Suppose that $f$ is continuous and $h\neq 0$. Let $D_hf(x)=\frac{3}{2h^3}\int_{-h}^{h}tf(x+t)dt$. (a) Show that $D_hp(x)=p'(x)$ for all polynomials Suppose that $f$ is continuous and $h\neq 0$. Let 
$$D_hf(x)=\frac{3}{2h^3}\int_{-h}^{h}tf(x+t)dt$$ 
(a) Show that $D_hp(x)=p'(x)$ for all polynomials to some degree. 
(b) Prove that if $f$ has a second continuous derivative then 
$$\lim_{h\to 0}D_hf(x)=f'(x)$$
(c) Suppose that $f$ has third continuous derivatives. Find the approximation order of $f'(x)$ by $D_hf(x)$ with respect to $h$. That is, show that 
$$|D_hf(x)-f'(x)|=O(h^r)$$
for a certain $r$.
(a) I have tried to do the following $D_hp(x)=\frac{3}{2h^3}\int_{-h}^{h}tp(x+t)dt=\frac{3}{2h^3}\int_{-h}^{h}t\sum_{i=0}^{n}a_i(x+t)^idt$, where $p(x)=\sum_{i=0}^{n}a_i(x+t)^i$, but I do not know what to follow.
(b) Could you try to prove that $D_hf(x)=\frac{f(x+h)-f(x)}{h}$ and thus $\lim_{h\to 0}D_hf(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f'(x)$? Or how could I do this?
(c) I have thought about using part (c) but I do not know in what way to do it. Could anyone help me, please? Thank you very much.

b) Integration by parts gives
\begin{align*}
&\dfrac{2h^{3}}{3}D_{h}f(x)\\
&=\dfrac{h^{2}}{2}(f(x+h)-f(x-h))-\dfrac{1}{6}(h^{3}f(x+h)+h^{3}f'(x-h))+\dfrac{1}{6}\int_{-h}^{h}t^{3}f''(x+t)dt,
\end{align*}
so
\begin{align*}
&D_{h}f(x)\\
&=\dfrac{3}{4}\dfrac{f(x+h)-f(x-h)}{h}-\dfrac{1}{4}(f'(x+h)+f'(x-h))+\dfrac{1}{4}\dfrac{1}{h^{3}}\int_{-h}^{h}t^{3}f''(x+t)dt,
\end{align*}
where change of variable gives
\begin{align*}
\dfrac{1}{h^{3}}\int_{-h}^{h}t^{3}f''(x+t)dt=h\int_{-1}^{1}u^{3}f''(x+uh)du=2h\xi_{h}^{3}f''(x+\xi_{h}h),
\end{align*}
where the last follows by Mean Value Theorem for integrals, then taking $h\rightarrow 0$, one has
\begin{align*}
D_{h}f(x)\rightarrow\dfrac{3}{4}\cdot 2f'(x)-\dfrac{1}{4}\cdot 2f'(x)=f'(x).
\end{align*}
