If $x^y \cdot y^x=16$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $(2,2)$ is

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Aicharya · Accepted Answer

$x^y \cdot y^x=16$ $\Rightarrow y \log x+x \log y=\log 16$ [Taking log both sides] Differentiating we get $\frac{\mathrm{d} y}{\mathrm{~d} x} \cdot \log x+y \cdot \frac{1}{x}+\log y+x \cdot \frac{1}{y} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=0$ Putting $x=2$ and $y=0$ $\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}(1+\log 2)=-(1+\log 2) \ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-1 \end{aligned}$

If $x^y \cdot y^x=16$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $(2,2)$ is

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