Exercise 5.3, Question 1
\(2x + 3y = \sin x\)
Show solution
- \(2 + 3\dfrac{dy}{dx} = \cos x\).
Answer: \(\dfrac{dy}{dx} = \dfrac{\cos x - 2}{3}\)
NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.3
Exercise 5.3: Implicit functions; derivatives of inverse trigonometric functions. Differentiate both sides with respect to x, treating y as a function of x (so \(\tfrac{d}{dx}y^2 = 2y\tfrac{dy}{dx}\)), then collect the \(\tfrac{dy}{dx}\) terms. For inverse trigonometric expressions, first simplify with a substitution (\(x = \tan\theta\), \(\sin\theta\), \(\cos\theta\)) inside the stated interval; standard results: \((\sin^{-1}x)' = \tfrac{1}{\sqrt{1 - x^2}}\), \((\cos^{-1}x)' = -\tfrac{1}{\sqrt{1 - x^2}}\), \((\tan^{-1}x)' = \tfrac{1}{1 + x^2}\).
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Exercise 5.3 questions and solutions
Exercise 5.3, Question 2
\(2x + 3y = \sin y\)
Show solution
- \[\begin{aligned}&2 + 3y' = \cos y\,y' \\ \Rightarrow\ &y'(\cos y - 3) = 2\end{aligned}\]
Answer: \(\dfrac{dy}{dx} = \dfrac{2}{\cos y - 3}\)
Exercise 5.3, Question 3
\(ax + by^2 = \cos y\)
Show solution
- \[\begin{aligned}&a + 2by\,y' = -\sin y\,y' \\ \Rightarrow\ &y'(2by + \sin y) = -a\end{aligned}\]
Answer: \[\dfrac{dy}{dx} = -\dfrac{a}{2by + \sin y}\]
Exercise 5.3, Question 4
\(xy + y^2 = \tan x + y\)
Show solution
- \(y + xy' + 2yy' = \sec^2 x + y'\).
- \(y'(x + 2y - 1) = \sec^2 x - y\).
Answer: \[\dfrac{dy}{dx} = \dfrac{\sec^2 x - y}{x + 2y - 1}\]
Exercise 5.3, Question 5
\(x^2 + xy + y^2 = 100\)
Show solution
- \(2x + y + xy' + 2yy' = 0\).
Answer: \(\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}\)
Exercise 5.3, Question 6
\(x^3 + x^2y + xy^2 + y^3 = 81\)
Show solution
- \[3x^2 + (2xy + x^2y') + (y^2 + 2xyy') + 3y^2y' = 0\]
- \[y'(x^2 + 2xy + 3y^2) = -(3x^2 + 2xy + y^2)\]
Answer: \[\dfrac{dy}{dx} = -\dfrac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}\]
Exercise 5.3, Question 7
\(\sin^2 y + \cos xy = \kappa\)
Show solution
- \[2\sin y\cos y\,y' - \sin xy\,(y + xy') = 0\]
- \(y'(\sin 2y - x\sin xy) = y\sin xy\).
Answer: \[\dfrac{dy}{dx} = \dfrac{y\sin xy}{\sin 2y - x\sin xy}\]
Exercise 5.3, Question 8
\(\sin^2 x + \cos^2 y = 1\)
Show solution
- \(2\sin x\cos x - 2\cos y\sin y\,y' = 0\).
Answer: \[\dfrac{dy}{dx} = \dfrac{\sin 2x}{\sin 2y}\]
Exercise 5.3, Question 9
\(y = \sin^{-1}\dfrac{2x}{1 + x^2}\)
Show solution
- Put \(x = \tan\theta\): \(\dfrac{2x}{1 + x^2} = \sin 2\theta\).
- For \(|x| < 1\), \[2\theta \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\], so \(y = 2\theta = 2\tan^{-1}x\) and \(y' = \dfrac{2}{1 + x^2}\).
