Differentiate

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

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

Let us take u = logx and v = x

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

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

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)

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

$=\frac{1-\log x}{x^{2}}$

Ans) $=\frac{1-\log x}{x^{2}}$