ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.6

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.6

Question 1.
Find the domain of the differentiability of the following functions:
(i) sin^-1 x
(ii) cos^-1 x
(iii) sec^-1 x
(iv) cosec^-1 x
(v) tan^-1 x
(vi) cot^-1 x
Solution:
(i) Let y = sin^-1 x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}\)
For domain of \(\frac{d y}{d x}\) ;
\(\frac{1}{\sqrt{1-x^2}}\) should be a real number.
⇒ 1 – x² > 0
⇒ x² < 1
⇒ |x| < 1
⇒ – 1 < x < 1
∴ Domain of \(\frac{d y}{d x}\) = (- 1, 1)

(ii) Let y = cos^-1 x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=-\frac{1}{\sqrt{1-x^2}}\)
For domain of \(\frac{d y}{d x}\) ;
\(\frac{d y}{d x}\) must be a real number
⇒ \(-\frac{1}{\sqrt{1-x^2}}\) should be a real number.
⇒ 1 – x² > 0
⇒ x² < 1
⇒ |x| < 1
⇒ – 1 < x < 1
∴ Domain of \(\frac{d y}{d x}\) = (- 1, 1)

(iii) Let y = sec^-1 x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{1}{x \sqrt{x^2-1}}\)
For domain of \(\frac{d y}{d x}\) ;
\(\frac{d y}{d x}\) must be a real number
⇒ \(\frac{1}{x \sqrt{x^2-1}}\) must be a real number.
⇒ x² – 1 > 0 and x ≠ 0
⇒ x² > 1 and x ≠ 0
⇒ |x| > 1 and x ≠ 0
⇒ x > 1 or x < – 1 and x ≠ 0
∴ x ∈ (- ∞, – 1) ∪ (1, ∞)
Thus domain of \(\frac{d y}{d x}\) = (- ∞, – 1) ∪ (1, ∞).

(iv) Let y = cosec^-1 x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=-\frac{1}{|x| \sqrt{x^2-1}}\)
For doamin of \(\frac{d y}{d x}\);
\(\frac{d y}{d x}\) must be a real number
⇒ \(-\frac{1}{|x| \sqrt{x^2-1}}\) must be a real number
⇒ x² – 1 > 0 and |x| ≠ 0
⇒ |x| > 1 and |x| ≠ 0
⇒ x > 1 or x < – 1
⇒ x ∈ (- ∞, – 1) ∪ (1, ∞)
∴ domain of \(\frac{d y}{d x}\) = (- ∞, – 1) ∪ (1, ∞)

(v) Let y = tan^-1 x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{1}{1+x^2}\)
For domain of \(\frac{d y}{d x}\) ;
\(\frac{d y}{d x}\) must be a real number
⇒ \(\frac{1}{1+x^2}\) must be a real number
⇒ 1 + x² ≠ 0, which is true ∀ x ∈ R
Since 1 + x² ≥ 1 > 0
∴ Domain of \(\frac{d y}{d x}\) = R.

(vi) Let y = cot^-1 x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{-1}{1+x^2}\)
For domain of \(\frac{d y}{d x}\) ;
\(\frac{d y}{d x}\) must be a real number
⇒ \(\frac{-1}{1+x^2}\) must be a real number
⇒ 1 + x² ≠ 0 which is true ∀ x ∈ R
[∵ 1 + x² ≥ 1 > 0]
∴ Domain of \(\frac{d y}{d x}\) = R.

Question 2.
Find the derivatives of the following functions :
(i) tan^-1 (x√x)
(ii) cos^-1 3x
(iii) \(\sqrt{\sin ^{-1} 2 x}\)
(iv) sin (tan^-1 x)
(v) cot^-1 √x
(vi) (tan^-1 x)²
Solution:
(i) Let y = tan^-1 (x√x) ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{1}{1+\left(x^{3 / 2}\right)^2} \frac{d}{d x} x^{3 / 2}\)
[∵ \(\frac{d}{d x} \tan ^{-1} x=\frac{1}{1+x^2}\)]
= \(\frac{1}{1+x^3} \times \frac{3}{2} x^{\frac{3}{2}-1}\)
= \(\frac{3 \sqrt{x}}{2\left(1+x^3\right)}\)

(ii) Let y = cos^-1 3x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=-\frac{1}{\sqrt{1-(3 x)^2}} \frac{d}{d x}(3 x)\)
= \(\frac{-3}{1-9 x^2}\)

(iii) Let y = \(\sqrt{\sin ^{-1} 2 x}\)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{1}{2}\left(\sin ^{-1} 2 x\right)^{\frac{1}{2}-1} \frac{d}{d x} \sin ^{-1} 2 x\)
= \(\frac{1}{2 \sqrt{\sin ^{-1} 2 x}} \frac{1}{\sqrt{1-(2 x)^2}} \times 2\)
= \(\frac{1}{\sqrt{\sin ^{-1} 2 x} \sqrt{1-4 x^2}}\)

(iv) Let y = sin (tan^-1 x)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = cos (tan^-1 x) \(\frac{d}{d x}\) tan^-1 x
= cos (tan^-1 x) . \(\frac{1}{1+x^2}\)

(v) Let y = cot^-1 √x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{-1}{1+(\sqrt{x})^2} \frac{d}{d x} x^{1 / 2}\)
= \(-\frac{1}{1+x} \frac{1}{2} x^{-1 / 2}\)
= \(-\frac{1}{2 \sqrt{x}(1+x)}\)

