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

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Aicharya · Accepted Answer

$x^y \cdot y^x=16$ Taking $\log$ on both sides, $	herefore \mathrm{y} \log \mathrm{x}+\mathrm{x} \log \mathrm{y}=\log 16$ Differentiating w.r.t. $x, 	herefore \frac{y}{x}+(\log x) \frac{d y}{d x}+\left(\frac{x}{y}ight) \frac{d y}{d x}+\log y=0$ $\begin{aligned} & 	herefore\left[\log x+\frac{x}{y}ight] \frac{d y}{d x}=-\left[\frac{y}{x}+\log yight] \ & 	herefore \frac{d y}{d x}=\frac{-\left[\frac{y}{x}+\log yight]}{\left[\log x+\frac{x}{y}ight]} \Rightarrow\left(\frac{d y}{d x}ight)_{(2,2)}=-\left[\frac{(1+\log 2)}{\log 2+1}ight]=-1 \end{aligned}$

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

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