- (For \(|x| > 1\), \(y = \pm\pi - 2\tan^{-1}x\) and \(y' = -\dfrac{2}{1 + x^2}\); the textbook answer assumes \(|x| < 1\).)
Answer: \(\dfrac{dy}{dx} = \dfrac{2}{1 + x^2}\) for \(|x| < 1\)
Exercise 5.3, Question 10
\(y = \tan^{-1}\dfrac{3x - x^3}{1 - 3x^2}\), \(-\tfrac{1}{\sqrt3} < x < \tfrac{1}{\sqrt3}\)
Show solution
- Put \(x = \tan\theta\), \[\theta \in \left(-\tfrac{\pi}{6}, \tfrac{\pi}{6}\right)\]: the fraction is \(\tan 3\theta\) with \(3\theta\) in the principal branch.
- So \(y = 3\tan^{-1}x\).
Answer: \(\dfrac{dy}{dx} = \dfrac{3}{1 + x^2}\)
Exercise 5.3, Question 11
\(y = \cos^{-1}\dfrac{1 - x^2}{1 + x^2}\), \(0 < x < 1\)
Show solution
- \(x = \tan\theta\), \[\theta \in \left(0, \tfrac{\pi}{4}\right)\]: the fraction is \(\cos 2\theta\) with \(2\theta \in [0, \pi]\).
- \(y = 2\tan^{-1}x\).
Answer: \(\dfrac{dy}{dx} = \dfrac{2}{1 + x^2}\)
Exercise 5.3, Question 12
\(y = \sin^{-1}\dfrac{1 - x^2}{1 + x^2}\), \(0 < x < 1\)
Show solution
- \(\sin^{-1}u = \tfrac{\pi}{2} - \cos^{-1}u\), and by Q11 \[\cos^{-1}\dfrac{1 - x^2}{1 + x^2} = 2\tan^{-1}x\]
- \(y = \tfrac{\pi}{2} - 2\tan^{-1}x\).
Answer: \(\dfrac{dy}{dx} = -\dfrac{2}{1 + x^2}\)
Exercise 5.3, Question 13
\(y = \cos^{-1}\dfrac{2x}{1 + x^2}\), \(-1 < x < 1\)
Show solution
- \(\cos^{-1}u = \tfrac{\pi}{2} - \sin^{-1}u\), and \[\sin^{-1}\dfrac{2x}{1 + x^2} = 2\tan^{-1}x\] for \(|x| < 1\).
- \(y = \tfrac{\pi}{2} - 2\tan^{-1}x\).
Answer: \(\dfrac{dy}{dx} = -\dfrac{2}{1 + x^2}\)
Exercise 5.3, Question 14
\(y = \sin^{-1}\left(2x\sqrt{1 - x^2}\right)\), \(-\tfrac{1}{\sqrt2} < x < \tfrac{1}{\sqrt2}\)
Show solution
- \(x = \sin\theta\), \[\theta \in \left(-\tfrac{\pi}{4}, \tfrac{\pi}{4}\right)\]: \(2x\sqrt{1 - x^2} = \sin 2\theta\) with \(2\theta\) in the branch.
- \(y = 2\sin^{-1}x\).
Answer: \[\dfrac{dy}{dx} = \dfrac{2}{\sqrt{1 - x^2}}\]
Exercise 5.3, Question 15
\(y = \sec^{-1}\dfrac{1}{2x^2 - 1}\), \(0 < x < \tfrac{1}{\sqrt2}\)
Show solution
- \(y = \cos^{-1}(2x^2 - 1)\). Put \(x = \cos\theta\), \[\theta \in \left(\tfrac{\pi}{4}, \tfrac{\pi}{2}\right)\]: \(2x^2 - 1 = \cos 2\theta\) with \[2\theta \in \left(\tfrac{\pi}{2}, \pi\right)\]
- \(y = 2\theta = 2\cos^{-1}x\).
Answer: \[\dfrac{dy}{dx} = -\dfrac{2}{\sqrt{1 - x^2}}\]
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