Derivative of Integral with chain rule I need to find the derivative (respect to $x$) of the function
$$
G(x)=\int _{ { e }^{ x } }^{ sin(x) }{ ln( } { y }^{ 2 })dy
$$
By FTC, $A'(x)=ln({ x }^{ 2 })$.
So far I have
$$
G'(x)=\frac { d }{ dx } (A(\sin { x } ))-A({ e }^{ x }))\\ G'(x)=A'(\sin { x } )(\sin { x } )'-A'({ e }^{ x })({ e }^{ x })'\\ G'(x)=ln(\sin ^{ 2 }{ x } )(\cos { x } )-ln({ e }^{ 2x })({ e }^{ x })
$$
Can I apply the same method to compute the derivative of
$$
F(x)=\int _{ sin(x) }^{ cos(x) }{ \frac { 1 }{ \sqrt { 1-{ u }^{ 2 } }  }  } du
$$
Any help would be appreciated. 

The answer is yes.  It's also worth pointing out that
$$
    \int_0^y \frac{1}{\sqrt{1-u^2}}\,du = \arcsin y
$$
So there ought to be considerable simplification in your derivative.
