If $x y= an ^{-1}(x y)+\cot ^{-1}(x y)$, then $\left(\frac{\mathrm{d} y}{\mathrm{~d} x} ight)_{(4,2)}=($ where $x, y \in I R)$

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If $x y=	an ^{-1}(x y)+\cot ^{-1}(x y)$, then $\left(\frac{\mathrm{d} y}{\mathrm{~d} x}ight)_{(4,2)}=($ where $x, y \in I R)$

Aicharya · Accepted Answer

$\begin{aligned} & x y=	an ^{-1}(x y)+\cot ^{-1}(x y) \ & \Rightarrow x y=\frac{\pi}{2} \end{aligned}$ diff. we get $1 \cdot y+x \cdot \frac{d y}{d x}=0$ $\begin{aligned} & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-y}{x} \ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x} 	ext { at, }(4,2)=\frac{-2}{4}=\frac{-1}{2} \end{aligned}$

If $x y=\tan ^{-1}(x y)+\cot ^{-1}(x y)$, then $\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(4,2)}=($…

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