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If $x^{2} y^{2}=\sin ^{-1} \sqrt{x^{2}+y^{2}}+\cos ^{-1} \sqrt{x^{2}+y^{2}}$, then $\frac{d y}{d x}=$
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The correct answer is:
$\frac{-y}{x}$
(D)
We know, $\sin ^{-1} \theta+\cos ^{-1} \theta=\frac{\pi}{2} \Rightarrow x^{2} y^{2}=\frac{\pi}{2}$ Differentiating w.r.t. $\mathrm{x}$
$\begin{array}{l}
x^{2} \cdot 2 y \frac{d y}{d x}+y^{2} \cdot 2 x=0 \Rightarrow x^{2} \cdot 2 y \frac{d y}{d x}=-y^{2} \cdot 2 x \\
\frac{d y}{d x}=\frac{-y^{2} \cdot 2 x}{x^{2} \cdot 2 y}=\frac{-y}{x}
\end{array}$
We know, $\sin ^{-1} \theta+\cos ^{-1} \theta=\frac{\pi}{2} \Rightarrow x^{2} y^{2}=\frac{\pi}{2}$ Differentiating w.r.t. $\mathrm{x}$
$\begin{array}{l}
x^{2} \cdot 2 y \frac{d y}{d x}+y^{2} \cdot 2 x=0 \Rightarrow x^{2} \cdot 2 y \frac{d y}{d x}=-y^{2} \cdot 2 x \\
\frac{d y}{d x}=\frac{-y^{2} \cdot 2 x}{x^{2} \cdot 2 y}=\frac{-y}{x}
\end{array}$
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