(vi) Let y = (tan^-1 x)² ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 2 tan^-1 x \(\frac{d}{d x}\) (tan^-1 x)
= \(\frac{2 \tan ^{-1} x}{1+x^2}\)

Question 3.
Find the derivative of the following functions :
(i) sec^-1 (cosec x)
(ii) tan^-1 (cot x)
Solution:
(i) Let y = sec^-1 (cosec x)
= sec^-1 (sec (\(\frac{\pi}{2}\) – x))
⇒ y = \(\frac{\pi}{2}\) – x
Differentiating w.r.t x, we have
\(\frac{d y}{d x}\) = 0 – 1
= – 1

(ii) Let y = tan^-1 (cot x)
= tan^-1 (tan (\(\frac{\pi}{2}\) – x))
⇒ y = \(\frac{\pi}{2}\) – x
Differentiating w.r.t x, we get
\(\frac{d y}{d x}\) = 0 – 1
= – 1.

Question 4.
If f(x) = cot^-1 (tan x), show that f'(x) = – 1.
Solution:
Given f(x) = cot^-1 (tan x)
= cot^-1 (cot (\(\frac{\pi}{2}\) – x))
⇒ f(x) = \(\frac{\pi}{2}\) – x
Differentiating both sides w.r.t. x, we have
f'(x) = 0 – 1 = – 1

Question 5.
If y = sin^-1 x + cos^-1 x, find \(\frac{d y}{d x}\).
Solution:
Given y = sin^-1 x + cos^-1 x
= \(\frac{\pi}{2}\)
∴ \(\frac{d y}{d x}\) = 5 (0 – 1) = – 5.

Question 6.
If f(x) = sin^-1 x + sec^-1 \(\frac{1}{x}\), find f'(x).
Solution:
Given f(x) = sin^-1 x + sec^-1 \(\frac{1}{x}\)
= sin^-1 x + cos^-1 x
[∵ cos^-1 x = sec^-1 \(\frac{1}{x}\)]
⇒ f(x) = \(\frac{\pi}{2}\)
[∵ sin^-1 x + cos^-1 x = \(\frac{\pi}{2}\) ∀ x ∈ [- 1, 1]]
Differentiating both sides w.r.t x, we have
f'(x) = 0

Question 7.
If y = tan^-1 x + cot^-1 \(\frac{1}{x}\), x > 0, find \(\frac{d y}{d x}\).
Solution:
Given y = tan^-1 x + cot^-1 \(\frac{1}{x}\)
⇒ y = tan^-1 x + tan^-1 x
[∵ cot^-1 x = tan^-1 x if x > 0]
⇒ y = 2 tan^-1 x ;
Diff. both sides w.r.t x, we have
\(\frac{d y}{d x}\) = \(\frac{2}{1+x^2}\).

Question 8.
If y = sin^-1 (cos x) + 3 cosec^-1 (sec x), find \(\frac{d y}{d x}\).
Solution:
Given y = sin^-1 (cos x) + 3 cosec^-1 (sec x)
= sin^-1 (sin (\(\frac{\pi}{2}\) – x)) + 3 cosec^-1 (cosec (\(\frac{\pi}{2}\) – x))
⇒ y = \(\frac{\pi}{2}\) – x + 3 (\(\frac{\pi}{2}\) x)
⇒ y = 2π – 4x ;
Differentiating both sides w.r.t x, we have
\(\frac{d y}{d x}\) = 0 – 4
= – 4.

Question 9.
If y = 2 cos^-1 (sin x) + 3 cot^-1 (tan x), find \(\frac{d y}{d x}\).
Solution:
Given y = 2 cos^-1 (sin x) + 3 cot^-1 (tan x)
= 2 cos^-1 (cos (\(\frac{\pi}{2}\) – x)) + 3 cot\(\frac{\pi}{2}\) (cot (\(\frac{\pi}{2}\) – x))
= 2 (\(\frac{\pi}{2}\) – x) + 3 (\(\frac{\pi}{2}\) – x)
= 5 (\(\frac{\pi}{2}\) – x)
∴ \(\frac{d y}{d x}\) = 5 (0 – 1) = – 5.

Question 10.
Differentiate the following functions w.r.t. x:
(i) x tan^-1 x
(ii) x cos^-1 √x
(iii) \(\frac{\sin ^{-1} x}{x}\)
Solution:
(i) Let y = x tan^-1 x ;
Diff. both sides w.r.t x, we have
\(\frac{d y}{d x}\) = x \(\frac{d}{d x}\) tan^-1 x + tan^-1 \(\frac{d}{d x}\) x
= x² + tan^-1 x

(ii) Let y = x cos^-1 √x ;
Diff. both sides w.r.t x, we have

(iii) Let y = \(\frac{\sin ^{-1} x}{x}\) ;
Diff. both sides w.r.t x, we have
\(\frac{d y}{d x}=\frac{x \frac{d}{d x} \sin ^{-1} x-\sin ^{-1} x \frac{d}{d x} x}{x^2}\)
= \(\frac{\frac{x}{\sqrt{1-x^2}}-\sin ^{-1} x}{x^2}\)
= \(\frac{x-\sqrt{1-x^2} \sin ^{-1} x}{x^2 \sqrt{1-x^2}}\).

