# Class 12 RD Sharma Solutions- Chapter 11 Differentiation – Exercise 11.8 | Set 1

Last Updated : 13 Jan, 2021

### 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-1âˆš(1-x2) with respect to cos-1 x, if

(i) x âˆˆ (0, 1)

(ii) x âˆˆ (-1, 0)

Solution:

(i) Let u = sin-1âˆš(1-x2)

Substitute x = cos Î¸, in above equation â‡’ Î¸ = cos-1x

u = sin-1âˆš(1-cos2Î¸)

u = sin-1(sin Î¸) —–(i) [ sin2Î¸ + cos2Î¸ = 1 ]

And, v = cos-1x —–(ii)

Now, x âˆˆ (0,1)

â‡’ cos Î¸ âˆˆ (0,1)

â‡’ Î¸ âˆˆ (0,Ï€/2)

So, from equation (i),

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

u = cos-1x [ d(cos-1x)/dx = -1/âˆš(1-x2) ]

Differentiating above equation with respect to x

du/dx = -1/âˆš(1-x2) —–(iii)

Differentiating equation (ii) with respect to x

dv/dx = -1/âˆš(1-x2) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 1 (Ans)

(ii) Let u = sin-1âˆš(1-x2)

Substitute x = cos Î¸, in above equation â‡’ Î¸ = cos-1x

u = sin-1âˆš(1-cos2Î¸)

u = sin-1(sin Î¸) —–(i)

And, v = cos-1x —–(ii)

Now, x âˆˆ (-1,0)

â‡’ cos Î¸ âˆˆ (-1,0)

â‡’ Î¸ âˆˆ (Ï€/2, Ï€)

So, from equation (i),

u = Ï€ – Î¸ [ sin-1(sin Î¸) = Ï€ – Î¸, Î¸ âˆˆ (Ï€/2, 3Ï€/2) ]

u = Ï€ – cos-1x [ d(cos-1x)/dx = -1/âˆš(1-x2) ]

Differentiating above equation with respect to x

du/dx = +1/âˆš(1-x2) —–(iii)

Differentiating equation (ii) with respect to x

dv/dx = -1/âˆš(1-x2) —–(iv)

Dividing equation (iii) by (iv)

du/dv = -1 (Ans)

### Question 5: Differentiate sin-1(4xâˆš(1-4x2)) with respect to âˆš(1-4x2)

(i) x âˆˆ (-1/2, -1/2âˆš2)

(ii) x âˆˆ (1/2âˆš2, 1/2)

Solution:

(i) Let u = sin-1(4xâˆš(1-4x2))

Substitute 2x = cos Î¸ â‡’ Î¸ = cos-1(2x)

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

u = sin-1(2 cos Î¸ sin Î¸)

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

Let, v = âˆš(1-4x2)

dv/dx = 1/(2 * âˆš(1-4x2)) * (-8x) = -4x/âˆš(1-4x2) —–(ii)

Here, x âˆˆ (-1/2,-1/2âˆš2)

â‡’ 2x âˆˆ (-1, -1/âˆš2)

â‡’ Î¸ âˆˆ (Â¾ Ï€, Ï€)

So, from equation (i), u = Ï€ – 2Î¸ [ sin-1(sin Î¸) = Ï€ – Î¸, Î¸ âˆˆ (Ï€/2, 3Ï€/2) ]

u = Ï€ – 2cos-1(2x)

du/dx = 0 – 2* (-1/âˆš(1-4x2)) * 2 = 4/âˆš(1-4x2) —–(iii) [ d(cos-1x)/dx = -1/âˆš(1-x2) ]

Dividing equation (iii) by (ii)

du/dv = -(1/x) (Ans.)

(ii) Let u = sin-1(4xâˆš(1-4x2))

Substitute 2x = cos Î¸ â‡’ Î¸ = cos-1(2x)

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

u = sin-1(2 cos Î¸ sin Î¸)

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

Let, v = âˆš(1-4x2)

dv/dx = 1/(2 * âˆš(1-4x2)) * (-8x) = -4x/âˆš(1-4x2) —–(ii)

Here, x âˆˆ (1/2âˆš2, 1/2)

â‡’ 2x âˆˆ (1/âˆš2, 1)

â‡’ Î¸ âˆˆ (0, Ï€/4)

So, from equation (i), u = 2Î¸ [ sin-1(sin Î¸) = Î¸, Î¸ âˆˆ (-Ï€/2, Ï€/2) ]

u = 2cos-1(2x)

du/dx = 2* (-1/âˆš(1-4x2)) * 2 = -4/âˆš(1-4x2) —–(iii) [ d(cos-1x)/dx = -1/âˆš(1-x2) ]

Dividing equation (iii) by (ii)

du/dv = (1/x) (Ans.)

