Q. If $F(x) = \int_{x^{2}}^{x^{3}}{\log{t\ dt,\ (x > 0),\ }}$ then $F^{'}(x)$ is equal to
A) $\left( 9x^{2} - 4x \right)\log x$
B) $\left( 4x - 9x^{2} \right)\log x$
C) $\left( 9x^{2} + 4x \right)\log x$
D) None of these
Explanation
(a)
$F(x) = \int_{x^{2}}^{x^{3}}{\log{t\ dt}}$
Applying Leibnitz theorem, we get
$F^{'}(x) = \log{x^{3} \bullet \frac{d}{dx}x^{3} - \log{x^{2} \bullet \frac{d}{dx}x^{2}}}$
$= 3\log{x \bullet 3x^{2} - 2\log{x \bullet 2x}}$
$= \left( 9x^{2} - 4x \right)\log x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.