Question

If $f\left(x\right)={\int }_{x^2}^{x^4}\sin \sqrt{t}dt$, then $f^{\prime }\left(x\right)$ is equal to

• A. $\sin x^2-\sin x$
• B. $4x^3\sin x^2-2x\sin x$
• C. $x^2\sin x^2-x\sin x$
• D. $\sin x-\cos x$

Answer 1

Answer: (B)

$4x^3\sin x^2-2x\sin x$

According Leibniz integral rule-

$\frac{d}{dx}{\int }_{a\left(x\right)}^{b\left(x\right)}f\left(x\right)dx$

$=f\left(b\right(x\left)\right)\frac{d}{dx}b\left(x\right)-f\left(a\right(x\left)\right)\frac{d}{dx}a\left(x\right)$

Hence,

$f^{{}^{\prime }}\left(x\right)=\sin \sqrt{x^4}\times \frac{d}{dx}{x}^4-\sin \sqrt{x^2}\times \frac{d}{dx}{x}^2$

$⟹f^{{}^{\prime }}\left(x\right)=4x^3\sin x^2-2x\sin x$

Hence, answer is option-(B).