If $y=1+x e^y$, then $\frac{d y}{d x}=$

If $y=1+x e^y$, then $\frac{d y}{d x}=$
  1. $\frac{\mathrm{e}^{\mathrm{y}}}{2-\mathrm{y}}$
  2. $\frac{e^y}{2+y}$
  3. $\frac{\mathrm{e}^{\mathrm{y}}}{1-\mathrm{e}^{\mathrm{y}}}$
  4. $\frac{\mathrm{e}^{\mathrm{y}}}{1+\mathrm{e}^{\mathrm{y}}}$

Solution

$\begin{aligned} & \mathrm{y}=1+\mathrm{xe}^{\mathrm{y}} \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=0+\mathrm{xe}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{e}^{\mathrm{y}} \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{xe}^{\mathrm{y}}-1\right)=-\mathrm{e}^{\mathrm{y}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{e}^{\mathrm{y}}}{\mathrm{xe}^{\mathrm{y}}-1} \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{e}^{\mathrm{y}}}{\left(1+\mathrm{xe}^{\mathrm{y}}\right)-2}=\frac{-\mathrm{e}^{\mathrm{y}}}{\mathrm{y}-2}=\frac{\mathrm{e}^{\mathrm{y}}}{2-\mathrm{y}}\end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 2)

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