How to approach integrals as a function? I'm trying to solve the following question involving integrals, and can't quite get what am I supposed to do:
$$f(x) = \int_{2x}^{x^2}\root 3\of{\cos z}~dz$$
$$f'(x) =\ ?$$
How should I approach such integral functions? Am I just over-complicating a simple thing?
 A: For this problem, you will ultimately use a version of the Fundamental Theorem of Calculus: If $f$ is continuous, then the function $F$ defined by $F(x)=\int_a^x f(z)\,dz$ is differentiable and $F'(x)=f(x)$.
So for instance, for $F(x)=\int_0^x\root3\of{\cos z}\,dz$, we have $F'(x)=\root3\of{\cos x}$.
One can combine this with the chain rule, when it applies, to differentiate a function whose rule is of the form $F(x)=\int_a^{g(x)} f(z)\,dz$. Here, we recognize that $F$ is a composition of the form $F=G\circ g$ with $G(x)=\int_a^x f(z)\,dz$.  The  derivative is $F'(x)=\bigl[ G(g(x))\bigr]'=G'(g(x))\cdot g'(x)=f(g(x))\cdot g'(x)$.
For example, for $F(x)=\int_0^{x^2}\root3\of{\cos z}\,dz$, we have $F'(x)=\root3\of{\cos x^2}\cdot(x^2)'=2x\root3\of{\cos x^2} $.
Now to tackle your problem proper and take advantage of these rules, we just "split the integral":
$$\tag{1}
\int_{2x}^{x^2}\root3\of{\cos z}\,dz=
\int_{2x}^{0}\root3\of{\cos z}\,dz+
\int_{0}^{x^2}\root3\of{\cos z}\,dz.
$$
But wait! We can only use the aforementioned differentiation rules for functions defined by an integral when it's the upper limit of integration that is the variable.  The first integral in the right hand side of $(1)$ does not satisfy this.  Things are easily remedied, though; write the right hand side of $(1)$ as:
$$
-\int_{0}^{2x}\root3\of{\cos z}\,dz+
\int_{0}^{x^2}\root3\of{\cos z}\,dz;
$$
and now things are set up to use our rule (of course, you'll also use the rule $[cf+g]'=cf'+g'$).
A: Not to give it away completely. Using the Fundamental Theorem of Calculus, $f(x) = C(x^2)-C(2x)$, where $C(x)$ is the anti-derivative of the integrand. Now, use the Chain rule to compute $f'(x)$, which will depend only on the $C'(x)$, which is the integrand itself, evaluated at $x$.
A: \begin{eqnarray}
f'(x)&=&(x^2)'(\sqrt[s]{\cos z})|_{z=x^2}-(2x)'(\sqrt[s]{\cos z})|_{z=2x}\cr
&=&2x\sqrt[s]{\cos x^2}-2\sqrt[s]{\cos 2x}
\end{eqnarray}
A: Generally, to differentiate an integral of the form:
$$\int_{g_1(x)}^{g_2(x)}f(z)dz$$
we use Leibniz rule.  First assume that $F(x)$ is the anti-derivative of $f(x)$.  That is $F'(x) = f(x)$. Then it follows that,
$$\int_{g_1(x)}^{g_2(x)}f(z)dz = F(z)|_{z=g_2(x)} - F(z)|_{z=g_1(x)} = F(g_2(x)) - F(g_1(x))$$
Now if we differentiate this result using the chain rule we get: 
$$\frac{d}{dx}\left(F(g_2(x)) - F(g_1(x))\right) = f(g_2(x))g_2'(x) - f(g_1(x))g_1'(x).$$
Note that it is not necessary to find the anti-derivative $F()$.
A: Using the Leibnitz rule of differentiation of integrals, 
which states that if
\begin{align}
f(x) = \int_{a(x)}^{b(x)} g(y) \ dy,
\end{align}
then
\begin{align}
f^{\prime}(x) = g(b(x)) b^{\prime}(x) - g(a(x)) a^{\prime}(x).
\end{align}
Thus, for your problem $a^{\prime}(x) = 2$ and $b^{\prime}(x) = 2x$ and, therefore, 
\begin{align}
f^{\prime}(x) = \int_{2x}^{x^2} \sqrt[3]{\cos z} dz = \sqrt[3]{\cos (x^2)} (2 x) - \sqrt[3]{\cos (2x)} (2).
\end{align}
