RBSE Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions Chapter 5 Continuity and Differentiability Ex 5.3

Find \(\frac{d y}{d x}\) in the following:

Question 1.
2x + 3y = sin x
Answer:
Given, 2x + 3y = sin x
Differentiating both sides w.r.t. x, we get
2.1 + 3\(\frac{d y}{d x}\) = cos x ⇒ 3\(\frac{d y}{d x}\) = cos x - 2
Thus, \(\frac{d y}{d x}\) = \(\frac{\cos x-2}{3}\)

Question 2.
2x + 3y = sin y
Answer:
Given, 2x + 3y = sin y
Differentiating both sides w.r.t. x, we get
2.1 + 3\(\frac{d y}{d x}\) = cos y \(\frac{d y}{d x}\)
⇒ 2 = cos y \(\frac{d y}{d x}\) - 3\(\frac{d y}{d x}\) ⇒ 2 = (cos y - 3) \(\frac{d y}{d x}\)
Thus, \(\frac{d y}{d x}\) = \(\frac{2}{\cos y-3}\)

Question 3.
ax + by² = cos y
Answer:
Given, ax + by² = cos y
Differentiating both sides w.r.t. x, we get
a.1 = 2by\(\frac{d y}{d x}\) = - sin y \(\frac{d y}{d x}\)
⇒ (sin y + 2by) \(\frac{d y}{d x}\) = - a
Thus, \(\frac{d y}{d x}\) = \(\frac{-a}{\sin y+2 b y}\)

Question 4.
xy + y² = tan x + y
Answer:
Given xy + y² = tan x + y
Differentiating both sides w.r.t. x, we get
x. \(\frac{d y}{d x}\) + 1.y + 2y \(\frac{d y}{d x}\) = sec² x + \(\frac{d y}{d x}\)

Question 5.
x² + xy + y² = 100
Answer:
Given, x² + xy + y² = 100
Differentiating both sides w.r.t. x, we get
2x + x. \(\frac{d y}{d x}\) + 1.y + 2y \(\frac{d y}{d x}\) = 0
⇒ (x + 2y) \(\frac{d y}{d x}\) = - y - 2x = - (y + 2x)
Thus, \(\frac{d y}{d x}\) = -\(\frac{(y+2 x)}{(x+2 y)}\)

Question 6.
x³ + x²y + xy² + y³ = 81
Answer:
Given x³ + x²y + xy² + y³ = 81
Differentiating both sides w.r.t. x, we get
3x² + 2xy + x²\(\frac{d y}{d x}\) + 1.y² + x.2y\(\frac{d y}{d x}\) + 3y²\(\frac{d y}{d x}\) = 0
⇒ 3x² + 2xy + y² + (x² + 2xy + 3y²) \(\frac{d y}{d x}\) = 0
⇒ (x² + 2xy + 3y²)\(\frac{d y}{d x}\) = - (3x² + 2xy + y²)
Thus, \(\frac{d y}{d x}\) = - \(\frac{\left(3 x^{2}+2 x y+y^{2}\right)}{\left(x^{2}+2 x y+3 y^{2}\right)}\)

Question 7.
sin² y + cos xy = k
Answer:
Given, sin² y + cos xy = k
Differentiating both sides w.r.t. x, we get
2 sin y cos y \(\frac{d y}{d x}\) + (- sin xy) (x \(\frac{d y}{d x}\) + 1.y) = 0
⇒ sin 2y \(\frac{d y}{d x}\) - x sin xy \(\frac{d y}{d x}\) - y sin xy = 0
[∵ 2 sin y cos y = sin2y]
⇒ (sin 2y - x sin xy)\(\frac{d y}{d x}\) = y sin xy
Thus, \(\frac{d y}{d x}\) = \(\frac{y \sin x y}{(\sin 2 y-x \sin x y)}\)

Question 8.
sin² x + cos² y = 1
Answer:
Given sin² x + cos² y = 1
Differentiating both sides w.r.t. x, we get
2 sin x cos x - 2 cos y sin y \(\frac{d y}{d x}\) = 0
⇒ sin 2x - sin 2y \(\frac{d y}{d x}\) = 0
⇒ - sin 2y \(\frac{d y}{d x}\)= - sin 2x
⇒ \(\frac{d y}{d x}\) = \(\frac{-\sin 2 x}{-\sin 2 y}=\frac{\sin 2 x}{\sin 2 y}\)
Thus, \(\frac{d y}{d x}\) = \(\frac{-\sin 2 x}{-\sin 2 y}=\frac{\sin 2 x}{\sin 2 y}\)

Question 9.
y = sin^-1\(\left(\frac{2 x}{1+x^{2}}\right)\)
Answer:
Given y = sin^-1\(\left(\frac{2 x}{1+x^{2}}\right)\)
Differentiating both sides w.r.t. x, we get

Alternatively:
y = sin^-1 \(\left(\frac{2 x}{1+x^{2}}\right)\)

let x = tan θ,
Then θ = tan^-1 x
∴ y = sin^-1 \(\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)\)
⇒ y = sin^-1 (sin 2θ)
⇒ y = 2θ (∵ sin 2θ = \(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\))
⇒ y = 2 tan^-1 x
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{2}{1+x^{2}}\)

Question 10.
y = tan^-1\(\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)\), - \(\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}\)
Answer:
Given y = tan^-1\(\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)\)
Let x = tan θ then θ = tan^-1 x
and 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
Differentiating both sides w.r.t x, we get
\(\frac{d y}{d x}\)= \(\frac{3}{1+x^{2}}\)

Question 11.
y = cos^-1 \(\left(\frac{1-x^{2}}{1+x^{2}}\right)\), 0 < x < 1
Answer:
Given, y = cos^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\), 0 < x < 1
Let x = tan θ then θ = tan^-1x
and y = cos^-1\(\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)\)
⇒ y = cos^-1 (cos 2θ) = 2θ = 2 tan^-1x
Differentiating both sides w.r.t x, we get
\(\frac{d y}{d x}\) = \(\frac{2}{1+x^{2}}\)

Question 12.
y = sin^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\), 0 < x < 1
Answer:
Given y = sin^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Let x = tan θ, then θ = tan^-1 x

Question 13.
y = cos^-1\(\left(\frac{2 x}{1+x^{2}}\right),\) - 1 < x < 1
Answer:

Question 14.
y = sin^-1 (2x\(\sqrt{1-x^{2}}\)), -\(\frac{1}{\sqrt{2}}\) < x < \(\frac{1}{\sqrt{2}}\)
Answer:
Given y = sin^-1 (2x\(\sqrt{1-x^{2}}\))
Let x = sin θ, then θ = sin^-1 x
and y = sin^-1 (2 sin θ \(\sqrt{1-\sin ^{2} \theta}\))
⇒ y = sin^-1 (2 sin θ cos θ)
⇒ y = sin^-1 (sin 2θ) = 2θ = 2 sin^-1 x
Differentiating both sides w.r.t x, we get
\(\frac{d y}{d x}\) = 2 \(\frac{d}{d x}\) (sin^-1 x)
Thus, \(\frac{d}{d x}\) = \(\frac{2}{\sqrt{1-x^{2}}}\)

Question 15.
y = sec^-1\(\left(\frac{1}{2 x^{2}-1}\right)\), 0 < x < \(\frac{1}{\sqrt{2}}\)
Answer: