How to find the derivative of $F(x)=\int_{x^2}^{4x^2} \sin \sqrt t\;\;dt$? For a real number $t&gt;0$, let $\sqrt t$ denote the positive square root of t.  For a real number $x&gt;0$, let $F(x)=\int_{x^2}^{4x^2} \sin \sqrt t\;\;dt$.  If $F'$ is the derivative of $F$, then 


  
    (a). $F'(\frac \pi 2)=0$
    
    (b). $F'(\frac \pi 2)=\pi$
    
    (c). $F'(\frac \pi 2)=-\pi$
    
    (d). $F'(\frac \pi 2)=2\pi$
  


I am guessing that I could use Fundamental theorem of integral calculus. But I couldn't construct a function whose derivative is $\sin \sqrt t$. If I through I could find the answer.
Hint: 

$$
F(x)=\int_{x^2}^{4x^2} \sin \sqrt t\;\;dt = \int_{0}^{4x^2} \sin \sqrt t\;\;dt - \int_{0}^{x^2} \sin \sqrt t\;\;dt
$$

We could define $$G(x) = \int_0^ x \sin \sqrt t\;\;dt$$ and rewrite as
$$
F(x) = G(4x^2) - G(x^2).
$$

By Chain Rule,
$$
F^\prime(x) = G^\prime(4x^2)(8x) - G^\prime(x^2)(2x).
$$
By FTC,
$$G^\prime(x) = \sin \sqrt x.$$

Is it clear how to find $F^\prime(\frac{\pi}{2})$ from here?
You do not need to solve the integral because you can use the Second Fundamental Theorem of Calculus.

With that the integral is cancelled by the derivative but you need to use the composition of functions.

If we call the function that is inside the integral f(t) and the upper limit v(x) and the lower limit u(x). The solution is

((derivative v(x)) f(v(x)) ) -  ((derivative u(x)) f(u(x)) )=

(8x)(sin sqrt(4x*2)) - (2x)(sin sqrt(x*2))=

8x sin(2x) - 2x sin(x) 

The solution works for any x value.

If you need to evaluate the result at pi/2, you just plug it instead of the x.

4pi sin(pi) - pi sin(pi/2)= 

negative pi