Question

$x^{y} \cdot y^{x}=1$, then $\frac{d y}{d x}$ 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
$\frac{y}{x}+\log x \frac{d y}{d x}+\frac{x}{y} \frac{d y}{d x}+\log y=0$
$\Rightarrow \frac{d y}{d x}\left(\log x+\frac{x}{y}\right)=-\left(\log y+\frac{y}{x}\right)$
$\Rightarrow \frac{d y}{d x}=\frac{-y(x \log y+y)}{x(y \log x+x)}$