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If $x=\sqrt{\mathrm{e}^{\sin ^{-1} \mathrm{t}}}$ and $y=\sqrt{\mathrm{e}^{\cos ^{-1} \mathrm{t}}}$, then $\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}$ is
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Verified Answer
The correct answer is:
$\frac{2y}{x^2}$
$\begin{aligned}
x y & =\sqrt{\mathrm{e}^{\sin ^{-1} \mathrm{t}}} \cdot \sqrt{\mathrm{e}^{\cos ^{-1} \mathrm{t}}} \\
& =\sqrt{\mathrm{e}^{\sin ^{-1} \mathrm{t}+\cos ^{-1} \mathrm{t}}} \\
\therefore \quad x y & =\sqrt{\mathrm{e}^{\frac{\pi}{2}}}
\end{aligned}$
Differentiating both sides w.r.t. $x$, we get
$\begin{aligned}
& x \frac{\mathrm{d} y}{\mathrm{~d} x}+y \cdot 1=0 \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{y}{x}...(i)
\end{aligned}$
$\begin{aligned} \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2} & =-\left(\frac{x \frac{\mathrm{d} y}{\mathrm{~d} x}-y \cdot 1}{x^2}\right) \\ \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d}^2} & =-\left(\frac{x\left(-\frac{y}{x}\right)-y}{x^2}\right)... [From (i)] \\ & =\frac{2 y}{x^2}\end{aligned}$
x y & =\sqrt{\mathrm{e}^{\sin ^{-1} \mathrm{t}}} \cdot \sqrt{\mathrm{e}^{\cos ^{-1} \mathrm{t}}} \\
& =\sqrt{\mathrm{e}^{\sin ^{-1} \mathrm{t}+\cos ^{-1} \mathrm{t}}} \\
\therefore \quad x y & =\sqrt{\mathrm{e}^{\frac{\pi}{2}}}
\end{aligned}$
Differentiating both sides w.r.t. $x$, we get
$\begin{aligned}
& x \frac{\mathrm{d} y}{\mathrm{~d} x}+y \cdot 1=0 \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{y}{x}...(i)
\end{aligned}$
$\begin{aligned} \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2} & =-\left(\frac{x \frac{\mathrm{d} y}{\mathrm{~d} x}-y \cdot 1}{x^2}\right) \\ \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d}^2} & =-\left(\frac{x\left(-\frac{y}{x}\right)-y}{x^2}\right)... [From (i)] \\ & =\frac{2 y}{x^2}\end{aligned}$
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