If φ(x)=∫limits_1 / x^√x sin ≤ft(t^2) d t, then φ '(1) is equal to

By Leibnitz's rule, we have

$\phi'(x) =\frac{1}{2 \sqrt{x}} \sin x-\left(-\frac{1}{x^2}\right) \sin \frac{1}{x^2}$

$\phi'(1)=\frac{1}{2} \sin 1+\sin 1=\frac{3}{2} \sin 1$