Students appreciate clear and concise ISC Maths Class 12 Solutions Chapter 5 Continuity and Differentiability Ex 5.10 that guide them through exercises.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.10

Find $$\frac{d y}{d x}$$ in the following (1 to 6) questions when :

Question 1.
(i) x = at2, y = 2 at (NCERT)
(ii) x = 2at2, y = at4 (NCERT)
Solution:
(i) Given x =at2 …………(1)
and y = 2at
Differentiate (1) and (2) w.r.t. ‘t’, we have
$$\frac{d x}{d t}$$ = 2at
and $$\frac{d y}{d t}$$ = 2a
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{2 a}{2 a t}=\frac{1}{t}$$, t ≠ 0

(ii) Given, x = 2at2 ………….(1)
and y = at4 ………….(2)
Differentiate (1) and (2) w.r.t. ‘t’, we have
$$\frac{d x}{d t}$$ = 4at
and $$\frac{d y}{d t}$$ = 4at3
∴ $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$$
= $$\frac{4 a t^3}{4 a t}$$
= t2

Question 2.
(i) x = a cos θ, y = b sin θ (NCERT)
(ii) x = a cos θ, y = a sin θ (NCERT)
Solution:
(i) Given, x = a cos θ ;
y = b sin θ
Differentiating both sides w.r.t. θ ; we get
∴ $$\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$$
= $$\frac{b \cos \theta}{-a \sin \theta}$$
= – $$\frac{b}{a}$$ cot θ

(ii) Given, x = a cos θ ………..(1)
and y = a sin θ ………….(2)
Diff. (1) and (2) w.r.t. ‘θ’, we have
$$\frac{d x}{d \theta}$$ = – a sin θ
and $$\frac{d y}{d \theta}$$ = a cos θ
∴ $$\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$$
= $$\frac{a \cos \theta}{-a \sin \theta}$$
= – cot θ.

Question 3.
(i) x = a sec θ, y = b tan θ (NCERT)
(ii) x = sin t, y = cos 2t (NCERT)
Solution:
(i) Given x = a sec θ
and y = b tan θ
Diff. both given eqn. w.r.t. θ, we have
∴ $$\frac{d x}{d \theta}$$ = a sec θ tan θ
and $$\frac{d y}{d \theta}$$ = b sec2 θ
Thus $$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$$
= $$\frac{b \sec ^2 \theta}{a \sec \theta \tan \theta}$$
= $$\frac{b}{a}$$ cosec θ

(ii) Given x = sin t
and y = cos 2t ………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. t, we have
$$\frac{d x}{d t}$$ = cos t
$$\frac{d y}{d t}$$ = – sin 2t . 2
Thus, $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{-2 \sin 2 t}{\cos t}$$
= $$\frac{-4 \sin t \cos t}{\cos t}$$
= – 4 sin t

Question 4.
(i) x = 4t, y = (NCERT)
(ii) x = t + $$\frac{1}{t}$$, y = t – $$\frac{1}{t}$$ (NCERT Exampler)
(iii) x = a (t – sin t), y = a (1 + cos t)
Solution:
(i) Given x = 4t ……….(1)
and y = 4/t ………..(2)
Diff. (1) and (2) w.r.t. ‘t’, we have
$$\frac{d x}{d t}$$ = 4
and $$\frac{d y}{d t}$$ = – 4/t2
∴ $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$$
= $$\frac{-\frac{4}{t^2}}{4}$$
= – $$\frac{1}{t^2}$$

(ii) Given x = (t + $$\frac{1}{t}$$)
and y = (t – $$\frac{1}{t}$$) ……………(1)
Differentiating both sides of eqn. (1) w.r.t. x ; we have
$$\frac{d x}{d t}=\left(1-\frac{1}{t^2}\right)$$ ;
$$\frac{d y}{d t}=\left(1+\frac{1}{t^2}\right)$$
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{\left(t^2+1\right)}{\left(t^2-1\right)}$$
= $$\frac{\left(t+\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)}=\frac{x}{y}$$ [using eqn. (1)]

