Question

Find $\frac{dy}{dx}$ for $x^y=e^{y-x}$.

Answer 1

Given,

$x^y=e^{y-x}$

Now, tacking logarithm both sides

w.r.t to the base 'e' we get,

y log x = y - x

Now differentiating both sides

w.r.t to 'x' we gets,

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

or, $(\frac{y}{x}+1)=(1-logx)\frac{dy}{dx}$

or, $\frac{(y+x)}{x(1-logx)}=\left(\frac{dy}{dx}\right)$;