Question 11.
Differentiate the following functions w.r.t. x:
(i) cos^-1 \(\left(\frac{1-x}{1+x}\right)\)
(ii) sec^-1 \(\left(\frac{1}{x-1}\right)\)
(iii) sin^-1 \(\left(\frac{1}{\sqrt{x+1}}\right)\) (NCERT Exampler)
Solution:
(i) Let y = cos^-1 \(\left(\frac{1-x}{1+x}\right)\)
Diff. both sides w.r.t x, we have

(ii) Let y = sec^-1 \(\left(\frac{1}{x-1}\right)\) ;
Differentiating both sides w.r.t x, we have
\(\frac{d y}{d x}=\frac{1}{\frac{1}{x-1} \sqrt{\left(\frac{1}{x-1}\right)^2-1}} \frac{d}{d x}\left(\frac{1}{x-1}\right)\)
= \(\frac{(x-1)^2}{\sqrt{1-(x-1)^2}}\left\{\frac{-1}{(x-1)^2}\right\}\)
= \(\frac{-1}{\sqrt{2 x-x^2}}\)

(iii) Let y = sin^-1 \(\left(\frac{1}{\sqrt{x+1}}\right)\)
Diff. both sides w.r.t x, we get
\(\frac{d y}{d x}=\frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{x+1}}\right)^2}} \frac{d}{d x}\left(\frac{1}{\sqrt{x+1}}\right)\)
= \(\frac{1}{\sqrt{1-\frac{1}{x+1}}} \frac{d}{d x}(x+1)^{-\frac{1}{2}}\)
= \(\frac{\sqrt{x+1}}{\sqrt{x+1-1}}\left(-\frac{1}{2}\right)(x+1)^{-\frac{1}{2}-1} \frac{d}{d x}(x+1)\)
= \(-\frac{1}{2} \frac{(x+1)^{-1}}{\sqrt{x}}\)
= \(-\frac{1}{2 \sqrt{x}(1+x)}\)

Question 12.
(i) sin^-1 x + sin^-1 \(\sqrt{1-x^2}\) (NCERT)
(ii) x (sin^-1 x)² + 2 \(\sqrt{1-x^2}\) sin^-1 x – 2x.
Solution:
(i) Let y = sin^-1 x + sin^-1 \(\sqrt{1-x^2}\)
Diff. both sides w.r.t x, we get

Aliter:
Let y = sin^-1 x + sin^-1 \(\sqrt{1-x^2}\) ; |x| ≤ 1
put x = sin θ
θ = sin^-1 x
∴ y = sin^-1 (sin θ ) + sin^-1 \(\sqrt{1-\sin ^2 \theta}\)
= θ + sin^-1 (cos θ)
⇒ y = θ + sin^-1 [sin (\(\frac{\pi}{2}\) – θ)]
⇒ y = θ + \(\frac{\pi}{2}\) – θ
⇒ y = \(\frac{\pi}{2}\)
Diff. both sides w.r.t x, we get
∴ \(\frac{d y}{d x}\) = 0

(ii) Let y = x (sin^-1 x)² + 2 \(\sqrt{1-x^2}\) sin^-1 x – 2x
Diff. both sides w.r.t x, we have
\(\frac{d y}{d x}\) = \(\frac{d y}{d x}\) {x (sin^-1 x)²} + 2 \(\frac{d}{d x}\) {\(\sqrt{1-x^2}\) sin^-1 x} – 2
= x \(\frac{d}{d x}\) (sin^-1 x)² + (sin^-1 x)² \(\frac{d}{d x}\) (x) + 2 [\(\sqrt{1-x^2}\) \(\frac{d}{d x}\) sin^-1 x + sin^-1 x \(\frac{d}{d x}\) \(\sqrt{1-x^2}\)] – 2
= 2x sin^-1 x \(\frac{d}{d x}\) (sin^-1 x) + (sin^-1)² . 1 + 2 \(\left[\sqrt{1-x^2} \times \frac{1}{\sqrt{1-x^2}}+\frac{\sin ^{-1} x}{2 \sqrt{1-x^2}}(-2 x)\right]\) – 2
= \(\frac{2 x \sin ^{-1} x}{\sqrt{1-x^2}}\) + (sin^-1)² + 2 \(\left[1-\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}\right]\) – 2
= (sin^-1)².

Question 13.
(i) tan^-1 (sin x + cos x)
(ii) tan^-1 \(\left(\frac{\sin x}{1+\cos x}\right)\) (NCERT)
(iii) cot^-1 \(\left(\frac{1+\cos x}{\sin x}\right)\)
(iv) tan^-1 \(\left(\frac{\cos x}{1+\sin x}\right)\)
Solution:
(i) Let y = tan^-1 (sin x + cos x)
Diff. both sides w.r.t x, we have
\(\frac{d y}{d x}=\frac{1}{1+(\sin x+\cos x)^2}\)
\(\frac{d}{d x}(\sin x+\cos x)=\frac{\cos x-\sin x}{1+\sin ^2 x+\cos ^2 x+2 \sin x \cos x}\)
= \(\frac{\cos x-\sin x}{2+\sin 2 x}\)

(ii) Let y = tan^-1 \(\left(\frac{\sin x}{1+\cos x}\right)\)
= tan^-1 \(\left(\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\right)\)
⇒ y = tan^-1 (tan \(\frac{x}{2}\))
= \(\frac{x}{2}\) ;
Diff. bothsides w.r.t x, we have
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2}\)

