(1 + tan2Î¸)]

Now, let v = tan-1(2x/(1 – x2))

v = tan-1(2 tan Î¸/(1 – tan2Î¸))

v = tan-1(tan 2Î¸) —–(ii) [tan 2Î¸ = 2 tan Î¸/(1 – tan2Î¸)]

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-1x

du/dx = 2/(1 + x2) —–(iii)

From equation (ii), v = 2Î¸ [tan-1(tan Î¸) = Î¸, Î¸ âˆˆ [- Ï€/2, Ï€/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(âˆš(1-x2)/x) if 1/2<x<1.

Solution:

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

Substitute x = cos Î¸ â‡’ Î¸ = cos-1x

u = cos-1(4 cos3Î¸ – 3cos Î¸)

u = cos-1(cos3Î¸) —–(i) [cos 3Î¸ = 4 cos3Î¸ – 3cos Î¸]

Now, let v = tan-1(âˆš(1 – x2)/x)

v = tan-1(âˆš(1 – cos2Î¸)/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-1x

du/dx = -3/âˆš(1-x2) —–(iii)

From equation (ii), v = Î¸ [tan-1(tan Î¸) = Î¸, Î¸ âˆˆ [- Ï€/2, Ï€/2]]

v = cos-1x

dv/dx = -1/âˆš(1-x2) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 3 (Ans)

### Question 17: Differentiate tan-1(x/âˆš(1-x2)) with respect to sin-1(2xâˆš(1-x2)) if -1/âˆš2<x<1/âˆš2.

Solution:

Let u = tan-1(x/âˆš(1-x2))

Substitute x = sin Î¸ â‡’ Î¸ = sin-1x

u = tan-1(sinÎ¸/âˆš(1 – sin2Î¸))

u = tan-1(sin Î¸/cos Î¸)

u = tan-1(tan Î¸) —–(i)

And, v = sin-1(2xâˆš(1-x2))

v = sin-1(2 sin Î¸ âˆš(1-sin2Î¸))

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-1x

du/dx = 1/ âˆš(1-x2) —–(iii)

From equation (ii), v = 2Î¸ [sin-1(sin Î¸) = Î¸, Î¸ âˆˆ (-Ï€/2, Ï€/2)]

v = 2sin-1x

dv/dx = 2/âˆš(1-x2) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 1/2 (Ans)

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

Solution:

Let u = sin-1âˆš(1 – x2 )

Substitute x = cos Î¸ â‡’ Î¸ = cos-1x

u = sin-1âˆš(1 – cos2Î¸)

u = sin-1(sin Î¸) —–(i)

And, v = cot-1(x/âˆš(1-x2))

v = cot-1(cos Î¸/âˆš(1 – cos2Î¸))

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-1x

du/dx = -1/ âˆš(1-x2) —–(iii)

And, from equation (ii), v = Î¸ [tan-1(tan Î¸) = Î¸, Î¸ âˆˆ [- Ï€/2, Ï€/2]]

v = cos-1x

dv/dx = -1/ âˆš(1-x2) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 1 (Ans)

### Question 19: Differentiate sin-1(2axâˆš(1-a2x2)) with respect to âˆš(1-a2x2) if -1/âˆš2<ax<1/âˆš2

Solution:

Let u = sin-1(2axâˆš(1-a2x2))

Substitute ax = sin Î¸ â‡’ Î¸ = sin-1(ax)

u = sin-1(2sin Î¸ âˆš(1 – sin2Î¸))

u = sin-1(2 sin Î¸ cos Î¸)

u = sin-1(sin 2Î¸) —–(i)

And, let v = âˆš(1-a2x2)

dv/dx = -a2x/ âˆš(1 – a2x2) —–(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 – a2x2) —–(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-x2) if -1<x<1 .

Solution:

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

Substitute x = tan Î¸ â‡’ Î¸ = tan-1x

u = 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-1x

du/dx = -1/(1 + x2) —–(ii)

And, v = âˆš(1 – x2)

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

Dividing equation (ii) by (iii)

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