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

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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 = e^t (sin t + cos t), y = e^t (sin t – cos t)
(ii) x = cos θ – cos 2θ, y = sin θ – sin 2θ (NCERT)
Solution:
(i) Given x = e^t (sin t + cos t)
and y = e^t (sin t – cos t)
Diff. both eqn.’s w.r.t. t, we have
\(\frac{d x}{d t}\) = e^t (cos t – sin t) + (sin t + cos t) e^t
= e^t (2 cos t)
and \(\frac{d y}{d t}\) = e^t (cos t + sin t) + (sin t – cos t) e^t
= e^t (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 = ae^θ (sin θ – cos θ) and y = ae^θ (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 e^θ (sin θ – cos θ) ……………..(1)
and y = a e^θ (sin θ + cos θ)
diff. eqn. (1) w.r.t. θ ; we have
\(\frac{d x}{d \theta}\) = a e^θ (sin θ – cos θ) + a e^θ (cos θ + sin θ)
= a e^θ (sin θ – cos θ + cos θ + sin θ)
= 2ae^θ sin θ
and y = ae^θ (sin θ + cos θ) ………….(2)
diff. eqn. (2) w.r.t. θ ; we have
\(\frac{d y}{d \theta}\) = ae^θ (sin θ + cos θ) + ae^θ (cos θ – sin θ)
= ae^θ (sin θ + cos θ + cos θ – sin θ)
= 2ae^θ 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. e^{cos 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 = e^{cos 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 ;
\(\frac{d z}{d x}\) = e^{cos 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 x^x w.r.t. x log x.
Solution:
Let y = x^x ………..(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
⇒ \(\frac{d y}{d x}\) = x^x [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
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(\frac{x^x(1+\log x)}{1+\log x}\)
= x^x.