Differentiate
$\frac{\cos x}{\log x}$
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 $)=\frac{-x \sin x(\log x)-(\cos x)}{x(\log x)^{2}}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.