(iii) x = a (t – sin t)
and y = a (1 + cos t)
∴ $$\frac{d x}{d t}$$ = a (1 – cos t)
and $$\frac{d y}{d t}$$ = – a sin t
Thus $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$$
= $$\frac{-a \sin t}{a(1-\cos t)}$$
= $$\frac{-2 a \sin \frac{t}{2} \cos \frac{t}{2}}{2 a \sin ^2 \frac{t}{2}}$$
= – cot $$\frac{t}{2}$$

Question 5.
(i) x = a cos3 t, y = b sin t
(ii) x = a (1 – sin t), y = a (1 + cos t)
Solution:
Given x = a cos3 t ………..(1)
and y = b sin3 t ………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. t, we have
$$\frac{d x}{d t}$$ = 3a cos2 t (- sin t)
and $$\frac{d y}{d t}$$ = 3b sin2 t (cos t)
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{3 b \sin ^2 t \cos t}{3 a \cos ^2 t \sin t}$$
= – $$\frac{b}{a}$$ tan t

(ii) Given x = a (1 – sin t) ……….(1)
and y = a (1 + cos t) ………..(2)
Diff. both eqns. w.r.t. t, we have
$$\frac{d x}{d t}$$ = – a cos t ;
$$\frac{d y}{d t}$$ = – a sin t
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{-a \sin t}{-a \cos t}$$
= tan t.

Question 5 (old).
(i) x = a sin3 t, y = a cos3 t (ISC 2004)
Solution:
Given x = a sin3 t
and y = a cos3 t
Diff. eqn. (1) and eqn. (2) w.r.t. ‘t’, we have
$$\frac{d x}{d t}$$ = 3a sin2 t cos t ;
$$\frac{d y}{d t}$$ = 3a cos2 t (- sin t)
Thus, $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{-3 a \cos ^2 t \sin t}{3 a \sin ^2 t \cos t}$$
= – cos t

Question 6.
(i) x = a (θ + sin θ), y = a (1 – cos θ)
(ii) x = a (1 – cos θ), y = (θ + sin θ)
Solution:
(i) Given x = a (θ + sin θ) ………..(1)
and y = a (1 – cos θ) …………..(2)
Diff. eqn. (1) and (2) w.r.t. θ, we have
$$\frac{d x}{d \theta}$$ = a (1 + cos θ)
$$\frac{d x}{d \theta}$$ = a sin θ
Thus, $$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$$
= $$\frac{a \sin \theta}{a(1+\cos \theta)}$$
= $$\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}$$
= tan $$\frac{\theta}{2}$$

(ii) Given x = a (1 – cos θ) ;
y = a (θ + sin θ)
Diff. both sides w.r.t. θ ; we get
$$\frac{d x}{d \theta}$$ = a sin θ;
$$\frac{d x}{d \theta}$$ = a (1 + cos θ)
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$$
= $$\frac{a(1+\cos \theta)}{a \sin \theta}$$
= $$\frac{1+\cos \theta}{sin \theta}$$
= $$\frac{2 \cos ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$$
= 2 cot $$\frac{\theta}{2}$$

Question 7.
(i) x = et (sin t + cos t), y = et (sin t – cos t)
(ii) x = cos θ – cos 2θ, y = sin θ – sin 2θ (NCERT)
Solution:
(i) Given x = et (sin t + cos t)
and y = et (sin t – cos t)
Diff. both eqn.’s w.r.t. t, we have
$$\frac{d x}{d t}$$ = et (cos t – sin t) + (sin t + cos t) et
= et (2 cos t)
and $$\frac{d y}{d t}$$ = et (cos t + sin t) + (sin t – cos t) et
= et (2 sin t)
∴ $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$$
= $$\frac{2 e^t \sin t}{2 e^t \cos t}$$
= tan t

(ii) Given x = cos θ – cos 2θ ……….(1)
and y = sin θ – sin 2θ ………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. θ, we have
$$\frac{d x}{d \theta}$$ = – sin θ + 2 sin 2θ
and $$\frac{d y}{d \theta}$$ = cos θ – 2 cos 2θ
∴ $$\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$$
= $$\frac{\cos \theta-2 \cos 2 \theta}{-2 \sin \theta+2 \sin 2 \theta}$$

