Question

If $f\left(x\right)={\int }_{2x}^{\sin x}\cos \left(t^3\right)dt,$ then $f^{\prime }\left(x\right)$ is equal to :

• A. $\cos \left({\sin }^{3}x\right)\cos x-1\cos \left(8x^3\right)$
• B. $\sin \left({\sin }^{3}x\right)\sin x-2\sin \left(8x^3\right)$
• C. $\cos \left({\cos }^{3}x\right)\cos x-2\cos \left(8x^3\right)$
• D. $\cos \left({\sin }^{3}x\right)-\cos \left(8x^3\right)$
• E. $\sin \left({\sin }^{3}x\right)\cos x-2\sin \left(8x^3\right)$

Answer 1

The correct option is A $\cos \left({\sin }^{3}x\right)\cos x-1\cos \left(8x^3\right)$

Given, $f\left(x\right)={\int }_{2x}^{\sin x}\cos t^{3}dt$
Using Leibnitz's rule, we get
$f^{\prime }\left(x\right)=\cos \left({\sin }^{3}x\right)\frac{d}{dx}\left(\sin x\right)-\cos \left(2x{)}^3\frac{d}{dx}\right(2x)$
$=\cos \left({\sin }^{3}x\right)-\cos 8x^3\left(2\right)$