(iii) Let y = cot^-1 \(\left(\frac{1+\cos x}{\sin x}\right)\)
= cot^-1 \(\left(\frac{2 \cos ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right)\)
= cot^-1 (cot \(\frac{x}{2}\))
⇒ y = \(\frac{x}{2}\) ;
Diff. bothsides w.r.t x, we have
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2}\)

(iv) Let y = tan^-1 \(\left(\frac{\cos x}{1+\sin x}\right)\)
= tan^-1 \(\left[\frac{\sin \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}\right]\)
= tan^-1 \(\left[\frac{2 \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right]\)
⇒ y = tan^-1 [tan (\(\frac{\pi}{4}\) – \(\frac{x}{2}\))
= \(\frac{\pi}{4}\) – \(\frac{x}{2}\)
Diff. both sides w.r.t x, we have
⇒ \(\frac{d y}{d x}\) = – \(\frac{1}{2}\)

Question 14.
(i) cot^-1 (cosec x + cot x)
(ii) tan^-1 \(\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\) (NCERT Exampler)
(iii) cot^-1 \(\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)\)
Solution:
(i) Let y = cot^-1 (cosec x + cot x)
⇒ y = cot^-1 \(\left(\frac{1}{\sin x}+\frac{\cos x}{\sin x}\right)\)
= cot^-1 \(\left(\frac{1+\cos x}{\sin x}\right)\)
⇒ y = cot^-1 \(\left\{\frac{2 \cos ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right\}\)
= cot^-1 (cot \(\frac{x}{2}\))
= \(\frac{x}{2}\)
On differentiating both sides w.r.t x, we have
\(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x}{2}\right)=\frac{1}{2}\)

(ii) Let y = tan^-1 \(\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\)
= tan^-1 \(\left(\sqrt{\frac{2 \sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}}\right)\)
⇒ y = tan^-1 (tan \(\frac{x}{2}\))
= \(\frac{x}{2}\)
On differentiating bothsides w.r.t x, we have
∴ \(\frac{d y}{d x}\) = – \(\frac{1}{2}\)

(iii) Let y = cot^-1 \(\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)\)

Diff. both sides w.r.t x, we have
\(\frac{d y}{d x}\) = 0 + \(\frac{1}{2}\)
= \(\frac{1}{2}\)

Question 15.
(i) If y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\), prove that (1 – x²) \(\frac{d y}{d x}\) – xy = 1. (ISC 2006)
(ii) If y = sin^-1 \(\left(\frac{2 x}{1+x^2}\right)\), prove that (1 – x²) \(\frac{d y}{d x}\) – xy = 2.
Solution:
(i) Given y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\)
⇒ \(\left(\sqrt{1-x^2}\right)\) y = sin^-1 x ……………….(1)
Diff. eqn. (1) w.r.t. x ; we have

⇒ (1 – x²) \(\frac{d y}{d x}\) – xy = 1

(ii) Given y = sin^-1 \(\left(\frac{2 x}{1+x^2}\right)\) ……………..(1)
put x = tan θ
i.e. θ = tan^-1 x
∴ from (1) ; we have
y = sin^-1 \(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\)
= sin^-1 (sin 2θ)
⇒ y = 2θ
= 2 tan^-1 x
Diff. both sides (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = \(\frac{2}{1+x^2}\)
⇒ (1 + x²) \(\frac{d y}{d x}\) = 2

Differentiating the following functions w.r.t. x :

Question 16.
(i) tan^-1 \(\left(\frac{2 x}{1-x^2}\right)\)
(ii) cos^-1 \(\left(\frac{1-x^2}{1+x^2}\right)\) (NCERT)
(iii) cos^-1 (1 – 2x²)
(iv) sin^-1 (2x \(\sqrt{1-x^2}\)) (NCERT)
Solution:
(i) Let y = tan^-1 \(\left(\frac{2 x}{1-x^2}\right)\)
putting x = tan θ
⇒ θ = tan^-1 x
∴ from (1) ;
y = tan^-1 \(\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)\)
= tan^-1 (tan 2θ) = 2θ
⇒ y = 2 tan^-1 x ;
Diff. both sides (1) w.r.t. x ; we get
\(\frac{d y}{d x}=\frac{2}{1+x^2}\)

(ii) Let y = cos^-1 \(\left(\frac{1-x^2}{1+x^2}\right)\)
put x = tan θ
⇒ θ = tan^-1 x
∴ y = cos^-1 \(\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\)
⇒ y = cos^-1 (cos 2θ)
= 2θ
= 2 tan^-1 x
Diff. w.r.t. x, we have

(iii) Let y = cos^-1 (1 – 2x²)
put x = sin θ then θ = sin^-1 x
Then y = cos^-1 (1 – 2 sin² θ)
= cos^-1 (cos 2θ)
= 2θ
⇒ y = 2 sin^-1 x ;
On Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{2}{\sqrt{1-x^2}}\)

(iv) Let y = sin^-1 (2x \(\sqrt{1-x^2}\)) …………..(1)
put x = sin θ
⇒ θ = sin^-1 x
∴ from (1) ;
y = sin^-1 (2 sin θ cos θ)
= sin^-1 (sin 2θ) = 2θ
⇒ y = 2 sin^-1 x ;
Differentiating w.r.t. x ; we get
\(\frac{d y}{d x}=\frac{2}{\sqrt{1-x^2}}\)

