RD Sharma Class 12 Ex 11.8 Solutions Chapter 11 Differentiation

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter11
Exercise11.8
CategoryRD Sharma Solutions

Table of Contents

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 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-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 + x)]

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 – x) / (1 + x))

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 – x)

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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