Question

If $x^y·y^x=1$, prove that $\frac{dy}{dx}=-\frac{y\left(y+x\log y\right)}{x\left(y\log x+x\right)}$

Answer 1

$\mathrm{We}\mathrm{have},x^y\times y^x=1$
Taking log on both sides,
$$\begin{gathered} \log \left(x^y\times y^x\right)=\log \left(1\right) \\ \Rightarrow y\log x+x\log y=\log 1 \end{gathered}$$
Differentiating with respect to x ,
$$\begin{gathered} \frac{d}{dx}\left(y\log x\right)+\frac{d}{dx}\left(x\log x\right)=\frac{d}{dx}\left(\log 1\right) \\ \Rightarrow \left[y\frac{d}{dx}\left(\log x\right)+\log x\frac{dy}{dx}\right]+\left[x\frac{d}{dx}\left(\log y\right)+\log y\frac{d}{dx}\left(x\right)\right]=0 \\ \Rightarrow \left[y\left(\frac{1}{x}\right)+\log x\frac{dy}{dx}\right]+\left[x\left(\frac{1}{y}\frac{dy}{dx}\right)+\log y\left(1\right)\right]=0 \\ \Rightarrow \frac{y}{x}+\log x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}+\log y=0 \\ \Rightarrow \frac{dy}{dx}\left(\log x+\frac{x}{y}\right)=-\left[\log y+\frac{y}{x}\right] \\ \Rightarrow \frac{dy}{dx}\left[\frac{y\log x+x}{y}\right]=-\left[\frac{x\log y+y}{x}\right] \\ \Rightarrow \frac{dy}{dx}=-\frac{y}{x}\left[\frac{x\log y+y}{y\log x+x}\right] \end{gathered}$$