Derivative of the Prox / Proximal Operator Consider a proximal operator,
$$ \operatorname{Prox}_{ \lambda f( u ) } \left( x \right) = \arg \min_{u} \lambda f \left( u \right) + \frac{1}{2} {\left\| u - \mu x \right\|}_{2}^{2}.$$
What is the partial derivative of the proximal operator w.r.t. $\lambda$ and $\mu$, i.e.
$$\frac{\partial\operatorname{Prox}_{ \lambda f( u ) } \left( x \right)}{\partial\lambda}, \quad \frac{\partial\operatorname{Prox}_{ \lambda f( u ) } \left( x \right)}{\partial\mu}?$$
If the general case is not solvable, is it possible to compute the derivative if we restrict $f$ to be an $L_p$ norm?
 A: The prox operator takes a point (vector) and maps it into a subset of your vector space, this mapping might be empty, a singleton or a set. Therefore the prox operator is not differentiable.
The following example is from the book by Beck. Consider the following functions:
\begin{align}
g_1(x) &=0, \\
g_2(x)&=\begin{cases} 
0 & x \neq 0\\
- c & x=0,
\end{cases}\\
g_3(x)&=\begin{cases} 
0 & x \neq 0\\
 c & x=0,
\end{cases}
\end{align}
then the prox of the previous functions is:
\begin{align}
\text{prox}_{g_1}(x)&=\{x\}.\\
\text{prox}_{g_2}(x)&=\begin{cases} 
\{0\}, & |x| < \sqrt{2c},\\
 \{x\}, & |x| > \sqrt{2c}, \\
\{0,x\}, & |x| = \sqrt{2c}.
\end{cases}\\
\text{prox}_{g_3}(x)&=\begin{cases} 
\{0\} & x \neq 0,\\
 \emptyset & x=0.
\end{cases}
\end{align}
On the other hand, the Moreau envelope, defined as
$$M^{\mu}_f(x) = \inf_{y}\bigg\{f(y)+\frac{1}{2\mu} ||x-y||^2 \bigg\},$$
is a smooth map (in fact $\mu$ is called the smoothing parameter), therefore it makes sense to talk about differentiability. The derrivate of the Moreau envelope is
$$\nabla M^{\mu}_f(x) = \frac{1}{\mu}(x - \text{prox}_{\mu f}(x)).$$
You can read more on the excellent books by Beck (Ch. 6) and Bauschke & Combettes (Ch. 12).
A: For the restricted case where $f$ is differentiable one can derive a solution. First, the derivative w.r.t. to $\lambda$ is
$$\frac{\partial\operatorname{Prox}_{ \lambda f( u ) } \left( x \right)}{\partial\lambda} = \lim_{\epsilon\to 0}\frac{1}{\epsilon}\left[\operatorname{Prox}_{ (\lambda + \epsilon) f( u ) } \left( x \right) - \operatorname{Prox}_{ \lambda f( u ) } \left( x \right)\right]$$
The solution to $\operatorname{Prox}_{ (\lambda + \epsilon) f( u ) } \left( x \right)$ can be computed from a simple Taylor expansion. In particular, any solution has to fulfill
$$(\lambda + \epsilon) \nabla f(u) + (u - \mu x) = 0$$
$$\Leftrightarrow (\lambda + \epsilon) \nabla f(u^{*} + du) + u^{*} + du - \mu x = 0$$
where $u^{*} = \operatorname{Prox}_{ \lambda f( u ) } \left( x \right)$. Then, with $H_f(u^{*})$ being the Hessian of $f$,
$$\Leftrightarrow (\lambda + \epsilon) (\nabla f(u^{*}) + H_f(u^{*}) du) + u^{*} + du - \mu  x = 0$$
$$\Leftrightarrow \epsilon \nabla f(u^{*}) + (\lambda + \epsilon) H_f(u^{*}) du + du = 0$$
Hence,
$$du = -\epsilon\left[(\lambda + \epsilon)H_f(u^{*}) + I\right]^{-1}\nabla f(u^{*})$$
$$\Rightarrow \frac{\partial\operatorname{Prox}_{ \lambda f( u ) } \left( x \right)}{\partial\lambda} = -\left[\lambda H_f(u^{*}) + I\right]^{-1}\nabla f(u^{*})$$
In a very similar way we can find
$$\frac{\partial\operatorname{Prox}_{ \lambda f( u ) } \left( x \right)}{\partial\mu} = \left[\lambda H_f(u^{*}) + I\right]^{-1} x$$
