Q66 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
Given $y=\tan ^{-1} x$, then $\dfrac{d^2 y}{d x^2}$ in terms of ' $y$ ' is
- A$-2 \sin y \cos ^2 y$
- B$-\cos 2 y \sin ^2 y$
- C$-\sin 2 y \cos y$
- Correct answerD$-\sin 2 y \cos ^2 y$
Solution
Given $y = \tan^{-1} x$, we have $x = \tan y$.
Differentiating both sides with respect to $x$, we get $\dfrac{dy}{dx} = \dfrac{1}{1 + x^2}$.
Substituting $x = \tan y$, we obtain $\dfrac{dy}{dx} = \dfrac{1}{1 + \tan^2 y} = \dfrac{1}{\sec^2 y} = \cos^2 y$.
Now, differentiating $\dfrac{dy}{dx} = \cos^2 y$ with respect to $x$ using the chain rule:
$\dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}(\cos^2 y) = \dfrac{d}{dy}(\cos^2 y) \times \dfrac{dy}{dx}$.
$\dfrac{d^2 y}{dx^2} = (2 \cos y \times (-\sin y)) \times \cos^2 y$.
$\dfrac{d^2 y}{dx^2} = -2 \sin y \cos y \times \cos^2 y$.
Using the identity $\sin 2y = 2 \sin y \cos y$, we get $\dfrac{d^2 y}{dx^2} = -\sin 2y \cos^2 y$.
Answer: $-\sin 2 y \cos ^2 y$
Differentiating both sides with respect to $x$, we get $\dfrac{dy}{dx} = \dfrac{1}{1 + x^2}$.
Substituting $x = \tan y$, we obtain $\dfrac{dy}{dx} = \dfrac{1}{1 + \tan^2 y} = \dfrac{1}{\sec^2 y} = \cos^2 y$.
Now, differentiating $\dfrac{dy}{dx} = \cos^2 y$ with respect to $x$ using the chain rule:
$\dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}(\cos^2 y) = \dfrac{d}{dy}(\cos^2 y) \times \dfrac{dy}{dx}$.
$\dfrac{d^2 y}{dx^2} = (2 \cos y \times (-\sin y)) \times \cos^2 y$.
$\dfrac{d^2 y}{dx^2} = -2 \sin y \cos y \times \cos^2 y$.
Using the identity $\sin 2y = 2 \sin y \cos y$, we get $\dfrac{d^2 y}{dx^2} = -\sin 2y \cos^2 y$.
Answer: $-\sin 2 y \cos ^2 y$
Hence, the correct answer is option: 'D'.