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If $x y=\tan ^{-1}(x y)+\cot ^{-1}(x y)$, then $\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(4,2)}=($ where $x, y \in I R)$
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Verified Answer
The correct answer is:
$\frac{-1}{2}$
$\begin{aligned}
& x y=\tan ^{-1}(x y)+\cot ^{-1}(x y) \\
& \Rightarrow x y=\frac{\pi}{2}
\end{aligned}$
diff. we get $1 \cdot y+x \cdot \frac{d y}{d x}=0$
$\begin{aligned}
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-y}{x} \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x} \text { at, }(4,2)=\frac{-2}{4}=\frac{-1}{2}
\end{aligned}$
& x y=\tan ^{-1}(x y)+\cot ^{-1}(x y) \\
& \Rightarrow x y=\frac{\pi}{2}
\end{aligned}$
diff. we get $1 \cdot y+x \cdot \frac{d y}{d x}=0$
$\begin{aligned}
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-y}{x} \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x} \text { at, }(4,2)=\frac{-2}{4}=\frac{-1}{2}
\end{aligned}$
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