Find this Calculus question Confusing Given that $F=\int _{2x}^{x²} f(x)$\, dx 
What is the value of $F'(x)$
I thought this should just Be $f(x²)-f(2x)...$But my Class' toppper got;
 $2xf(x²)+2f(2x)...$Think he did something about implicit differentiation.
Can anyone please help me out here?
 A: Suppose you know an anti-derivative $G$ for $f$.  Then $G' = f$.  So then
$$F(x) = \int_{2x}^{x^2} f(t) \; dt = G(x^2) - G(2x).$$
Then by chain rule
$$F'(x) = G'(x^2)(2x) - G'(2x)(2) = f(x^2)2x - f(2x)2.$$
A: The notation is rather bad (invalid, even). It should have different variables for the limits and the inside, e.g. 
$$F(x)=\int _{2x}^{x^2} f(t)\, dt $$
For simplicity, I'll consider 
$$F(x)=\int _{a}^{x^2} f(t)\, dt $$
Then F(x) is the area from $a$ to $x^2$ under the curve of f(t). If you increase $x$ by $h$, then you're now going to $(x+h)^2$, or $x^2+2xh+h^2$. We can ignore the $h^2$ term as $h$ goes to zero. So this is an increase of $2xh$ in your width, at a point where the height is $f((x+h)^2)$. Again, we're taking the limit as $h$ goes to zero, so we can take the height to be $f(x^2)$. Area is height times width, so the additional area is $(2xh)f(x^2)$. The derivative is change in $y$ over change in $x$. $y$, or the area, changed by $(2xh)f(x^2)$, and $x$ changed by $h$. So the derivative is $2xf(x^2)$.
A similar analysis applies to the lower limit of $2x$.
A: I guess you mean this :
$$F=\int _{2x}^{x^2} f(t)\, dt$$
Then apply Leibniz rule
$$F'(x)=f(x^2)(x^2)'-f(2x)(2x)'$$
$$F'(x)=2xf(x^2)-2f(2x)$$
A: If $$F=\int_a^x f(x)\ dx$$ then $F'=f(x).$  This means that by the chain rule, if  $$F=\int_a^{g(x)} f(x)\ dx$$ we get $F'=f(g(x))g'(x).$  For example this will mean  if  $$F=\int_a^{x^2} f(x)\ dx$$ we get $F'=f(x^2)2x.$  This explains half the answer.  To get the full answer above, split the integral into two parts using the fact that $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx $ and use the chain rule twice.
