Differentiate

To find: Differentiation of $\frac{e^{x}}{(1+x)}$

Formula used: (i) $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

(ii) $\frac{d e^{x}}{d x}=e^{x}$

Let us take $u=e^{x}$ and $v=(1+x)$

$u^{\prime}=\frac{d u}{d x}=\frac{d\left(e^{x}\right)}{d x}=e^{x}$

$v^{\prime}=\frac{d v}{d x}=\frac{d(1+x)}{d x}=1$

Putting the above obtained values in the formula:-

$\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

$\left(\frac{e^{x}}{(1+x)}\right)^{\prime}=\frac{e^{x} \times(1+x)-e^{x} \times 1}{(1+x)^{2}}$

$=\frac{x e^{x}}{(1+x)^{2}}$

Ans $)=\frac{x e^{x}}{(1+x)^{2}}$