Differentiate

To find: Differentiation of $\frac{x^{4}}{\sin x}$

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

(iii) $\frac{d \sin x}{d x}=\cos x$

Let us take $u=\left(x^{4}\right)$ and $v=(\sin x)$

$u^{\prime}=\frac{d u}{d x}=\frac{d\left(x^{4}\right)}{d x}=4 x^{3}$

$v^{\prime}=\frac{d v}{d x}=\frac{d(\sin x)}{d x}=\cos x$

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)

$\left[\frac{x^{4}}{\sin x}\right]=\frac{4 x^{3} \times(\sin x)-\left(x^{4}\right) \times(\cos x)}{(\sin x)^{2}}$

$=\frac{x^{3}[4(\sin x)-x(\cos x)]}{(\sin x)^{2}}$

Ans $)=\frac{x^{3}[4(\sin x)-x(\cos x)]}{(\sin x)^{2}}$