If $\phi(x)=\int\limits_{1 / x}^{\sqrt{x}} \sin \left(t^2\right) d t$, then $\phi '(1)$ is equal to

sin 1 2 sin 1 (3/2) sin 1 none of these

(3/2) sin 1

By Leibnitz's rule, we have $\phi'(x) =\frac{1}{2 \sqrt{x}} \sin x-\left(-\frac{1}{x^2}\right) \sin \frac{1}{x^2}$ $\phi'(1)=\frac{1}{2} \sin 1+\sin 1=\frac{3}{2} \sin 1$