Finding second derivative of integral Here is the problem I'm looking at:
Given $f: \mathbb{R} \to \mathbb{R}$ is differentiable, define the function
$$ H(x) = \int_{-x}^x [f(t)+f(-t)] dt \text{ } \text{ } \text{  for all x}$$
Find $H''(x)$

Now here's my crack at the solution. Is this right?
$H'(x) = \displaystyle\frac{d}{dx} \displaystyle\int_{-x}^x f(t) + f(-t) dt = [x+(-x)] - [(-x) +x] = 0$
$H''(X) = \displaystyle\frac{d}{dx} H'(X) = \displaystyle\frac{d}{dx} 0 = 0$
 A: By the Second Fundamental Theorem of Calculus,
$$\frac{d}{dx}\int_a^x g(t)\,dt = g(x).$$
Using the Chain Rule,
$$\frac{d}{dx}\int_a^{h(x)}g(t)\,dt = g(h(x))h'(x).$$
So
$$\begin{align*}
H'(x) &= \frac{d}{dx}\int_{-x}^x (f(t)+f(-t))\,dt\\
&= \frac{d}{dx}\left(\int_0^x(f(t)+f(-t))\,dt - \int_0^{-x}(f(t)+f(-t))\,dt\right)\\
&= \left(f(x) + f(-x)\right) -\Bigl(\bigl( f(-x) + f(-(-x))\bigr)(-x)'\Bigr)\\
&= f(x) + f(-x) + f(-x) + f(x) \\
&= 2f(x) + 2f(-x).
\end{align*}$$
Therefore,
$$H''(x) = \frac{d}{dx}(2f(x)+2f(-x)) = 2f'(x) + 2f'(-x)(-x)' = 2(f'(x) - f'(-x)).$$
A: It is definitively not right. When you compute $H'(x)$, you get
$$
\begin{align}
H'(x) & = \lim_{h \to 0} \frac{H(x+h) - H(x)}{h} \\
& = \lim_{h \to 0} \frac{ \int_{-x-h}^{x+h} [f(t)+f(-t)] \, dx - \int_{-x}^x [f(t)+f(-t)] \, dx }{h} \\
& = \lim_{h \to 0} \frac{ \int_{x}^{x+h} [f(t)+f(-t)] \, dx + \int_{-x-h}^{-x} [f(t)+f(-t)] \, dx }h  \\
& = [f(x) + f(-x)] + [f(-x) + f(-(-x))] = 2f(x) + 2f(-x). \\
\end{align}
$$
by the fundamental theorem of calculus. Thus 
$$
H''(x) = 2f'(x) -2f'(-x).
$$
Hope that helps,