### Question 6: Differentiate tan-1((âˆš(1+x2)-1)/x) with respect to sin-1(2x/(1+x2)) if -1<x<1.

Solution:

Let u = tan-1((âˆš(1+x2)-1) / x)

Substitute x = tan Î¸ â‡’ Î¸ = tan-1x

u = tan-1((âˆš(1+tan2Î¸)-1) / tan Î¸)

u = tan-1((sec Î¸ -1) / tan Î¸) [sec2Î¸ = 1 + tan2Î¸]

u = tan-1((1 – cos Î¸) / sin Î¸) [sec Î¸ = 1/cos Î¸]

u = tan-1((2sin2(Î¸/2) / 2 sin(Î¸/2) cos(Î¸/2)) [1- cos2Î¸ = 2 sin2Î¸, and sin2Î¸ = 2sinÎ¸cosÎ¸]

u = tan-1((sin(Î¸/2) / cos(Î¸/2))

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

Now, let v = sin-1(2x/1+x2)

v = sin-1(2 tanÎ¸ / (1 + tan2Î¸))

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

Here, -1<x<1

â‡’ -1<tan Î¸ <1

â‡’ – Ï€/4 < Î¸ < Ï€/4

Therefore, from (i), u = Î¸/2 [ tan-1(tan Î¸) = Î¸, Î¸ âˆˆ [ – Ï€/2, Ï€/2] ]

u = 1/2 * tan-1x

Differentiating it with respect to x,

du/dx = Â½*(1 + x2)) —–(iii) [ d(tan-1x)/dx = 1/(1 + x2 )]

From equation (ii), v = 2Î¸ [ sin-1(sin Î¸) = Î¸, Î¸ âˆˆ [ – Ï€/2, Ï€/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(2xâˆš(1-x2)) with respect to sec-1(1/âˆš(1-x2)) if

(i) x âˆˆ (0,1/âˆš2)

(ii) x âˆˆ (1/âˆš2,1)

Solution:

(i) Let u = sin-1(2x âˆš(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 2 Î¸) —–(i)

And, let v = sec-1(1/âˆš(1-x2))

v = sec-1(1/âˆš(1-sin2Î¸))

v = sec-1(1/ cos Î¸) [sin2Î¸ + cos2Î¸ = 1]

v = sec-1(sec Î¸) [sec Î¸ = 1/ cos Î¸]

v = cos-1(1/sec Î¸)

v = cos-1(cos Î¸) —-(ii) [sec-1x=cos-1(1/x)]

Here, x âˆˆ (0,1/âˆš2)

sin Î¸ âˆˆ (0,1/âˆš2)

Î¸ âˆˆ (0, Ï€ / 4)

From equation (i), u = 2Î¸ [ sin-1(sin Î¸) = Î¸, Î¸ âˆˆ [ – Ï€ / 2, Ï€ / 2] ]

u = 2sin-1x

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

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

v = sin-1x

dv/dx = 1/âˆš(1-x2) —-(iv)

Dividing equation (iii) by (iv)

du/dv = 2 (Ans)

(ii) Let u = sin-1(2x âˆš(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 2 Î¸) —–(i)

And, let v = sec-1(1/âˆš(1-x2))

v = sec-1(1/âˆš(1-sin2Î¸))

v = sec-1(1/ cos Î¸) [sin2Î¸ + cos2Î¸ = 1]

v = sec-1(sec Î¸) [sec Î¸ = 1/ cos Î¸]

v = cos-1(1/sec Î¸)

v = cos-1(cos Î¸) —-(ii) [sec-1x=cos-1(1/x)]

Here, x âˆˆ (1/âˆš2, 1)

sin Î¸ âˆˆ (1/âˆš2, 1)

Î¸ âˆˆ (Ï€ / 4, Ï€/ 2)

From equation (i), u = 2Î¸ [ sin-1(sin Î¸) = Î¸, Î¸ âˆˆ [ – Ï€ / 2, Ï€ / 2] ]

u = 2sin-1x

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

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

v = sin-1x

dv/dx = 1/âˆš(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 Î¸ â‡’ Î¸ = tan-1x

u = sin-1(2 tan Î¸ / (1 + tan2Î¸))

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

Let v = cos-1((1 – x2 ) / (1 + x2 ))

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

v = cos-1(cos 2Î¸) —–(ii) [ cos2Î¸ = (1 – tan2Î¸) / (1 + tan2Î¸) ]

Here, 0 < x <1

â‡’ 0 < tan Î¸ < 1

â‡’ 0 < Î¸ < Ï€/4

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

u = 2 tan-1x

Differentiating above equation with respect to x,

du/dx = 2/(1 + x2) —–(iii)

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

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 âˆš(1 + a2x2).

Solution:

Let u = tan-1((1 + ax) / (1 – ax))

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

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

u = tan-1((tan Ï€/4 + tan Î¸) / (1 – tan Ï€/4 * tan Î¸))

u = tan-1(tan (Ï€/4 + Î¸) [ tan(A + B) = (tan A + tan B) / (1 – tan A * tan B) ]

u = Ï€/4 + Î¸

u = Ï€/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 = âˆš(1 + a2x2)

dv/dx = 1/(2*âˆš(1 + a2x2)) * a2 * 2x = a2x / âˆš(1 + a2x2) —–(ii)

Dividing equation (i) by (ii)

du/dv = 1/(ax*âˆš(1 + a2x2)) (Ans)

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