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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Aicharya · Accepted Answer

We know, $\sin ^{-1} 	heta+\cos ^{-1} 	heta=\frac{\pi}{2} \Rightarrow x^{2} y^{2}=\frac{\pi}{2}$ Differentiating w.r.t. $\mathrm{x}$ $\begin{array}{l} x^{2} \cdot 2 y \frac{d y}{d x}+y^{2} \cdot 2 x=0 \Rightarrow x^{2} \cdot 2 y \frac{d y}{d x}=-y^{2} \cdot 2 x \ \frac{d y}{d x}=\frac{-y^{2} \cdot 2 x}{x^{2} \cdot 2 y}=\frac{-y}{x} \end{array}$

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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