Question 8.
(i) x = a (cos θ + cos 2θ), y = b (sin θ + sin 2θ)
(ii) x = $$\frac{1+\log t}{t^2}$$, y = $$\frac{3+ 2\log t}{t^2}$$ (NCERT Exampler)
Solution:
Given x = a (cos θ + cos 2θ) …………(1)
and y = b (sin θ + sin 2θ) ………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. θ, we have
$$\frac{d x}{d \theta}$$ = a (- sin θ – 2 sin 2θ) ……….(3)
and $$\frac{d y}{d \theta}$$ = b (cos θ + 2 cos 2θ) …………..(4)
∴ $$\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$$
= $$\frac{b(\cos \theta+2 \cos 2 \theta)}{a(-\sin \theta-2 \sin 2 \theta)}$$
= $$-\frac{b(\cos \theta+2 \cos 2 \theta)}{a(\sin \theta+2 \sin 2 \theta)}$$
[using eqn. (3) and eqn. (4)]

(iv) Given x = $$\frac{1+\log t}{t^2}$$ ……….(1)
and y = $$\frac{3+ 2\log t}{t^2}$$ …………..(2)
Differentiate eqn. (1) and (2) both sides w.r.t. ‘t’ ; we have

Question 9.
x = 3 cos θ – 2 cos3 θ, y = 3 sin θ – 2 sin3 θ
Solution:
Given x = 3 cos θ – 2 cos3 θ
and y = 3 sin θ – 2 sin3 θ
Diff. both given eqn’s w.r.t. θ, we have
$$\frac{d x}{d \theta}$$ = – 3 sin θ – 6 cos2 θ (- sin θ)
= – 3 sin θ + 6 cos2 θ sin θ ………..(1)
and $$\frac{d y}{d \theta}$$ = 3 cos θ – 6 sin2 θ cos 2 …………….(2)
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$$
= $$\frac{3 \cos \theta-6 \sin ^2 \theta \cos \theta}{-3 \sin \theta+6 \cos ^2 \theta \sin \theta}$$
[using (1) and (2)]
= $$\frac{3 \cos \theta\left[1-2 \sin ^2 \theta\right]}{-3 \sin \theta\left[1-2 \cos ^2 \theta\right]}$$
= $$\frac{3 \cos \theta\left(1-2 \sin ^2 \theta\right)}{-3 \sin \theta\left[1-2\left(1-\sin ^2 \theta\right)\right]}$$
= $$\frac{3 \cos \theta\left(1-2 \sin ^2 \theta\right)}{3 \sin \theta\left(1-2 \sin ^2 \theta\right)}$$
= cos θ.

Question 10.
(i) If x = 2 cos t – cos 2t and y = 2 sin t – sin 2t, find $$\frac{d y}{d x}$$ at t = $$\frac{\pi}{4}$$.
(ii) If x = 3 sin t – sin 3t, y = 3 cos t – cos 3t, find $$\frac{d y}{d x}$$ at t = $$\frac{\pi}{3}$$. (NCERT Exampler)
Solution:
(i) Given x = 2 cos t – cos 2t
and y = 2 sin t – sin 2t
Diff. both sides w.r.t. ‘t’ ; we have
$$\frac{d x}{d t}$$ = – 2 sin t + 2 sin 2t ;
$$\frac{d y}{d t}$$ = 2 cos t – 2 cos 2t
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{2(\cos t-\cos 2 t)}{2(\sin 2 t-\sin t)}$$
at t = $$\frac{\pi}{4}$$;
$$\frac{d y}{d x}=\frac{\cos \left(\frac{\pi}{4}\right)-\cos \frac{\pi}{2}}{\sin \frac{\pi}{2}-\sin \frac{\pi}{4}}$$
= $$\frac{\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}$$
= $$\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}$$
= √2 + 1

