If \( x=\ \sqrt{a^{\tan^{-1}t}} \) and \( y=\sqrt{a^{\cot^{-1}t}} \) , then \( \displaystyle\frac{d^2y}{dx^2} \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{2y}{x^2} \)
Since \( \tan^{-1}t+\cot^{-1}t=\frac{\pi}{2} \), the product xy is a constant:
\( \displaystyle xy=\sqrt{a^{\tan^{-1}t+\cot^{-1}t}} \)
\( \displaystyle =\sqrt{a^{\pi/2}}=c\ \text{(a constant)} \)
Differentiating xy = c with respect to x:
\( \displaystyle x\frac{dy}{dx}+y=0 \)
\( \displaystyle \frac{dy}{dx}=-\frac{y}{x} \)
Differentiating again by the quotient rule:
\( \displaystyle \frac{d^2y}{dx^2}=-\frac{x\frac{dy}{dx}-y}{x^2} \)
\( \displaystyle =-\frac{-y-y}{x^2} \)
\( \displaystyle =\frac{2y}{x^2} \)
Why the other options are incorrect
Option (1) \( -\displaystyle\frac{y}{x} \) — Incorrect. This is only the first derivative \( \frac{dy}{dx} \).
Option (2) \( \displaystyle\frac{y}{x^2} \) — Incorrect. The value is \( \frac{2y}{x^2} \); the factor 2 comes from the two equal terms \( -y-y \).
Option (4) 0 — Incorrect. The value is \( \frac{2y}{x^2} \), which is not zero since y > 0.