Question

If $y^x=x^y$, then find $\frac{dy}{dx}$.

• A. $\frac{x(y\log y-y)}{y(x\log x-x)}$
• B. $\frac{y(x\log y-y)}{x(y\log x-x)}$
• C. $y(x\log y-y)$
• D. $\frac{y(x\log x-y)}{x(y\log y-x)}$

Answer 1

Answer: (B)

$\frac{y(x\log y-y)}{x(y\log x-x)}$

$y^x=x^y$

Taking log on both sides

$\Rightarrow$ $\log y^x=\log x^y$

$\Rightarrow$ $x\log y=y\log x$

$\Rightarrow$ $x\frac{1}{y}\frac{dy}{dx}+\log y\times 1=\frac{y}{x}+\log x\frac{dy}{dx}$

$\Rightarrow$ $(\frac{x}{y}-\log x)\frac{dy}{dx}=\frac{y}{x}-\log y$

$\Rightarrow$ $\frac{dy}{dx}=\frac{\frac{y}{x}-\log y}{\frac{x}{y}-\log x}=\frac{y(y-x\log y)}{x(x-y\log x)}=\frac{y(x\log y-y)}{x(y\log x-x)}$