Question

If $f\left(x\right)={\int }_{x^2}^{x^3}\log tdt$ $(x>0)$, then $f’\left(x\right)$is equal to.

• A. $(9x^2–4x)\log x$
• B. $(4x–9x^2)\log x$
• C. $(9x^2+4x)\log x$
• D. $(3x^2–2x)\log x$

Answer 1

The correct option is A

$(9x^2–4x)\log x$

Explanation for correct answer:

Find the value of $f’\left(x\right)$:

Given that,

$f\left(x\right)={\int }_{x^2}^{x^3}\log tdt$

By applying Leibnitz's theorem.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \log x^3\frac{d}{dx}\left(x^3\right)-\log x^2\frac{d}{dx}\left(x^2\right)\left[\because \frac{d}{dx}\left[{\int }_{a\left(x\right)}^{b\left(x\right)}f\left(x,t\right)dt\right]=f\left(x,b\left(x\right)\right)\frac{db\left(x\right)}{dx}-f\left(x,a\left(x\right)\right)\frac{da\left(x\right)}{dx}\right]\\ & =& 3x^2\log x^3-2x\log x^2\\ & =& 9x^2\log x-4x\log x\\ & =& \left(9x^2-4x\right)\log x\end{array}$

Hence, the correct option is A.