Differentiate:
$x^{n} \cot x$
To find: Differentiation of $x^{n} \cot x$
Formula used: (i) (uv)' = u'v + uv' (Leibnitz or product rule)
(ii)
$\frac{d x^{n}}{d x}=n x^{n-1}$
(iii)
$\frac{d \cot x}{d x}=-\operatorname{cosec}^{2} x$
Let us take $u=x^{n}$ and $v=\cot x$
$u^{\prime}=\frac{d u}{d x}=\frac{d x^{n}}{d x}=n x^{n-1}$
$v^{\prime}=\frac{d v}{d x}=\frac{d \cot x}{d x}=-\operatorname{cosec}^{2} x$
Putting the above obtained values in the formula :-
$(u v)^{\prime}=u^{\prime} v+u v^{\prime}$
$\left(x^{n} \cot x\right)^{\prime}=n x^{n-1} \times \cot x+x^{n} x-\operatorname{cosec}^{2} x$
$=n x^{n-1} \cot x-x^{n} \operatorname{cosec}^{2} x$
$=x^{n}\left(n x^{-1} \cot x-\operatorname{cosec}^{2} x\right)$
Ans) $x^{n}\left(n x^{-1} \cot x-\operatorname{cosec}^{2} x\right)$