$F$ be a function of $x,$ $y,$ $z.$ $z$ depends on $x$ and $y.$ Then when evaluating $\partial F/\partial x,$ do I treat $z$ as a constant? I am under the impression that if I don't, what I would get is the total derivative $dF/dx.$


I think you are saying , given a function of three variables $G(x,y,z)$ and another function of two variables $g(x,y)$ we define a function of two variables $F(x,y)=G(x,y,g(x,y))$, yes?

Assume that is what you meant. Then, what is $\frac {\partial F}{\partial x}$? Well, it is $$\frac {\partial F}{\partial x}=\lim_{h\to 0}\frac {F(x+h,y)-F(x,y)}{h}=\lim_{h\to 0}\frac {G(x+h,y,g(x+h,y))-G(x,y,g(x,y))}h$$

So you can see that the $z-$component of $G$ is not constant. The derivative will involve both $\frac {\partial G}{\partial x}$ and $\frac {\partial G}{\partial z}$. Indeed, it is not difficult to show from this that $$\frac {\partial F}{\partial x}=\frac {\partial G}{\partial x}+\frac {\partial G}{\partial z}\frac {\partial g}{\partial x}$$

Example: say $G(x,y,z)=x+y +z^2$ and let $g(x,y)=2x+3y$. Then define $$F(x,y)=G(x,y,g(x,y))=x+y+(2x+3y)^2=4x^2+x+y+12xy +9y^2$$

Of course we have $$\frac {\partial F}{\partial x}=8x+1+12y$$

We have the partials for $G$, namely $$\frac {\partial G}{\partial x}=1\quad \frac{\partial G}{\partial y}=1 \quad \frac {\partial G}{\partial z}=2z$$

We claim that $$\frac {\partial F}{\partial x}=\frac {\partial G}{\partial x}+\frac {\partial G}{\partial z}\frac {\partial g}{\partial x}$$

But the left hand is $$1+ 2z\times 2=1+4(2x+3y)=8x+12y+1$$ which is consistent.

  • $\begingroup$ That doesn't sound like what I was asking. Maybe I was ambiguous. I meant to say that F is a function of x,y, and z (say). And z in turn is a function of x and y. So when evaluating ∂F(x,y,z)/∂x, do I treat z as a constant like I do with y? $\endgroup$ – Hrit Roy Sep 11 '17 at 20:28
  • $\begingroup$ Well perhaps I misinterpreted, though I do not think I did. Please edit your question to explain precisely what you meant. As it stands, it is not clear. (I note that the other responder has the same confusion I have). $\endgroup$ – lulu Sep 11 '17 at 20:30
  • $\begingroup$ Note: I believe that you are only quibbling as to whether the implicit function of three variables ought to be called $F$ or $G$. I called it $G$ to make it clear that it is different from your $F$ which, after all, is a function of only two variables. $\endgroup$ – lulu Sep 11 '17 at 20:31
  • $\begingroup$ I do sense that I am wrong somewhere. Could you please point out? I'll explain how I see things : F be a function of x and y and z. z depends on x and y (is saying z is a function of x and y incorrect?). I do not know the precise nature of the dependency. I am not trying to introduce any new function G. I am taking the partial derivative of F w.r.t. x. My question is whether or not I treat z as a constant. $\endgroup$ – Hrit Roy Sep 11 '17 at 20:37
  • $\begingroup$ Right. You are saying exactly the same thing I am saying only you call my function $G$ by the same name as you call your function of two variables, $F$. That isn't "wrong" as such but I think it is needlessly confusing. Anyway, read my post carefully. Or read the post from @TedShifrin which is more or less identical to mine. We use different words so I'll leave mine up, but we are saying exactly the same thing. $\endgroup$ – lulu Sep 11 '17 at 20:48

You need more careful notation. You start with a function $F(x,y,z)$ and you form a new function $\phi(x,y) = F(x,y,g(x,y))$. Do you mean $\partial F/\partial x$ or do you mean $\partial\phi/\partial x$? [The latter is what you're referring to as the total derivative.] As it stands, your question really is ill-defined, but given the way you phrased it, I would vote unequivocally for the latter; otherwise, you must define the function $F$ with three independent variables.

  • $\begingroup$ Of course when I say ∂F/∂x I mean ∂F(x,y,g)/∂x. $\endgroup$ – Hrit Roy Sep 11 '17 at 20:24
  • $\begingroup$ That makes no sense, though. $g$ is a function, not a variable. Start with $F(x,y,z) = x^2+y^2+z^2$. Once you put $z=g(x,y)$, you have a function of $x,y$ only. You can't even tell any more what the original $F(x,y,z)$ was once I write $\phi(x,y) = x^2+y^2+(x+y+3e^x)^2$. $\endgroup$ – Ted Shifrin Sep 11 '17 at 20:29
  • $\begingroup$ No, I'm not. We've explained it to you and you're not listening. $\endgroup$ – Ted Shifrin Sep 11 '17 at 20:31
  • $\begingroup$ By the way, I have taught multivariable calculus in some form or other for approximately 48 years and written a textbook on the subject. So you don't need to be rude. $\endgroup$ – Ted Shifrin Sep 11 '17 at 20:36
  • $\begingroup$ Oops. Never meant to sound rude! $\endgroup$ – Hrit Roy Sep 11 '17 at 20:46

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