Is discrete cosine transform differentiable?
The Discrete Cosine Transform of a discrete sequence $x$ can be defined as:

$$
{\tt DCT(}x{\tt )} = X_k = \frac{1}{2} (x_0 + (-1)^k x_{N-1}) 
 + \sum_{n=1}^{N-2} x_n \cos \left[\frac{\pi}{N-1} n k \right] \quad \quad k = 0, \dots, N-1.
$$

If, by differentiable, you mean:
$$
\frac{\partial {\tt DCT(}x{\tt )}}{\partial x}
$$
then isn't it just
$$
\frac{\partial {\tt DCT(}x{\tt )}}{\partial x} = \frac{\partial X_k}{\partial x} =\left [ 
\begin{array}{ccccc}
\frac{1}{2} &amp; \cdots &amp; \cos \left[\frac{\pi}{N-1} n k \right] &amp; \cdots &amp; (-1)^k\frac{1}{2}
\end{array}
\right ]
$$ 
where $x = [ x_0, x_1, \ldots, x_{N-2}, x_{N-1} ]^T$ ?

What do you mean by a "mathematical proof"?