Question

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

Answer 1

$y=e^{x+y}$

Taking log

$log\left(y\right)=(x+y)lo{g}_e^e$ $lo{g}_e^e=1$

Differentiating w.r.t. x

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

$\therefore \left(\frac{dy}{dx}\right)(\frac{1}{y}-1)=1$

$\therefore \left(\frac{dy}{dx}\right)=\frac{y}{1-y}=\frac{e^{x+y}}{1-e^{x+y}}$.