Question

If $e^{x+y}-x=0$, prove that $\frac{dy}{dx}=\frac{1-x}{x}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},e^{x+y}-x=0 \\ \Rightarrow e^{x+y}=x...\left(1\right) \end{gathered}$$
Differentiating with respect to x using chain rule,
$$\begin{gathered} \frac{d}{dx}\left(e^{x+y}\right)=\frac{d}{dx}\left(x\right) \\ \Rightarrow e^{x+y}\frac{d}{dx}\left(x+y\right)=1 \\ \Rightarrow x\left[1+\frac{dy}{dx}\right]=1\left[\mathrm{Using}\mathrm{equation}\left(\mathrm{i}\right)\right] \\ \Rightarrow 1+\frac{dy}{dx}=\frac{1}{x} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{x}-1 \\ \Rightarrow \frac{dy}{dx}=\frac{1-x}{x} \end{gathered}$$