(ii) Given x = 3 sin t – sin 2t
= 3 sin t – (3 sin t – 4 sin3 t)
⇒ x = 4 sin3 t …….(1)
and y = 3 cos t – cos 3t ………..(2)
Diff. eqn. (1) and (2) w.r.t. t ; we get
$$\frac{d x}{d t}$$ = 12 sin2 t cos t
$$\frac{d y}{d t}$$ = – 3 sin t + 3 sin 3t
∴ $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$$
= $$\frac{-\sin t+\sin 3 t}{4 \sin ^2 t \cos t}$$
at t = $$\frac{\pi}{3}$$ ;
$$\frac{d y}{d x}=\frac{-\frac{\sqrt{3}}{2}+0}{4 \times \frac{3}{4} \times \frac{1}{2}}$$
= $$\frac{-\sqrt{3} \times 2}{2 \times 3}$$
= $$-\frac{1}{\sqrt{3}}$$

Question 11.
(i) If x = $$\frac{2 b t}{1+t^2}$$ and y = $$\frac{a\left(1-t^2\right)}{1+t^2}$$, find $$\frac{d y}{d x}$$ at t = 2.
(ii) If x = aeθ (sin θ – cos θ) and y = aeθ (sin θ + cos θ), find $$\frac{d y}{d x}$$ at θ = $$\frac{\pi}{4}$$.
Solution:
Given x = $$\frac{2 b t}{1+t^2}$$
and y = $$\frac{a\left(1-t^2\right)}{1+t^2}$$
Diff. both equations w.r.t. t ; we have

(ii) Given x = a eθ (sin θ – cos θ) ……………..(1)
and y = a eθ (sin θ + cos θ)
diff. eqn. (1) w.r.t. θ ; we have
$$\frac{d x}{d \theta}$$ = a eθ (sin θ – cos θ) + a eθ (cos θ + sin θ)
= a eθ (sin θ – cos θ + cos θ + sin θ)
= 2aeθ sin θ
and y = aeθ (sin θ + cos θ) ………….(2)
diff. eqn. (2) w.r.t. θ ; we have
$$\frac{d y}{d \theta}$$ = aeθ (sin θ + cos θ) + aeθ (cos θ – sin θ)
= aeθ (sin θ + cos θ + cos θ – sin θ)
= 2aeθ cos θ
∴ $$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$$
= $$\frac{2 a e^\theta \cos \theta}{2 a e^\theta \sin \theta}$$
= cot θ
at θ = $$\frac{\pi}{4}$$ ;
$$\frac{d y}{d x}$$ = cot $$\frac{\pi}{4}$$ = 1.

Question 12.
(i) $$\frac{x^2}{1-x^2}$$ w.r.t. x2
(ii) sin x2 w.r.t. x3 (ISC 2009)
(iii) cot3 (2x + 1) w.r.t. x2 + 1
(iv) sin2 x w.r.t. ecos x
Solution:
(i) Let y = $$\frac{x^2}{1-x^2}$$
and z = x2
So we want to differentiate y w.r.t. i.e. we want to find $$\frac{d y}{d z}$$
Diff. given eqns w.r.t. x, we have
$$\frac{d y}{d x}=\frac{\left(1-x^2\right) 2 x-x^2(-2 x)}{\left(1-x^2\right)^2}$$
= $$\frac{2 x}{\left(1-x^2\right)^2}$$
and $$\frac{d z}{d x}$$ = 2x
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{2 x}{\frac{\left(1-x^2\right)^2}{2 x}}$$
= $$\frac{1}{\left(1-x^2\right)^2}$$

(ii) Let y = sin x2
and z = x3
So we want to diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$
Diff. both eqn’s w.r.t. x : we have
$$\frac{d y}{d x}$$ = cos x2 . 2x ;
$$\frac{d z}{d x}$$ = 3x2
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{2 x \cos x^2}{3 x^2}$$
= $$\frac{2}{3 x}$$ cos x2

(iii) Let y = cot3 (2x + 1)
and z = x2 + 1
So, we want to diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$.
Diff. given eqn’s both sides w.r.t. x, we have
$$\frac{d y}{d x}$$ = 3 cot2 (2x + 1) {- cosec2 (2x + 1)} . 2
and $$\frac{d z}{d x}$$ = 2x
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{-6 \cot ^2(2 x+1) \ {cosec}^2(2 x+1)}{2 x}$$
= $$\frac{-3 \cot ^2(2 x+1) \ {cosec}^2(2 x+1)}{x}$$

