Determinate $d^{k}f_{c}$ for function $f(x,y)$ with k and c given In my Calculus 3 class I have been given following problem:
Determinate: $d^{k}f_{c}$ for function $f(x,y)$ where $c = (c1,c2)$ and $k = 2$.
My problem is that I have no clue what does $d^{k}f_{c}$ notation mean, since I have never seen it before.
 A: If $M$ is a real smooth manifold of dimension $n$, you can define a linear application on the space of $r$-forms
$$d^r:\Omega^r(M)\to\Omega^{r+1}(M)$$
called $\textit{exterior derivative}$. In local coordinates, for a smooth chart $(U_{\alpha},\phi_{\alpha})$, the general expression of $d^r$ is given by
$$d^r(\omega\vert_{U_{\alpha}})=d^r\left(\sum_I\omega_{\alpha,I}dx_{\alpha}^I \right)=\sum_Id^r(\omega_{\alpha,I})\wedge dx_{\alpha}^I.$$
In your case you have a smooth function $f_c(x,y)$ which is a zero-form, so an element of $\Omega^0(\mathbb R^2)$ (or $\Omega^0(U)$, with $U\subset \mathbb R^2$). In this case you'll have a unique representation of the $k$-th exterior derivative of the function since $\mathbb R^2$ has a global chart given by the coordinates $(x,y)$. The first exterior derivative maps $f_c\mapsto d^1f_c\in\Omega^1(\mathbb R^2)$, in particular
$$d^1f_c(x,y)=\frac{\partial f_c(x,y)}{\partial x}dx+\frac{\partial f_c(x,y)}{\partial y}dy\text{ and }\\
\Omega^2(\mathbb R^2)\ni d^2f_c(x,y)=d^1(d^1f_c(x,y))=d^1\left (\frac{\partial f_c(x,y)}{\partial x}dx\right )+d^1\left (\frac{\partial f_c(x,y)}{\partial y}dy\right)=\dots$$
Is the problem more clear now?
