Question

If $x=e^{x/y}$, prove that $\frac{dy}{dx}=\frac{x-y}{x\log x}$

Answer 1

$x=e^{x/y}$

$lnx=\frac{x}{y}$

$y$ $lnx=x$

differentiate both sides $\Rightarrow$

$y\left(\frac{1}{x}dx\right)+\left(lnx\right)\left(dy\right)=dx$

$lnx$ $dy=\left(\frac{x-y}{x}\right)dx$

$\therefore \frac{dy}{dx}=\frac{x-y}{xlnx}$