NCERT Solutions Class-12-Chapter-5-Differentiability and Continuity
Excercise-5.3
Note:
Implicit Differentiation: When \( y \) cannot be easily expressed as \( y = f(x) \), differentiate both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \) (using Chain Rule, e.g., \( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \)), and then solve for \( \frac{dy}{dx} \).
Inverse Trig Derivatives: Standard formulas like \( \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}} \) are often used after simplifying the expression using trigonometric substitutions.
Q1
Find \( \frac{dy}{dx} \) in the following: \( 2x + 3y = \sin x \)
Solution:
Differentiating both sides with respect to \( x \):
\( \frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin x) \)
\( 2 + 3\frac{dy}{dx} = \cos x \)
\( 3\frac{dy}{dx} = \cos x - 2 \)
\( \frac{dy}{dx} = \frac{\cos x - 2}{3} \)
Q2
Find \( \frac{dy}{dx} \) in the following: \( 2x + 3y = \sin y \)
Solution:
Differentiating both sides with respect to \( x \):
\( 2 + 3\frac{dy}{dx} = \frac{d}{dx}(\sin y) \)
\( 2 + 3\frac{dy}{dx} = \cos y \cdot \frac{dy}{dx} \)
\( 2 = (\cos y - 3)\frac{dy}{dx} \)
\( \frac{dy}{dx} = \frac{2}{\cos y - 3} \)
Q3
Find \( \frac{dy}{dx} \) in the following: \( ax + by^2 = \cos y \)
Solution:
Differentiating with respect to \( x \):
\( \frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y) \)
\( a + b(2y)\frac{dy}{dx} = -\sin y \cdot \frac{dy}{dx} \)
\( a + 2by\frac{dy}{dx} = -\sin y \frac{dy}{dx} \)
\( a = -(\sin y + 2by)\frac{dy}{dx} \)
\( \frac{dy}{dx} = -\frac{a}{2by + \sin y} \)
Q4
Find \( \frac{dy}{dx} \) in the following: \( xy + y^2 = \tan x + y \)
Solution:
Differentiating both sides using Product Rule for \( xy \):
\( (x\frac{dy}{dx} + y \cdot 1) + 2y\frac{dy}{dx} = \sec^2 x + \frac{dy}{dx} \)
\( x\frac{dy}{dx} + 2y\frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x - y \)
\( \frac{dy}{dx}(x + 2y - 1) = \sec^2 x - y \)
\( \frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1} \)
Q5
Find \( \frac{dy}{dx} \) in the following: \( x^2 + xy + y^2 = 100 \)
Solution:
Differentiating with respect to \( x \):
\( 2x + (x\frac{dy}{dx} + y) + 2y\frac{dy}{dx} = 0 \)
\( 2x + y + (x + 2y)\frac{dy}{dx} = 0 \)
\( \frac{dy}{dx} = -\frac{2x + y}{x + 2y} \)
Q6
Find \( \frac{dy}{dx} \) in the following: \( x^3 + x^2y + xy^2 + y^3 = 81 \)
Solution:
Differentiating term by term:
\( 3x^2 + (x^2\frac{dy}{dx} + y \cdot 2x) + (x \cdot 2y\frac{dy}{dx} + y^2 \cdot 1) + 3y^2\frac{dy}{dx} = 0 \)
\( 3x^2 + x^2\frac{dy}{dx} + 2xy + 2xy\frac{dy}{dx} + y^2 + 3y^2\frac{dy}{dx} = 0 \)
Group terms with \( \frac{dy}{dx} \):
\( \frac{dy}{dx}(x^2 + 2xy + 3y^2) = -(3x^2 + 2xy + y^2) \)
\( \frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2} \)
Q7
Find \( \frac{dy}{dx} \) in the following: \( \sin^2 y + \cos xy = \kappa \)
Solution:
Differentiating with respect to \( x \):
\( 2\sin y \cdot \cos y \frac{dy}{dx} + (-\sin xy) \cdot \frac{d}{dx}(xy) = 0 \)
\( \sin 2y \frac{dy}{dx} - \sin xy (x\frac{dy}{dx} + y) = 0 \)
\( \sin 2y \frac{dy}{dx} - x\sin xy \frac{dy}{dx} - y\sin xy = 0 \)
\( \frac{dy}{dx}(\sin 2y - x\sin xy) = y\sin xy \)
\( \frac{dy}{dx} = \frac{y\sin xy}{\sin 2y - x\sin xy} \)
Q8
Find \( \frac{dy}{dx} \) in the following: \( \sin^2 x + \cos^2 y = 1 \)
Solution:
Differentiating with respect to \( x \):
\( 2\sin x \cos x + 2\cos y (-\sin y) \frac{dy}{dx} = 0 \)
\( \sin 2x - \sin 2y \frac{dy}{dx} = 0 \)
\( \sin 2y \frac{dy}{dx} = \sin 2x \)
\( \frac{dy}{dx} = \frac{\sin 2x}{\sin 2y} \)
Q9
Find \( \frac{dy}{dx} \) in the following: \( y = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)
Solution:
Let \( x = \tan \theta \). Then \( \theta = \tan^{-1} x \).
