Differentiate

To find: Differentiation of $\frac{\cos x}{\log x}$

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) $\frac{d \cos x}{d x}=-\sin x$

(iii) $\frac{d \log x}{d x}=\frac{1}{x}$

Let us take $u=(\cos x)$ and $v=(\log x)$

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

$v^{\prime}=\frac{d v}{d x}=\frac{d(\log x)}{d x}=\frac{1}{x}$

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)

$\left[\frac{\cos x}{\log x}\right]^{\prime}=\frac{-\sin x \times(\log x)-(\cos x) \times\left(\frac{1}{x}\right)}{(\log x)^{2}}$

$=\frac{-x \sin x(\log x)-(\cos x)}{x(\log x)^{2}}$

Ans $)=\frac{-x \sin x(\log x)-(\cos x)}{x(\log x)^{2}}$