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