If $\displaystyle p:\mathbb{R}^n\to\mathbb{R}^m$ is a linear map plus some constant and $\displaystyle f:A\subset\mathbb{R}^m\to\mathbb{R}^s$ is $\displaystyle k$ times differentiable, prove that:

$\displaystyle \textbf{D}^k(f\circ p)(x_0)(x_1,...,x_k)=\textbf{D}^k f(p(x_0))(\textbf{D}p(x_0)(x_1),...,\textbf{D}p(x_ 0)(x_k))$

So I started by induction. Using the chain rule,

$\displaystyle \textbf{D}(f\circ p)(x_0)(x_1)=(\textbf{D}f(p(x_0))\textbf{D}p(x_0)) (x_1)$

Terrific. So next I assumed that the $\displaystyle \textbf{D}^k$ statement was true. Proceeding for $\displaystyle \textbf{D}^{k+1}$, I looked at:

$\displaystyle \textbf{D}\bigg(\textbf{D}^k(f\circ p)(x_0)(x_1,...,x_{k+1})\bigg)$

Here's where I run into trouble though. I would like to use the induction hypothesis on the stuff inside the big parenthesis, but the fact that there are now $\displaystyle k+1$ components is throwing me off. Another thing that's not helping is the fact that I'm not really comfortable with this notation. I understand that $\displaystyle \textbf{D}^k(f\circ p)(x_0)$ is the $\displaystyle k$th derivative of $\displaystyle f\circ p$ evaluated at the point $\displaystyle x_0$, and I'm applying it to $\displaystyle (x_1,...,x_k)$ but is $\displaystyle (x_1,...,x_k)$ a vector, or are $\displaystyle x_1,...,x_k$ each vectors on their own?

So that's where I stand on this problem. Any help is appreciated.