Question

If $f\left(x\right)=\underset{x}{\overset{x^2}{\int }}\frac{1}{\log t}dt$ (where $x>0$), then the value of $f^{\prime }\left(x\right)$ is

• A. $\frac{x^2}{\log x}$
• B. $\frac{x}{\log x}$
• C. $\frac{x-1}{\log x}$
• D. $\frac{x+1}{\log x}$

Answer 1

The correct option is C $\frac{x-1}{\log x}$

Given : $f\left(x\right)=\underset{x}{\overset{x^2}{\int }}\frac{1}{\log t}dt$
Differentiating on both sides w.r.t. $x$(using newton leibnitz theorem)
$f^{\prime }\left(x\right)=\frac{d}{dx}\underset{x}{\overset{x^2}{\int }}\frac{1}{\log t}dt$
$\Rightarrow f^{\prime }\left(x\right)=\frac{d}{dx}\left(x^2\right)\cdot \frac{1}{\log x^2}-\frac{d}{dx}\left(x\right)\cdot \frac{1}{\log x}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{2x}{2\log x}-\frac{1}{\log x}$
$\therefore f^{\prime }\left(x\right)=\frac{x-1}{\log x}$