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?


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 $h\in\Bbb R$.

  • $\begingroup$ Kinda a new world to me...can you tell me how I would "read" your second expression $-\partial[-f(z)]z$. I get the first one and the last would be "the partial of $f(z)$ with respect to $z$", but the middle one is a bit foreign to me. I looked up Fréchet derivative and as I said, it's a new world to me. $\endgroup$ – ThatsRightJack Nov 18 '17 at 2:27
  • $\begingroup$ In the expression you can rename $g(z):=-f(z)$ and $y:=-z$ to have an easy reading as $\partial g(z)y$, where $\partial g(z)$ is the Fréchet derivative of $g$ at $z$ (assuming that it exists). When the Fréchet derivative exists at some point (in this case in the point $z$) then $\partial g(z)y=\partial_y g(z)$ where $\partial_y$ means partial or directional derivative, ... $\endgroup$ – Masacroso Nov 18 '17 at 3:01
  • $\begingroup$ ... where the concept of partial derivative is more general than the directional derivative, because the first doesnt necessarily rely in a "direction" of a vector space but the second yes. I assumed that in this case the domain of $f$ is a subset of some vector space (then $-z$ is a direction) so $\partial_y$ is a directional derivative also (not just a partial derivative). $\endgroup$ – Masacroso Nov 18 '17 at 3:01
  • $\begingroup$ Sorry, I need to says also that the concept of "partial derivative" depends on the context, so dont follow too much what I said above about it generality. Just follow what you learn in your multivariable calculus course. $\endgroup$ – Masacroso Nov 18 '17 at 3:20
  • 1
    $\begingroup$ yes, that correct. But in general, if you read some book about linear functions the usual notation is $Ax$ instead of $A(x)$, by example if $A$ is a matrix (it is also a costume to name linear functions with a capital letter but the case of a Fréchet derivative). Take a look at the wikipedia article (or some book) of Fréchet derivative. $\endgroup$ – Masacroso Nov 18 '17 at 3:43

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. $$\frac{\partial(-f(z))}{\partial (-z)}=-\frac{df(z)}{d(-z)}=-\frac{df(z)}{dz}\frac{dz}{d(-z)}=\frac{df}{dz}$$

  • $\begingroup$ O and just one more simple question with regards to your notation. You dropped the partial differential notation, yet $z$ can be written as $x+iy$. I ask because I always get directed back to Wirtinger derivative operators which are expressed as partials. Can you just make a comment on this. $\endgroup$ – ThatsRightJack Nov 18 '17 at 1:49
  • $\begingroup$ I suppose it may be a bad habit of mine to drop the partial but as I see it, let $\sigma=f^2+g-x^4+sin(y)$...if you have $f(\sigma)$ its derivative with respect to $\sigma$ is just $\frac{df}{d\sigma}$ $\endgroup$ – Keith Afas Nov 18 '17 at 1:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.