Question

Find $\frac{dy}{dx}$ when $x^y+y=x$.

Answer 1

Given,

$x^y+y=x$

or, $e^{y\log x}+y=x$

Now differentiating both sides with respect to $x$ we get,

or, $e^{y\log x}(\frac{y}{x}+\log x.\frac{dy}{dx})+\frac{dy}{dx}=1$

or, $y{x}^{y-1}+x^y.\log x.\frac{dy}{dx}+\frac{dy}{dx}=1$

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