(iv) Let y = sin2 x ………….(1)
and z = ecos x
We want to diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$
diff. eqn. (1) and eqn. (2) w.r.t. x ;
$$\frac{d y}{d x}$$ = 2 sin x cos x ;
$$\frac{d z}{d x}$$ = ecos x (- sin x)
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{2 \sin x \cos x}{e^{\cos x}(-\sin x)}$$
= – $$\frac{2 \cos x}{2^{\cos x}}$$

Question 8 (old).
(iv) log (sin x) w.r.t. $$\sqrt{cos x}$$
Solution:
Given, y = log (sin x)
and z = $$\sqrt{cos x}$$
i.e. we want diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$
Now, diff. both eqn’s w.r.t. x, we have
$$\frac{d y}{d x}$$ = $$\frac{1}{sin x}$$ cos x
= cot x
and $$\frac{d z}{d x}$$ = $$\frac{1}{2}$$ (cos x)$$-\frac{1}{2}$$ .(- sin x)
= – $$\frac{\sin x}{2 \sqrt{\cos x}}$$
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{2 \cot x}{-\sin x}$$ $$\sqrt{cos x}$$
= – 2 cot x cosec x $$\sqrt{cos x}$$

Question 13.
(i) Differentiate sin-1 $$\left(\frac{2 x}{1+x^2}\right)$$ w.r.t. tan-1 x.
(ii) Differentiate tan-1 $$\left(\frac{2 x}{1-x^2}\right)$$ w.r.t. tan-1 x.
Solution:
(i) Let y = sin-1 $$\left(\frac{2 x}{1+x^2}\right)$$ ……….(1)
and z = tan-1 x
i.e. we want to find $$\frac{d y}{d z}$$
put x = tan θ in eqn. (1) ; we have
y = sin-1 $$\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)$$
= sin-1 (sin 2θ)
= 2θ
= 2 tan-1 x
∴ $$\frac{d y}{d x}$$ = $$\frac{2}{1+x^2}$$
and $$\frac{d z}{d x}$$ = $$\frac{1}{1+x^2}$$
Thus, $$\frac{d y}{d z}=\frac{d y / d x}{d z / d x}$$
= $$\frac{2 / 1+x^2}{1 / 1+x^2}$$
= 2

(ii) Let y = tan-1 $$\left(\frac{2 x}{1-x^2}\right)$$ …………….(1)
and z = tan-1 x ………(2)
i.e. we want to find $$\frac{d y}{d z}$$
put x = tan θ
⇒ θ = tan-1 x in eqn. (1) ; we have
∴ y = tan-1 $$\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)$$
⇒ y = tan-1 (tan 2θ)
= 2θ
= 2 tan-1 x
Thus $$\frac{d y}{d x}$$ = $$\frac{2}{1+x^2}$$
and $$\frac{d z}{d x}$$ = $$\frac{1}{1+x^2}$$
∴ $$\frac{d y}{d x}$$ = 2.

Question 14.
Differentiate cos-1 $$\left(\frac{1}{\sqrt{1+t^2}}\right)$$ w.r.t. sin-1 $$\left(\frac{t}{\sqrt{1+t^2}}\right)$$.
Solution:
Let y = cos-1 $$\left(\frac{1}{\sqrt{1+t^2}}\right)$$ ………..(1)
and z = sin-1 $$\left(\frac{t}{\sqrt{1+t^2}}\right)$$
putting t = tan θ
i.e. θ = tan-1 t in eqn (1), we have
y = cos-1 $$\left(\frac{1}{\sqrt{1+\tan ^2 \theta}}\right)$$
= cos-1 (cos θ)
⇒ y = θ = tan-1 t
Diff. w.r.t. t ; we have
∴ $$\frac{d y}{d t}=\frac{1}{1+t^2}$$ ………..(3)
putting t = tan Φ i.e. Φ = tan-1 t, in eqn. (2) ; we have
z = sin-1 $$\left(\frac{\tan \phi}{\sec \phi}\right)$$
= sin-1 (sin Φ) = Φ
⇒ z = tan-1 t
Diff. both sides w.r.t. t, we get
∴ $$\frac{d z}{d t}=\frac{1}{1+t^2}$$ ……………..(4)
Now, we want to diff. y w.r.t. z to find $$\frac{d y}{d z}$$.
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d t}}{\frac{d z}{d t}}$$
= $$\frac{\frac{1}{1+t^2}}{\frac{1}{1+t^2}}$$
= 1 [using (3) and (4)]

