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If $x y=\tan ^{-1}(x y)+\cot ^{-1}(x y)$, then $\frac{d y}{d x}$ is equal
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Verified Answer
The correct answer is:
$-\frac{y}{x}$
$x y=\tan ^{-1}(x y)+\cot ^{-1}(x y)=\frac{\pi}{2}$
$\Rightarrow \quad x y=\frac{\pi}{2}$
$\therefore \quad x \frac{d y}{d x}+y=0$
$\Rightarrow \quad \frac{d y}{d x}=\frac{-y}{x}$
$\Rightarrow \quad x y=\frac{\pi}{2}$
$\therefore \quad x \frac{d y}{d x}+y=0$
$\Rightarrow \quad \frac{d y}{d x}=\frac{-y}{x}$
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