\( y = \sin^{-1}\left(\frac{2\tan \theta}{1+\tan^2 \theta}\right) \)
\( y = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1} x \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \)
Q10
Find \( \frac{dy}{dx} \) in the following: \( y = \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right), -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \)
Solution:
Let \( x = \tan \theta \). Then \( \theta = \tan^{-1} x \).
\( y = \tan^{-1}\left(\frac{3\tan \theta - \tan^3 \theta}{1-3\tan^2 \theta}\right) \)
\( y = \tan^{-1}(\tan 3\theta) = 3\theta = 3\tan^{-1} x \).
Differentiating:
\( \frac{dy}{dx} = \frac{3}{1+x^2} \)
Q11
Find \( \frac{dy}{dx} \) in the following: \( y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), 0 < x < 1 \)
Solution:
Let \( x = \tan \theta \).
\( y = \cos^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right) \)
\( y = \cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1} x \).
\( \frac{dy}{dx} = \frac{2}{1+x^2} \)
Q12
Find \( \frac{dy}{dx} \) in the following: \( y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right), 0 < x < 1 \)
Solution:
Let \( x = \tan \theta \).
\( y = \sin^{-1}(\cos 2\theta) \).
Use \( \sin^{-1} u = \frac{\pi}{2} - \cos^{-1} u \).
\( y = \frac{\pi}{2} - \cos^{-1}(\cos 2\theta) = \frac{\pi}{2} - 2\theta = \frac{\pi}{2} - 2\tan^{-1} x \).
\( \frac{dy}{dx} = 0 - \frac{2}{1+x^2} = -\frac{2}{1+x^2} \).
Q13
Find \( \frac{dy}{dx} \) in the following: \( y = \cos^{-1}\left(\frac{2x}{1+x^2}\right), -1 < x < 1 \)
Solution:
Let \( x = \tan \theta \).
\( y = \cos^{-1}(\sin 2\theta) \).
Use \( \cos^{-1} u = \frac{\pi}{2} - \sin^{-1} u \).
\( y = \frac{\pi}{2} - \sin^{-1}(\sin 2\theta) = \frac{\pi}{2} - 2\theta = \frac{\pi}{2} - 2\tan^{-1} x \).
\( \frac{dy}{dx} = -\frac{2}{1+x^2} \).
Q14
Find \( \frac{dy}{dx} \) in the following: \( y = \sin^{-1}(2x\sqrt{1-x^2}), -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \)
Solution:
Let \( x = \sin \theta \). Then \( \sqrt{1-x^2} = \cos \theta \).
\( y = \sin^{-1}(2\sin \theta \cos \theta) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1} x \).
\( \frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}} \).
Q15
Find \( \frac{dy}{dx} \) in the following: \( y = \sec^{-1}\left(\frac{1}{2x^2-1}\right), 0 < x < \frac{1}{\sqrt{2}} \)
Solution:
Let \( x = \cos \theta \).
\( y = \sec^{-1}\left(\frac{1}{2\cos^2 \theta - 1}\right) = \sec^{-1}\left(\frac{1}{\cos 2\theta}\right) \).
\( y = \sec^{-1}(\sec 2\theta) = 2\theta = 2\cos^{-1} x \).
\( \frac{dy}{dx} = 2 \cdot \frac{-1}{\sqrt{1-x^2}} = -\frac{2}{\sqrt{1-x^2}} \).