Question

$x^y\cdot y^x=1$, then $\frac{dy}{dx}$ is:

• A. $\frac{y(y+x\log y)}{x(y\log x+x)}$
• B. $\frac{y(x+y\log x)}{x(y+x\log y)}$
• C. $\frac{-y(y+x\log y)}{x(x+y\log x)}$
• D. None of these

Answer 1

Answer: (C)

We have $x^y\cdot y^x=1$
On taking log on both sides, we get
$y\log x+x\log y=0$
On differentiating w.r.t. $x$, we get
$\frac{y}{x}+\log x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}+\log y=0$
$\Rightarrow \frac{dy}{dx}\left(\log x+\frac{x}{y}\right)=-\left(\log y+\frac{y}{x}\right)$
$\Rightarrow \frac{dy}{dx}=\frac{-y(x\log y+y)}{x(y\log x+x)}$