Question

If $y=1+x.e^y$, show that $\frac{dy}{dx}=\frac{e^y}{2-y}$

Answer 1

Given $y=1+x.e^y$,

$⟹\frac{y-1}{e^y}=x$,

Now differentiating the equation with respect to $y$ on both sides gives,

$\frac{e^y-(y-1)e^y}{e^{2y}}=\frac{dx}{dy}⟹\frac{2-y}{e^y}=\frac{dx}{dy}⟹\frac{dy}{dx}=\frac{e^y}{2-y}$