Question:
If $e^{x+y}-x=0$, prove that $\frac{d y}{d x}=\frac{1-x}{x}$
Solution:
$e^{x+y}-x=0$
$e^{x+y}=x \ldots \ldots$ (i)
Differentiating it with respect to $x$ using chain rule,
$\frac{d}{d x}\left(e^{x+y}\right)=\frac{d}{d x}(x)$
$e^{x+y} \frac{d}{d x}(x+y)=1$
$1+\frac{d y}{d x}=\frac{1}{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x}}-1$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1-\mathrm{x}}{\mathrm{x}}$
Hence Proved.