soft thresholding derivation with two terms I am trying to minimize the following function
$$argmin_x||x-y||^2_2 + ||x-z||_2^2 + \lambda || \frac{x- w}{c}||_1$$
I have been doing it for while but i am not sure how to do it. I have already been through few of the examples for simple cases such as this one (Derivation of soft thresholding operator).
Taking the derivative w.r.t $x$ is given by
$$\frac{x{-w}}{\left|c\right|\left|x{-w}\right|}+4x{-2z}{-2y}$$,
i would like to know that how i would be able to go from here to the soft thresholding function described in the above link.
Any help regarding this would be quite useful

After applying @MichaelGrant 's variable substitution $u=(x-w)/c$ you end up with 
$$ \arg \min_u \Vert cu+w-y\Vert^2 + \Vert cu+w-z\Vert^2 + \lambda \Vert u \Vert_1$$ 
which basically looks like the standard soft-thresholding 
$$\arg \min_u (1/2)\Vert u-p\Vert^2 + \lambda \Vert u\Vert_1 $$
after some manipulation of the two other terms i.e.
$$\Vert cu+w-y\Vert^2 + \Vert cu+w-z\Vert^2 = 2 c^2 \left\Vert u-\frac{2w-y-z}{c} \right\Vert^2 + f(w,y,z) $$
and it becomes
$$
\begin{eqnarray}
\arg \min_u 2 c^2 \left\Vert u-\frac{2w-y-z}{c} \right\Vert^2 + \lambda \Vert u\Vert_1 & = &
4c^2 \arg \min_u \frac{1}{2}\left\Vert u-\frac{2w-y-z}{c} \right\Vert^2 + \frac{\lambda}{4c^2} \Vert u\Vert_1 \\ 
& = & 4c^2 S_{\lambda / 4c^2} \left( \frac{2w-y-z}{c} \right)
\end{eqnarray}
$$
where $S_\rho$ is the soft-thresholding function with parameter $\rho$.
