Calculate the (variational) derivative of the following equation; Consider $ E[u]= \int^1_0 \big(u'(x)\big)^2+\big(u(x)\big)^2-2f(x)u(x) dx.$
Calculate the variational derivation for a function $v$;
in other words,
calculate $\frac{d}{d\epsilon}E[u+\epsilon v]$ at $\epsilon=0$.
*$\frac{d}{d\epsilon}$ is the derivative with respect to $\epsilon$. 
My work:
$\frac{d}{d\epsilon}E[u+\epsilon x] = \frac{d}{d\epsilon}\int^1_0(u+\epsilon v)'(x)^2+(u+\epsilon v)(x)^2-2f(x)(u+\epsilon v)(x) dx$ .
I moved the derivative sign inside the integral and split the integral up, like so:
$\int^1_0\frac{d}{d\epsilon}(u+\epsilon v)'(x)^2 dx+\int^1_0\frac{d}{d\epsilon}(u+\epsilon v)(x)^2-2\int^1_0\frac{d}{d\epsilon}f(x)(u+\epsilon v)(x) dx$.
Here is where I get stuck; I'm not sure how to take the derivatives and simplify.
 A: It may be very unclear what stands for
$$
\frac{d}{d\epsilon}(u+\varepsilon v)'(x)^2
$$
Let's put the $x$ argument along with the derivative inside. Since the derivative is taken with respect to $x$, $\epsilon$ stands as a constant term.
$$
\frac{d}{d\epsilon}\big(u'(x)+\varepsilon v'(x)\big)^2.
$$
Now it is clear that
$$
\frac{d}{d\epsilon}(u'(x)+\varepsilon v'(x))^2 = 
\frac{d}{d\epsilon}\Big(\big(u'(x)\big)^2+2\varepsilon u'(x)v'(x) + \epsilon^2\big(v'(x)\big)^2\Big) = 2 u'(x) v'(x) + 2 \epsilon \big(v'(x)\big)^2.
$$
When $\epsilon = 0$ the last term vanishes. So 
$$
\left.\frac{d}{d\epsilon} \int_0^1 (u'(x) + \epsilon v'(x))^2 dx\right|_{\epsilon=0} = \int_0^1 2 u'(x) v'(x) dx
$$
This integral is often taken by parts to eliminate $v'(x)$:
$$
\int_0^1 2 u'(x) v'(x) dx = 2u'(x)v(x)\Big|_0^1 - 2\int_0^1 u''(x)v(x)dx.
$$
I hope that you can finish the exercise by yourself now. The final answer should have the form of
$$
\frac{d}{d\epsilon}E[u+\epsilon v] = 2u'(x)v(x)\Big|_0^1 + \int_0^1 [\bullet] v(x) dx
$$
Addition. The variational derivative is 
$$
\frac{d}{d\epsilon}E[u+\epsilon v] = 2u'(x)v(x)\Big|_0^1 + \int_0^1 [-2u''(x)+2u(x)-2f(x)] v(x) dx.
$$
Now we want to solve 
$$
\frac{d}{d\epsilon}E[u+\epsilon v]\big|_{\epsilon = 0} = 0
$$
for every test function $v(x)$. Let's start with $v(x) = \delta(x - a), 0 < a < 1$.
The $2u'(x)v(x)\Big|_0^1$ term vanishes since $v(0) = v(1) = 0$.
$$
\int_0^1 [-2u''(x)+2u(x)-2f(x)] \delta(x-a) dx = -2u''(a)+2u(a)-2f(a) = 0
$$
It means that for every inner point of $[0, 1]$ the following equation should hold
$$
u''(x) - u(x) = f(x),\qquad 0 < x < 1.
$$
If $u(x)$ satisfies this equation then the last integral in the variational derivatives is zero, i.e. for arbitrary $v(x)$
$$
\frac{d}{d\epsilon}E[u+\epsilon v]\big|_{\epsilon = 0} =  2u'(x)v(x)\Big|_0^1 = 2u'(1) v(1) - 2u'(0)v(0).
$$
By choosing $v(x) = x$ and $v(x) = 1 - x$ both of the $u'(0)$ and $u'(1)$ should be zero.
Result. 
$$
\frac{d}{d\epsilon}E[u+\epsilon v]\big|_{\epsilon = 0} = 0
$$
for every $v(x)$ is equivalent to following problem for $u(x)$:
$$
u''(x) - u(x) = f(x),\qquad 0 < x < 1\\
u'(0) = u'(1) = 0
$$
