Question

The value of $\frac{d}{dx}{\int }_{x^2}^x\sqrt{\cos t}dt$ at $x=0$ is

Answer 1

$\frac{d}{dx}{\int }_{x^2}^x\sqrt{\cos t}dt=\sqrt{\cos x}\frac{dx}{dx}-\sqrt{\cos x^2}\frac{d}{dx}\left(x^2\right)=\sqrt{\cos x}-2x\sqrt{\cos x^2}$
For $x=1$
$\sqrt{\cos x}-2x\sqrt{\cos x^2}=\sqrt{\cos 0}-2.0\sqrt{\cos 0^2}=1$