Question 17.
(i) sin^-1 (3x – 4x³)
(ii) tan^-1 \(\left(\frac{3 x-x^3}{1-3 x^2}\right) \) (NCERRT Exampler)
(iii) cos^-1 \(\left(\frac{2 x}{1+x^2}\right)\)
(iv) sec^-1 \(\left(\frac{1}{4 x^3-3 x}\right)\) (NCERT Exampler)
Solution:
(i) Let y = sin^-1 (3x – 4x³) …………..(1)
put x = tan θ
⇒ θ = tan^-1 x
∴ from (1) ;
y = sin^-1 (3 sin θ – 4 sin³ θ)
= sin^-1 (sin 3θ)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{3}{\sqrt{1-x^2}}\)

(ii) Let y = tan^-1 \(\left(\frac{3 x-x^3}{1-3 x^2}\right) \)
put x = tan θ
so that θ = tan^-1 x
∴ y = tan^-1 \(\left[\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right]\)
⇒ y = tan^-1 [tan 3θ] = 3θ
⇒ y = 3 tan^-1 x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{3}{1+x^2}\)

(iii) Let y = cos^-1 \(\left(\frac{2 x}{1+x^2}\right)\)
put x = tan θ
⇒ θ = tan^-1 x
∴ y = cos^-1 \(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\)
= cos^-1 (sin 2θ)
= cos^-1 [cos (\(\frac{\pi}{2}\) – 2θ)]
⇒ y = \(\frac{\pi}{2}\) – 2θ
⇒ y = \(\frac{\pi}{2}\) – 2 tan^-1 x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=-\frac{2}{1+x^2}\)

(iv) Let y = sec^-1 \(\left(\frac{1}{4 x^3-3 x}\right)\) …………(1)
put x = cos θ
⇒ θ = cos^-1 x
∴ from (1) ; we have
y = sec^-1 \(\left(\frac{1}{4 \cos ^3 \theta-3 \cos \theta}\right)\)
⇒ y = sec^-1 \(\left(\frac{1}{\cos 3 \theta}\right)\)
= sec^-1 (sec 3θ) = 3θ
⇒ y = 3 cos^-1 x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=-\frac{3}{\sqrt{1-x^2}}\)

Question 18.
(i) sec^-1 \(\left(\frac{x^2+1}{x^2-1}\right)\)
(ii) sin^-1 \(\left(\frac{1-x^2}{1+x^2}\right)\) (NCERT)
(iii) tan^-1 (\(\sqrt{1+x^2}\) – x)
(iv) cot^-1 (\(\sqrt{1+x^2}\) + x)
Solution:
(i) Let y = sec^-1 \(\left(\frac{x^2+1}{x^2-1}\right)\)
put x = tan θ
⇒ θ = tan^-1 x
Then y = sec^-1 \(\left(\frac{\tan ^2 \theta+1}{\tan ^2 \theta-1}\right)\)
= sec^-1 \(\left(-\frac{1}{\cos 2 \theta}\right)\)
[∵ cos 2θ = \(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\)]
= sec^-1 (- sec 2θ)
= sec^-1 (sec (π – 2θ))
= π – 2θ
= π – 2 tan^-1 x
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 0 – \(\frac{2}{1+x^2}\)
= \(-\frac{2}{1+x^2}\)

(ii) Let y = sin^-1 \(\left(\frac{1-x^2}{1+x^2}\right)\) ………..(1)
put x = tan θ
i.e. θ = tan^-1 x in eqn. (1) ; we have
∴ y = sin^-1 \(\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\)
= sin^-1 (cos 2θ)
= sin^-1 (sin (\(\frac{\pi}{2}\) – 2θ))
= \(\frac{\pi}{2}\) – 2θ
⇒ y = \(\frac{\pi}{2}\) – 2 tan^-1 x
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 0 – \(\frac{2}{1+x^2}\)
= \(-\frac{2}{1+x^2}\)

(iii) Let y = tan^-1 (\(\sqrt{1+x^2}\) – x) …………(1)
putting x = cot θ
⇒ θ = cot^-1 x in eqn. (1) ; we have
y = tan^-1 (\(\sqrt{1+\cot ^2 \theta}\) – cot θ)
= tan^-1 (cosec θ – cot θ)
= tan^-1 \(\left(\frac{1-\cos \theta}{\sin \theta}\right)\)
= tan^-1 \(\left\{\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right\}\)
= tan^-1 (tan \(\frac{\theta}{2}\))
∴ y = \(\frac{\theta}{2}\)
= \(\frac{1}{2}\) cot^-1 x
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = – \(\frac{1}{2\left(1+x^2\right)}\)

(iv) Let y = cot^-1 (\(\sqrt{1+x^2}\) + x)
putting x = cot θ
⇒ θ = cot^-1 x in eqn. (1) ; we have
∴ y = cot^-1 (\(\sqrt{1+\cot ^2 \theta}\) + cot θ)
= cot^-1 (cosec θ + cot θ)
⇒ y = cot^-1 \(\left(\frac{1+\cos \theta}{\sin \theta}\right)\)
= cot^-1 \(\left(\frac{2 \cos ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)\)
⇒ y = cot^-1 (cot \(\frac{\theta}{2}\))
= \(\frac{\theta}{2}\)
= \(\frac{1}{2}\) cot^-1 x
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = – \(\frac{1}{2\left(1+x^2\right)}\)

