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If $x^y \cdot y^x=16$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $(2,2)$ is
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The correct answer is:
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$x^y \cdot y^x=16$
$\Rightarrow y \log x+x \log y=\log 16$ [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 $y=0$
$\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}(1+\log 2)=-(1+\log 2) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-1
\end{aligned}$
$\Rightarrow y \log x+x \log y=\log 16$ [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 $y=0$
$\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}(1+\log 2)=-(1+\log 2) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-1
\end{aligned}$
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