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If \( \tan ^{-1}\left(x^{2}+y^{2}\right)=\alpha \) then \( \frac{d y}{d x} \) is equal to
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Verified Answer
The correct answer is:
\( -\frac{x}{y} \)
Given that, \( \tan ^{-1}\left(x^{2}+y^{2}\right)=\alpha \)
\( \Rightarrow x^{2}+y^{2}=\tan \alpha \)
Differentiating with respect to \( x \), we get
\[
\begin{array}{l}
2 x+2 y \frac{d y}{d x}=0 \\
\Rightarrow \frac{d y}{d x}=\frac{-x}{y}
\end{array}
\]
\( \Rightarrow x^{2}+y^{2}=\tan \alpha \)
Differentiating with respect to \( x \), we get
\[
\begin{array}{l}
2 x+2 y \frac{d y}{d x}=0 \\
\Rightarrow \frac{d y}{d x}=\frac{-x}{y}
\end{array}
\]
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