Question 19.
(i) tan^-1 \(\left(\frac{x}{\sqrt{a^2-x^2}}\right)\)
(ii) sin^-1 \(\left(\frac{1}{\sqrt{1+x^2}}\right)\)
(iii) cos^-1 \(\left(\frac{x}{\sqrt{a^2+x^2}}\right)\)
(iv) sin^-1 \(\left(\sqrt{\frac{1+x}{2}}\right)\)
(v) tan^-1 \(\sqrt{\frac{a-x}{a+x}}\) (ISC 2010)
Solution:
(i) Let y = tan^-1 \(\left(\frac{x}{\sqrt{a^2-x^2}}\right)\) ;
put x = a sin θ
⇒ θ = sin^-1 \(\frac{x}{a}\)
∴ y = tan^-1 \(\left[\frac{a \sin \theta}{\sqrt{a^2-a^2 \sin ^2 \theta}}\right]\)
= tan^-1 \(\left[\frac{a \sin \theta}{a \cos \theta}\right]\)
= tan^-1 (tan θ) = θ
⇒ y = sin^-1 \(\frac{x}{a}\) ;
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}=\frac{1}{\sqrt{1-\frac{x^2}{a^2}}} \frac{1}{a}\)
= \(\frac{1}{\sqrt{a^2-x^2}}\)

(ii) Let y = sin^-1 \(\left(\frac{1}{\sqrt{1+x^2}}\right)\)
put x = tan θ
so that θ = tan^-1 x
∴ y = sin^-1 \(\left(\frac{1}{\sqrt{1+\tan ^2 \theta}}\right)\)
= sin^-1 \(\left(\frac{1}{\sec \theta}\right)\)
= sin^-1 (cos θ)
⇒ y = sin^-1 [sin (\(\frac{\pi}{2}\) – θ)
= \(\frac{\pi}{2}\) – θ
⇒ y = \(\frac{\pi}{2}\) – tan^-1 x
Diff. w.r.t. x ; we have
\(\frac{d y}{d x}\) = – \(\frac{1}{1+x^2}\).

(iii) Let y = cos^-1 \(\left(\sqrt{\frac{1+x}{2}}\right)\) ; |x| < 1
put x = cos θ
⇒ θ = cos^-1 x
∴ y = cos^-1 \(\left(\sqrt{\frac{1+\cos \theta}{2}}\right)\)
= cos^-1 \(\left(\sqrt{\frac{2 \cos ^2 \frac{\theta}{2}}{2}}\right)\)
= cos^-1 (cos \(\frac{\theta}{2}\))
= \(\frac{\theta}{2}\)
⇒ y = \(\frac{1}{2}\) cos^-1 x
⇒ \(\frac{d y}{d x}\) = \(\frac{-1}{2 \sqrt{1-x^2}}\)

(iv) Let y = sin^-1 \(\left(\frac{x}{\sqrt{x^2+a^2}}\right)\)
putting x = a tan θ
θ = tan^-1 \(\frac{x}{a}\) in eqn. (1) ; we have
⇒ y = sin^-1 (sin θ)
= θ
= tan^-1 \(\frac{x}{a}\)
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}=\frac{1}{1+\frac{x^2}{a^2}} \frac{d}{d x}\left(\frac{x}{a}\right)\)
= \(\frac{a^2}{x^2+a^2} \times \frac{1}{a}\)
= \(\frac{a}{a^2+x^2}\)

(v) Let y = tan^-1 \(\sqrt{\frac{a-x}{a+x}}\)
putting x = a cos θ
θ = cos^-1 \(\frac{x}{a}\)
∴ from (1) ; we have
y = tan^-1 \(\sqrt{\frac{a-a \cos \theta}{a+a \cos \theta}}\)
= tan^-1 \(\left(\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\right)\)
= tan^-1 \(\left(\sqrt{\frac{2 \sin ^2 \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}}\right)\)
= tan^-1 (tan \(\left(\frac{\theta}{2}\right)\))
= \(\left(\frac{\theta}{2}\right)\)
⇒ y = \(\frac{1}{2}\) cos^-1 \(\frac{x}{a}\) ;
differentiating w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{2}\left(-\frac{1}{\sqrt{1-\frac{x^2}{a^2}}}\right) \cdot \frac{1}{a}\)
= – \(\frac{1}{2 \sqrt{a^2-x^2}}\)

Question 20.
(i) sin^-1 \(\left(\sqrt{\frac{1+x^2}{2}}\right)\)
(ii) tan^-1 \(\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\)
(iii) sin² (cot^-1 \(\sqrt{\frac{1+x}{1-x}}\))
(iv) sin^-1 \(\left(\frac{2 \sec x}{1+\sec ^2 x}\right)\)
Solution:
(i) Let y = sin^-1 \(\left(\sqrt{\frac{1+x^2}{2}}\right)\)
put x² = cos θ
⇒ θ = cos^-1 x²
⇒ y = sin^-1 \(\left(\sqrt{\frac{2 \cos ^2 \frac{\theta}{2}}{2}}\right)\)
= sin^-1 (cos \(\frac{\theta}{2}\))
⇒ y = sin^-1 [sin (\(\frac{\pi}{2}\) – \(\frac{\theta}{2}\))
= \(\frac{\pi}{2}\) – \(\frac{1}{2}\) cos^-1 x²
Diff. bothsides w.r.t. x,
\(\frac{d y}{d x}=-\frac{1}{2} \frac{-1}{\sqrt{1-\left(x^2\right)^2}} \times 2 x\)
= \(\frac{x}{\sqrt{1-x^4}}\)

