Differentiate:

Solution:

To find: Differentiation of $e^{x} \cot x$

Formula used: (i) (uv) $^{\prime}=u^{\prime} v+u v^{\prime}$ (Leibnitz or product rule)

(ii)

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

(iii)

$\frac{d \cot x}{d x}=-\operatorname{cosec}^{2} x$

Let us take $u=e^{x}$ and $v=\cot x$

$u^{\prime}=\frac{d u}{d x}=\frac{d e^{x}}{d x}=e^{x}$

$v^{\prime}=\frac{d v}{d x}=\frac{d \cot x}{d x}=-\operatorname{cosec}^{2} x$

Putting the above obtained values in the formula:-

$(u v)^{\prime}=u^{\prime} v+u v^{\prime}$

$\left(e^{x} \cot x\right)^{\prime}=e^{x} x \cot x+e^{x} x-\operatorname{cosec}^{2} x$

$=e^{x} \cot x-e^{x} \operatorname{cosec}^{2} x$

$=e^{x}\left(\cot x-\operatorname{cosec}^{2} x\right)$

Ans) $e^{x}\left(\cot x-\operatorname{cosec}^{2} x\right)$