Question

If $y=\underset{x^2}{\overset{x^3}{\int }}\frac{1}{\ln t}dt$ $(\text{where}x\in R^{+}-\{1\left\}\right)$, then the value of $\frac{dy}{dx}$ is

• A. $x(x-1)(\ln x{)}^2$
• B. $x(x-1)(\ln x{)}^{-2}$
• C. $x(x-1)\left(\ln x\right)$
• D. $x(x-1)(\ln x{)}^{-1}$

Answer 1

The correct option is D $x(x-1)(\ln x{)}^{-1}$

$y=\underset{x^2}{\overset{x^3}{\int }}\frac{1}{\ln t}dt$
$\Rightarrow \frac{dy}{dx}=\frac{1}{\ln x^3}\cdot \frac{d}{dx}\left(x^3\right)-\frac{1}{\ln x^2}\cdot \frac{d}{dx}\left(x^2\right)$
$=\frac{3x^2}{3\ln x}-\frac{2x}{2\ln x}=x(x-1)(\ln x{)}^{-1}$