(ii) Given y = tan^-1 \(\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}\) ;
putting x = cos θ ; we get
⇒ y = tan^-1 \(\left\{\frac{\sqrt{1+\cos \theta}-\sqrt{1-\cos \theta}}{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}\right\}\)
= tan^-1 \(\left\{\frac{\sqrt{2} \cos \theta / 2-\sqrt{2} \sin \theta / 2}{\sqrt{2} \cos \theta / 2+\sqrt{2} \sin \theta / 2}\right\}\)
= tan^-1 \(\left\{\frac{1-\tan \theta / 2}{1+\tan \theta / 2}\right\}\)
= tan^-1 [tan \(\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\)]
= \(\frac{\pi}{2}\) – \(\frac{\theta}{2}\)
= \(\frac{\pi}{2}\) – \(\frac{1}{2}\) cos^-1 x
[∵ x = cos θ
⇒ θ = cos^-1 x]
Diff. both sides w.r.t. x, we have
∴ \(\frac{d y}{d x}\) = 0 + \(\frac{1}{2} \frac{1}{\sqrt{1-x^2}}\)

(iii) Let y = sin² (cot^-1 \(\sqrt{\frac{1+x}{1-x}}\))
put x = cos 2t
i.e. \(\frac{1}{2}\) cos^-1 x = t
Then y = sin² (cot^-1 \(\sqrt{\frac{1+\cos 2 t}{1-\cos 2 t}}\))
= sin² (cot^-1 \(\sqrt{\frac{2 \cos ^2 t}{2 \sin ^2 t}}\))
⇒ y = sin² (cot^-1 (cot t))
= sin² t
⇒ y = \(\frac{1-\cos 2 t}{2}\)
= \(\frac{1-x}{2}\) ;
Differentiating both sides w.r.t. x, we have
∴ \(\frac{d y}{d x}\) = \(\frac{1}{2}\) (0 – 1) = \(\frac{1}{2}\)

(iv) Let y = sin^-1 \(\left(\frac{2 \sec x}{1+\sec ^2 x}\right)\)
put sec x = tan t
i.e. t = tan^-1 (sec x)
Then y = sin^-1 \(\left(\frac{2 \tan t}{1+\tan ^2 t}\right)\)
⇒ y = sin^-1 (sin 2t) = 2t
⇒ y = 2 tan^-1 (sec x)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{2}{1+\sec ^2 x} \frac{d}{d x}(\sec x)\)
∴ \(\frac{d y}{d x}=\frac{2 \sec x \tan x}{1+\sec ^2 x}\).

Question 21.
(i) cos^-1 \(\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)\) (NCERT Exampler)
(ii) cos^-1 \(\left(\frac{3 \cos x-4 \sin x}{5}\right)\)
(iii) sin^-1 \(\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)\)
(iv) cos^-1 \(\left(\frac{3 x+4 \sqrt{1-x^2}}{5}\right)\)
Solution:
(i) Let y = cos^-1 \(\left\{\frac{\cos x+\sin x}{\sqrt{2}}\right\}\)
⇒ y = cos^-1 {cos \(\frac{\pi}{4}\) cos x + sin \(\frac{\pi}{4}\) sin x}
⇒ y = cos^-1 {cos (\(\frac{\pi}{4}\) – x)}
= \(\frac{\pi}{4}\) – x
Diff. both sides w.r.t. x ; we have
∴ \(\frac{d y}{d x}\) = 0 – 1 = – 1.

(ii) Let y = cos^-1 \(\left(\frac{3 \cos x-4 \sin x}{5}\right)\)
put \(\frac{3}{5}\) = r cos α ………….(1)
\(\frac{4}{5}\) = r sin α ………….(2)
On squaring and adding (1) and (2) ; we have
r² = \(\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2\)
= \(\frac{9+16}{25}\) = 1
⇒ r = 1
On dividing eqn. (2) by eqn. (1) ; we have
tan α = \(\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\)
α = tan^-1 \(\frac{4}{3}\)
Then y = cos^-1 (r cos α cos x – r sin α sin x)
= cos^-1 (r cos (x + α)) [∵ r = 1]
⇒ y = cos^-1 (cos (x + α))
⇒ y = x + α
= x + tan^-1 \(\frac{4}{3}\)
On differentiating both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 1 + 0 = 1

(iii) Let y = sin^-1 \(\left\{\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right\}\) ;
putting x = sin θ ; we get
Thus y = sin^-1 \(\left\{\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right\}\)
= sin^-1 {sin θ cos \(\frac{\pi}{4}\) + cos θ sin \(\frac{\pi}{4}\)}
⇒ y = sin^-1 {sin (\(\frac{\pi}{4}\) + θ)}
= \(\frac{\pi}{4}\) + θ
∴ y = \(\frac{\pi}{4}\) + sin^-1 x
[∵ sin θ = x
⇒ θ = sin^-1 x]
Diff. both sides w.r.t. x ; we have
⇒ \(\frac{d y}{d x}\) = 0 + \(\frac{1}{\sqrt{1-x^2}}\)
= \(\frac{1}{\sqrt{1-x^2}}\)

