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

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

Find 

$\frac{dy}{dx}$ 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$\frac{dy}{dx}$ = cos x ⇒ 3$\frac{dy}{dx}$ = cos x - 2
Thus, $\frac{dy}{dx}$ = $\frac{\cos x-2}{3}$

Question 2.

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

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

Question 4.

xy + y2 = tan x + y
Answer:
Given xy + y2 = tan x + y
Differentiating both sides w.r.t. x, we get
x. $\frac{dy}{dx}$ + 1.y + 2y $\frac{dy}{dx}$ = sec2 x + $\frac{dy}{dx}$

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

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$\frac{dy}{dx}$ + 1.y2 + x.2y$\frac{dy}{dx}$ + 3y2$\frac{dy}{dx}$ = 0
⇒ 3x2 + 2xy + y2 + (x2 + 2xy + 3y2) $\frac{dy}{dx}$ = 0
⇒ (x2 + 2xy + 3y2)$\frac{dy}{dx}$ = - (3x2 + 2xy + y2)
Thus, $\frac{dy}{dx}$ = - $\frac{(3x^2+2xy+y^2)}{(x^2+2xy+3y^2)}$

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 $\frac{dy}{dx}$ + (- sin xy) (x $\frac{dy}{dx}$ + 1.y) = 0
⇒ sin 2y $\frac{dy}{dx}$ - x sin xy $\frac{dy}{dx}$ - y sin xy = 0
[∵ 2 sin y cos y = sin2y]
⇒ (sin 2y - x sin xy)$\frac{dy}{dx}$ = y sin xy
Thus, $\frac{dy}{dx}$ = $\frac{y\sin xy}{(\sin 2y-x\sin xy)}$

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 $\frac{dy}{dx}$ = 0
⇒ sin 2x - sin 2y $\frac{dy}{dx}$ = 0
⇒ - sin 2y $\frac{dy}{dx}$= - sin 2x
⇒ $\frac{dy}{dx}$ = $\frac{-\sin 2x}{-\sin 2y}=\frac{\sin 2x}{\sin 2y}$
Thus, $\frac{dy}{dx}$ = $\frac{-\sin 2x}{-\sin 2y}=\frac{\sin 2x}{\sin 2y}$

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

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

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
$\frac{dy}{dx}$ = $\frac{2}{1+x^2}$

Question 10.

y = tan-1$\left(\frac{3x-x^3}{1-3x^2}\right)$, - $\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$
Answer:
Given y = tan-1$\left(\frac{3x-x^3}{1-3x^2}\right)$
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
$\frac{dy}{dx}$= $\frac{3}{1+x^2}$

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
$\frac{dy}{dx}$ = $\frac{2}{1+x^2}$

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$\left(\frac{2x}{1+x^2}\right),$ - 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
$\frac{dy}{dx}$ = 2 $\frac{d}{dx}$ (sin-1 x)
Thus, $\frac{d}{dx}$ = $\frac{2}{\sqrt{1-x^2}}$

Question 15.

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

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