Question 15.
(i) Differentiate sin-1 $$\left(\frac{2 x}{1+x^2}\right)$$ w.r.t. cos-1 $$\left(\frac{1-x^2}{1+x^2}\right)$$.
(ii) Differentiate tan-1 $$\left(\frac{3 x-x^3}{1-3 x^2}\right)$$ w.r.t. tan-1 $$\frac{2 x}{1-x^2}$$.
Solution:
(i) Let y = sin-1 $$\left(\frac{2 x}{1+x^2}\right)$$ …………(1)
and z = cos-1 $$\left(\frac{1-x^2}{1+x^2}\right)$$ …………(2)
i.e. we want to diff. y w.r.t. z to find $$\frac{d y}{d z}$$
To simplify eqn. (1) ;
we put x = tan θ
i.e. θ = tan-1 x
y = sin-1 $$1\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)$$
= sin-1 (sin 2θ)
⇒ y = 2θ
= 2 tan-1 x
Diff. both sides w.r.t. x, we have
$$\frac{d y}{d x}=\frac{2}{1+x^2}$$ …………….(3)
To simplify eqn. (2)
we put x = tan Φ
i.e. Φ = tan-1 x
∴ z = cos-1 $$\left(\frac{1-\tan ^2 \phi}{1+\tan ^2 \phi}\right)$$
= cos-1 (cos 2Φ)
⇒ z = 2Φ = 2 tan-1 x
Diff. both sides w.r.t. x, we have
$$\frac{d z}{d x}=\frac{2}{1+x^2}$$ ………………(4)
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{\frac{2}{1+x^2}}{\frac{2}{1+x^2}}$$
= 1 [using eqn. (3) and eqn. (4)]

(ii) Let y = tan-1 $$\left(\frac{3 x-x^3}{1-3 x^2}\right)$$ …………….(1)
and z = tan-1 $$\frac{2 x}{1-x^2}$$ ……………(2)
i.e. we want to diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$.
Putting x = tan θ
⇒ θ = tan-1 x in eqn. (1)
and x = tan Φ
⇒ Φ = tan-1 x in eqn. (2) ; we have
∴ y = tan-1 $$\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)$$
On differentiating both eqn.’s w.r.t. x, we have
∴ $$\frac{d y}{d x}=\frac{3}{1+x^2}$$
and $$\frac{d z}{d x}=\frac{2}{1+x^2}$$
Thus, $$\frac{d y}{d z}=\frac{d y / d x}{d z / d x}$$
= $$\frac{3}{1+x^2} \times \frac{1+x^2}{2}$$
= $$\frac{3}{2}$$

Question 16.
(i) Differentiate tan-1 $$\left(\frac{\sqrt{1+x^2}-1}{x}\right)$$ w.r.t. tan-1 x.
(ii) Differentiate tan-1 $$\left(\frac{\sqrt{1-x^2}}{x}\right)$$ w.r.t. cos-1 (2x $$\sqrt{1-x^2}$$).
Solution:
(i) Let y = tan-1 $$\left(\frac{\sqrt{1+x^2}-1}{x}\right)$$ …………..(1)
and z = tan-1 x …………(2)
Now we want to find $$\frac{d y}{d z}$$
put x = tan θ in eqn. (1) ; we have
y = 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 = $$\frac{\theta}{2}$$
= $$\frac{1}{2}$$ tan-1 x
and z = tan-1 x
Diff. both given eqn’s w.r.t. x, we have
⇒ $$\frac{d y}{d x}=\frac{1}{2\left(1+x^2\right)}$$
and $$\frac{d z}{d x}=\frac{1}{1+x^2}$$
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{\frac{1}{2\left(1+x^2\right)}}{\frac{1}{1+x^2}}$$
= $$\frac{1}{2}$$

