Simplify partial derivative with negative sign in operator and operand I'm trying clarify my understanding of the following expression
$\displaystyle \frac{\partial (-f(z))}{\partial (-z)}$
I believe I can treat the negative sign in the operand as a constant $-1$ and pull it out in front of the expression to give
$\displaystyle -\frac{\partial f(z)}{\partial (-z)}$
What I'm not sure about is the negative sign in the operator. Can I also pull a negative sign out there? In other words, does my initial expression simplify to 
$\displaystyle \frac{\partial (-f(z))}{\partial (-z)} = \frac{\partial f(z)}{\partial z}$
or do I have to keep the operator as is? If $z$ is complex, does that place any restrictions on my ability to pull the negative sign out (assuming I can) or does it not matter if $z$ is real or complex? 
 A: If $f$ is Fréchet-differentiable at $z$ then
$$\frac{\partial[ -f(z)]}{\partial(-z)}=-\partial[-f(z)]z=\partial_z f(z)$$
where $\partial$ means Fréchet derivative and $\frac{\partial}{\partial x}\equiv \partial_x$. The above manipulation is the result of the linearity of $\partial$ and $\partial[-f(z)]$.
If $f$ is not Fréchet-differentiable at $z$ we need to justify the result using directional derivatives, that is
$$\frac{\partial [-f(z)]}{\partial(-z)}=\lim_{h\to 0}\frac{-f(z-hz)+f(z)}h=\lim_{r\to 0}\frac{f(z+rz)-f(z)}r=\partial_z f(z)$$
where we had set $r:=-h$ and the final step is just the result of the definition of directional derivative.
Note: if $z$ belongs to a complex vector space then $h\in\Bbb C$. If $z$ belongs to a real vector space then $h\in\Bbb R$.
A: I would imagine that there is no problem with doing that method! After all, even though its complex analysis, I imagine issues only arise when starting to deal with derivatives with respect to the conjugate...but aside from that, you have a single variable complex function, so its derivative would be the same as in the real case ie.
$$\frac{\partial(-f(z))}{\partial (-z)}=-\frac{df(z)}{d(-z)}=-\frac{df(z)}{dz}\frac{dz}{d(-z)}=\frac{df}{dz}$$
