second derivative and transformation Let $f : \mathbb R → \mathbb R$ be twice differentiable. Find $D^2 f (x_0)$ in terms of $f ′′(x_0)$.
I do not understand what does $D^2f(x_0)$ mean. If it means second derivative, then shouldn't it be $D^2 f (x_0) = f''(x_0)$? 
Thank you all. 
 A: As I say in the comments, the distinction between $D f(x_{0})$ and $f'(x_{0})$ is easier to explain, so I'll do that. I this doesn't make the case of $D^{2}f(x_{0})$ clear, then tell me and I'll explain it.   
The definition of $f'(x_{0})$ is the following limit, if it exists
$$\lim_{h \to 0} \frac{f(x_{0}+h)-f(x)}{h}$$
The definition of $D f(x_{0})$ is that it is a linear map $\mathbb{R} \to \mathbb{R}$ such that, for any $h \in \mathbb{R}$, 
$$ f(x+h)-f(x)=Df(x_{0})(h)+o(h)$$
where $o(h)/h \to 0$ as $h \to 0$. By linearity, $Df(x_{0})(h)=hDf(x_{0})(1)$, and it follows that 
$$f'(x_{0})=\lim_{h \to 0} \left[Df(x_{0})(1)+\frac{o(h)}{h}\right]$$
Hence $f'(x_{0})=Df(x_{0})(1)$. Is that clear?  
EDIT: Ok, so what is the definition of $D^{2}f(x_{0})$? It's probably easiest to think of this as a matrix of partial derivatives. In general (given a basis $\left\{e_{i}\right\}$), if $H$ is the Hessian defined by 
$$H_{ij}=\frac{\partial^{2}f}{\partial x_{i} \partial x_{j}}$$
then $D^{2}f(x_{0})(u)(v)=H_{ij}u_{i}v_{j}$. In particular, 
$$\frac{\partial^{2}f}{\partial x_{i} \partial x_{j}}=D^{2}f(x_{0})(e_{i},e_{j})$$
So, in the case $\mathbb{R} \to \mathbb{R}$, we have $f''(x_{0})=D^{2}f(x_{0})(1,1)$
