Solve this

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.

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