Differentiate

To find: Differentiation of $\frac{e^{x}}{\left(1+x^{2}\right)}$

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

(iii) $\frac{d x^{n}}{d x}=n x^{n-1}$

Let us take $u=e^{x}$ and $v=\left(1+x^{2}\right)$

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

$v^{\prime}=\frac{d v}{d x}=\frac{d\left(1+x^{2}\right)}{d x}=2 x$

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)

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

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

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

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