Q:

Rate of change taylor series

I am wondering how this is derived:
$$f(x+h)-f(x)=R(x,h)+\frac{1}{2} h^2 f''(x)+h f'(x)$$
I want to fully understand what is going on here, I would like to see the step by step derivation of this expression. 

A:

Start with the Taylor series for $f(x)$
$$f(x) = f(x_{0}) + f'(x_{0})(x-x_{0}) + \frac{1}{2!} f''(x_{0})(x-x_{0})^{2} + R(x,h)$$
with $R(x,h)$ your higher order terms. Now, you are expanding around the point $x_{0} = x$. Then, as the original argument for our function $f$ is actually $x+h$, we map $x \mapsto x+h$ and hence
\begin{align}
f(x+h) &= f(x) + f'(x)(x+h-x) + \frac{1}{2!} f''(x)(x+h-x)^{2} + R(x,h) \\
&= f(x) + hf'(x) + \frac{1}{2!} h^{2}f''(x) + R(x,h)
\end{align}
rearranging, we find
$$f(x+h) - f(x) = hf'(x) + \frac{1}{2!} h^{2}f''(x) + R(x,h)$$

