Existence of derivative of a given function Given $$F(x)=\frac{1}{2x}\int_{-x}^xf(t)dt,$$ where $f:\mathbb{R}\to\mathbb{R}$ is continuous and $f'(0)$ exists, how can I prove that $F'(0)$ also exists?
 A: We must define $F(0)$ to be able to discuss $F'(0)$. By the mean value theorem for integrals, we find that $\lim_{x\to 0}F(x)=f(0)$, so it is natural to define $F(0)=f(0)$. 
Using intuition
Since $f$ is differentiable at zero, we could draw a tangent line there. Now, $F(x)$ is given by the average of $f$ over the symmetric interval $[-x,x]$. Thus, for $x$ close to zero it must be close to $f(0)$ (draw a picture), i.e. $F$ is locally approximately constant around zero, and hence its derivative at zero exists and is zero. That is $F'(0)=0$.
A proof
Since $F$ is even, we study the right-derivative at zero only. We use the definition of derivative, and write
$$
\frac{F(x)-F(0)}{x}=\frac{F(x)-f(0)}{x}=\frac{1}{2x^2}\int_{-x}^xf(t)-f(0)\,dt.
$$
Next, we note that $\int_{-x}^x tf'(0)\,dt=0$ (since the integrand is odd), so we can safely add it inside the integral. Estimating with the triangle inequality, we get
$$
\begin{aligned}
\Bigl|\frac{F(x)-F(0)}{x}\Bigr|&\leq \frac{1}{2x^2}\int_{-x}^x|t|\cdot\Bigl|\frac{f(t)-f(0)}{t}-f'(0)\Bigr|\,dt\\
&\leq \frac{1}{2x^2}\int_{-x}^x|t|\,dt\cdot \sup_{t\in[-x,x]}\Bigl|\frac{f(t)-f(0)}{t}-f'(0)\Bigr|\\
&=\frac{1}{2}\sup_{t\in[-x,x]}\Bigl|\frac{f(t)-f(0)}{t}-f'(0)\Bigr|\to 0
\end{aligned}
$$
as $x\to 0^+$. In the last step we used the fact that $f$ is differentiable at zero.
We conclude that $F$ is differentiable at $0$ and $F'(0)=0$.
A: First, $F$ is defined everywhere except for $0$. Let us extend it by continuity to a function $\tilde F$ defined in $0$ too. In order to do this, we need to show that $\lim \limits _{x \to 0} F(x)$ exists. Using L'Hospital's theorem we get
$$\lim \limits _{x \to 0} F(x) = \lim \limits _{x \to 0} \frac {\int \limits _{-x} ^x f(t) \Bbb d t} {2x} = \lim \limits _{x \to 0} \frac {f(x) + f(-x)} 2 = f(0) ,$$
so me may define $\tilde F (x) = \left\{ \begin{matrix} F(x), & x \ne 0 \\ f(0), & x = 0 \end{matrix} \right.$. We want to show that $\tilde F '(0)$ exists.
By definition,
$$\tilde F ' (0) = \lim \limits _{x \to 0} \frac {\tilde F (x) - \tilde F (0)} {x - 0} = \lim \limits _{x \to 0} \frac {F (x) - f(0)} x = \lim \limits _{x \to 0} \frac {\frac 1 {2x} \int \limits _{-x} ^x f(t) \Bbb d t - f(0)} x = \lim \limits _{x \to 0} \frac {\frac 1 {2x} \int \limits _{-x} ^x f(t) \Bbb d t - \frac 1 {2x} \int \limits _{-x} ^x f(0) \Bbb d t} x = \lim \limits _{x \to 0} \frac {\frac 1 {2x} \int \limits _{-x} ^x f(t) - f(0) \Bbb d t} x = \frac 1 2 \lim \limits _{x \to 0} \frac {\int \limits _{-x} ^x f(t) - f(0) \Bbb d t} {x^2} .$$
Apply L'Hospital's theorem once to get
$$\frac 1 2 \lim \limits _{x \to 0} \frac {f(x) - f(0) + f(-x) - f(0)} {2x} = \frac 1 4 \lim \limits _{x \to 0} \left( \frac {f(x) - f(0)} x + \frac {f(-x) - f(0)} x \right) = \frac 1 4 \lim \limits _{x \to 0} \left( \frac {f(x) - f(0)} x - \frac {f(-x) - f(0)} {-x} \right) = \frac 1 4 (f'(0) - f'(0)) = 0 .$$
To conclude, $\tilde F'(0) = 0$.
A: Since $f'(0)$ exists, $f(x)=f(0)+xf'(0)+o(x)$, where $o(x^n)$ 
is a function such that $\lim_{x\to0}o(x^n)=0$ and 
$\lim_{x\to0}o(x^n)/x^n$ exists (is not infinite).
Therefore
$$F(x)=\frac1{2x}\int_{-x}^x\left[f(0)+tf'(0)+o(t)\right]\mathrm dt.$$
The integral gives
$$F(x)=\frac1{2x}\left[2x\,f(0)+0+o(x^2)\right]=f(0)+o(x)$$
from that we conclude that $F(0)=f(0)$ and $F'(0)$ exists and $F'(0)=0$. 