(ii) Let y = tan-1 $$\left(\frac{\sqrt{1-x^2}}{x}\right)$$ ………..(1)
and z = cos-1 (2x$$\sqrt{1-x^2}$$) ………………(2)
Put x =sin θ
i.e. θ = sin-1 x in eqn. (1) ; we have
y = tan-1 $$\left(\frac{\sqrt{1-\sin ^2 \theta}}{\sin \theta}\right)$$
= tan-1 (cot θ)
⇒ y = tan-1 (tan ($$\frac{\pi}{2}$$ – θ))
⇒ y = $$\frac{\pi}{2}$$ – θ
= $$\frac{\pi}{2}$$ – sin-1 x
Diff. both sides w.r.t. x, we have
$$\frac{d y}{d x}=-\frac{1}{\sqrt{1-x^2}}$$ ………..(3)
putting x = sin Φ i.e. Φ = sin-1 x in eqn. (2) ; we have
∴ z = cos-1 (2 sin Φ $$\sqrt{1-\sin ^2 \phi}$$)
= cos-1 (2 sin Φ cos Φ)
z = cos-1 (sin 2Φ)
= cos-1 (cos ($$\frac{\pi}{2}$$ – 2Φ))
z = $$\frac{\pi}{2}$$ – 2Φ
= $$\frac{\pi}{2}$$ – sin-1 x
Diff. both sides w.r.t. x ; we have
$$\frac{d z}{d x}=-\frac{2}{\sqrt{1-x^2}}$$ ……………..(4)
We want to diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{-\frac{1}{\sqrt{1-x^2}}}{\frac{-2}{\sqrt{1-x^2}}}$$
= $$\frac{1}{2}$$

Question 17.
Prove that the derivative of tan-1 $$\left(\frac{x}{1+\sqrt{1-x^2}}\right)$$ w.r.t. sin-1 x is independent of x.
Solution:
Let y = tan-1 $$\left(\frac{x}{1+\sqrt{1-x^2}}\right)$$ ………..(1)
and z = sin-1 x
Now we want to diff. y w.r.t. z
Put x = sin θ
∴ θ = sin-1 x in eqn. (1) ; we have
y = tan-1 $$\left(\frac{\sin \theta}{1+\cos \theta}\right)$$
= tan-1 $$\left(\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}\right)$$
⇒ y = tan-1 (tan $$\frac{\theta}{2}$$)
= $$\frac{\theta}{2}$$
= $$\frac{1}{2}$$ sin-1 x
Diff. both sides w.r.t. x, we have
Thus $$\frac{d y}{d x}$$ = $$\frac{1}{2} \frac{1}{\sqrt{1-x^2}}$$
Diff. eqn. (2) w.r.t. x
$$\frac{d z}{d x}=\frac{1}{\sqrt{1-x^2}}$$
∴ $$\frac{d y}{d z}=\frac{d y / d x}{d z / d x}$$
= $$\frac{1}{2 \sqrt{1-x^2}} \times \sqrt{1-x^2}$$
= $$\frac{1}{}2$$
∴ $$\frac{d y}{d z}$$ is independent of x.

Question 18.
Differentiate xx w.r.t. x log x.
Solution:
Let y = xx ………..(1)
and z = x log x ………..(2)
We want to diff. y w.r.t. z i.e. to find $$\frac{d y}{d z}$$
Taking logarithm on both sides of eqn. (1) we have
log y = x log x ;
diff. both sides w.r.t. x ; we have
$$\frac{1}{y} \frac{d y}{d x}$$ = x × $$\frac{1}{x}$$ + log x . 1
⇒ $$\frac{d y}{d x}$$ = xx [1 + log x] ………..(3)
Diff. eqn. (2) w.r.t. x ; we have
$$\frac{d z}{d x}$$ = x × $$\frac{1}{x}$$ + log x . 1
= 1 + log x
∴ $$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$$
= $$\frac{x^x(1+\log x)}{1+\log x}$$
= xx.