(iv) Given, y = cos^-1 \(\left(\frac{3 x+4 \sqrt{1-x^2}}{5}\right)\)
= cos^-1 \(\left(\frac{3}{5} x+\frac{4}{5} \sqrt{1-x^2}\right)\)
= cos^-1 \(\left(x \cdot \frac{3}{5}+\sqrt{1-\left(\frac{3}{5}\right)^2} \sqrt{1-x^2}\right)\)
= cos^-1 x – cos^-1 \(\frac{3}{5}\)
[∵ cos^-1 x – cos^-1 y = cos^-1 [xy + \(\sqrt{1-x^2} \sqrt{1-y^2}\)]]
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = – \(\frac{1}{\sqrt{1-x^2}}\)

Question 22.
(i) tan^-1 \(\left(\frac{4 \sqrt{x}}{1-4 x}\right)\)
(ii) tan^-1 \(\left(\frac{x^{1 / 3}+a^{1 / 3}}{1-x^{1 / 3} a^{1 / 3}}\right)\)
(iii) sin (2 sin^-1 x)
(iv) cot^-1 \(\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)\)
Solution:
(i) Let y = tan^-1 \(\left(\frac{4 \sqrt{x}}{1-4 x}\right)\)
= tan^-1 \(\left(\frac{2 \times 2 \sqrt{x}}{1-(2 \sqrt{x})^2}\right)\)
= 1
put 2√x = tan t
t = tan^-1 (2√x)
= tan^-1 \(\left(\frac{2 \tan t}{1-\tan ^2 t}\right)\)
= tan^-1 (tan 2t)
= 2t
= 2 tan^-1 (2√x)
∴ \(\frac{d y}{d x}\) = \(\frac{2}{1+(2 \sqrt{x})^2} \frac{d}{d x}(2 \sqrt{x})\)
= \(\frac{2}{1+4 x} \times \frac{2}{2 \sqrt{x}}\)
= \(\frac{2}{\sqrt{x}(1+4 x)}\)

(ii) Let y = tan^-1 \(\left[\frac{x^{1 / 3}+a^{1 / 3}}{1-x^{1 / 3} a^{1 / 3}}\right]\)
= tan^-1 x\(\frac{1}{3}\) + tan^-1 a\(\frac{1}{3}\) ……….(1)
[∵ tan^-1 x + tan^-1 y = tan^-1 \(\left(\frac{x+y}{1-x y}\right)\)]
Diff. both sides w.r.t. x, we have
∴ \(\frac{d y}{d x}=\frac{1}{1+\left(x^{1 / 3}\right)^2} \frac{d}{d x}\left(x^{1 / 3}\right)\)
= \(\frac{1}{1+x^{2 / 3}} \frac{1}{3} x^{-2 / 3}\)
= \(\frac{1}{3 x^{2 / 3}\left(1+x^{2 / 3}\right)}\)

(iii) Let y = sin (2 sin^-1 x) ………….(1)
put sin^-1 x = t
⇒ x = sin t
∴ from (1) ;
y = sin (2t) = 2 sin t cos t
= 2x \(\sqrt{1-x^2}\)
On differentiating w.r.t. x ; we have
\(\frac{d y}{d x}=2\left[\sqrt{1-x^2} \frac{d}{d x}(x)+x \frac{d}{d x}\left(1-x^2\right)^{1 / 2}\right]\)
= 2 \(\left[\sqrt{1-x^2} \cdot 1+\frac{x}{2}\left(1-x^2\right)^{-\frac{1}{2}}(-2 x)\right]\)
= 2 \(\left[\sqrt{1-x^2}-\frac{x^2}{\sqrt{1-x^2}}\right]\)
= \(\frac{2\left(1-x^2-x^2\right)}{\sqrt{1-x^2}}=\frac{2\left(1-2 x^2\right)}{\sqrt{1-x^2}}\)

(iv) Given y = cot^-1 \(\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}\) ………………..(1)

Differentiate both sides w.r.t. x, we have
∴ \(\frac{d y}{d x}\) = \(\frac{1}{2}\) and it is independent of x.

Question 22 (old).
(i) sin^-1 \(\left(\frac{6 x-4 \sqrt{1-4 x^2}}{5}\right)\)
Solution:
Let y = sin^-1 \(\left(\frac{6 x-4 \sqrt{1-4 x^2}}{5}\right)\)
putting 2x = sin θ
⇒ θ = sin^-1 2x
∴ from (1) ;
⇒ y = sin^-1 \(\left(\frac{3 \sin \theta-4 \cos \theta}{5}\right)\) …………(2)
putting \(\frac{3}{5}\) = r cos α ………….(3)
and \(\frac{4}{5}\) = r sin α ……………(4)
On squaring and adding (3) and (4) ; we have
r² = \(\frac{9+16}{25}\) = 1
⇒ r = 1
On dividing eqn. (4) by eqn. (3) ; we have
tan α = \(\frac{4}{3}\)
⇒ α = tan^-1 \(\frac{4}{3}\)
∴ from (2) ;
y = sin^-1 {r sin θ cos α – r sin α cos θ}
⇒ y = sin^-1 {sin (θ + α)}
= θ + α
⇒ y = sin^-1 2x + tan^-1 \(\frac{4}{3}\)
On differentiating w.r.t. x, we have
\(\frac{d y}{d x}=\frac{1}{\sqrt{1-4 x^2}} \frac{d}{d x}(2 x)+0\)
= \(\frac{2}{\sqrt{1-4 x^2}}\)