3

I'm reviewing the chain rule for partial derivatives, and since I've never actually learned it other than picking it up while learning other stuff in courses, I'm not 100% sure I'm doing it correctly when it comes to more complicated interdependencies.

So my question is what $\frac{\partial z}{\partial y}$, where $z = F(u, v, y), u = f(v, y)$ and $v = g(x, y)$ would be? My understanding is that it's:

$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial v}\frac{\partial v}{\partial y} + \frac{\partial z}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial y} + \frac{\partial z}{\partial y}$

Is this correct? I'm unsure about the first two terms and whether the third one would somehow be subsumed in the first one.

Ryker
  • 1,499
  • 2
    You might want to be more verbose with your notation. Partial derivatives act on functions. So instead of writing $(\partial/\partial y) z$ on both sides of the equation, try writing $(\partial/\partial y) F(u, v, y)$ versus $(\partial/\partial y) F(x, y)$. That should clear things up. – Chris Culter Jul 22 '13 at 18:20
  • @ChristopherCulter, you're right, of course, but I did this quickly and figured it was clear enough what I was after, so I didn't want to spend too much time on writing it up nicely in TeX. – Ryker Jul 22 '13 at 19:06

1 Answers1

4

It is always been helpful for me, to think of this as a tree. Then you walk to the leaves that contain to variable of your interest.

Chain rule tree

Derivative with respect to $x$ would be: $\frac{\partial F}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial v} \frac{\partial v}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x}$

Derivative with respect to $y$ would be: $\frac{\partial F}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial v} \frac{\partial v}{\partial y} + \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} + \frac{\partial z}{\partial y}$

Read about Leibniz notation, for further information on this notation.

The trick is to take ONLY the paths to the leaves with the variables you are interested in.

Sebastialonso
  • 228
  • Damn, I noticed I put the right thing in the thread title and my solution, but not the question itself (I edited it now). Basically, I was asking for $\frac{\partial z}{\partial y}$, not $\frac{\partial z}{\partial z}$. Could you perhaps verify if my solution is correct? Sorry about the confusion. – Ryker Jul 22 '13 at 18:39
  • 1
    @Ryker Yeap, your solution is correct. Also, you should take into consideration what C. Culter commented. I answered you with a focus on insight, but in rigour, what he says is the right thing to do. – Sebastialonso Jul 22 '13 at 18:44
  • For now I've only up-voted it, because it is helpful, but it doesn't actually answer the question. I feel that's how it's done on this website, so if you change your answer to the solution for $\frac{\partial z}{\partial y}$, I'll be glad to accept it! – Ryker Jul 22 '13 at 21:37
  • Oh! Of course, I'm sorry. I'll edit my answer. – Sebastialonso Jul 23 '13 at 00:57