1. Generalized chain rule

Hi! This problem is somewhat related to a thread that I have posted previously: Theorem considering multivariable calculus
The problem that I have now is the following: Say that I have an unknown function $f(x, y, z)$, but I know that there exist a known relation between $x$, $y$ and $z$ that can be expressed as $z = z(x, y)$ (or, equivalently, a known function $g$ satisfying that $g(x, y, z) = 0$ with $\partial g/\partial z = 1$). Is there a way to write the derivative $\partial f/\partial z$ in terms of $\partial f/\partial x$, $\partial f/\partial y$, $z(x, y)$, $\partial z/\partial x$ and $\partial z/\partial y$? If not, why not?

Thanks for any help,
Simon

2. Re: Generalized chain rule

It has been a long times since I have done multivariable calculus, so I may have this wrong, but I believe:

$\dfrac{df}{dz} = 0 = \dfrac{\partial f}{\partial x} \dfrac{\partial x}{\partial z} + \dfrac{\partial f}{\partial y} \dfrac{\partial y}{\partial z} + \dfrac{\partial f}{\partial z}$

Solving for $\dfrac{\partial f}{\partial z}$ gives

$\dfrac{\partial f}{\partial z} = -\dfrac{ \left( \dfrac{\partial f}{\partial x} \right) }{ \left( \dfrac{\partial z}{\partial x} \right) } - \dfrac{ \left( \dfrac{\partial f}{\partial y} \right) }{ \left( \dfrac{\partial z}{\partial y} \right) }$

3. Re: Generalized chain rule

Originally Posted by SlipEternal
It has been a long times since I have done multivariable calculus, so I may have this wrong, but I believe:

$\dfrac{df}{dz} = 0 = \dfrac{\partial f}{\partial x} \dfrac{\partial x}{\partial z} + \dfrac{\partial f}{\partial y} \dfrac{\partial y}{\partial z} + \dfrac{\partial f}{\partial z}$

Solving for $\dfrac{\partial f}{\partial z}$ gives

$\dfrac{\partial f}{\partial z} = -\dfrac{ \left( \dfrac{\partial f}{\partial x} \right) }{ \left( \dfrac{\partial z}{\partial x} \right) } - \dfrac{ \left( \dfrac{\partial f}{\partial y} \right) }{ \left( \dfrac{\partial z}{\partial y} \right) }$
pretty sure you have to put in the condition that the Jacobian is non-singular to be able to do this.

4. Re: Generalized chain rule

Originally Posted by romsek
pretty sure you have to put in the condition that the Jacobian is non-singular to be able to do this.
Ah, it has been a long time... You are correct. My bad.

5. Re: Generalized chain rule

By the time I got it, it was done. But anyhow, with apologies, and a little more detail:

df=fxdx+fydy+fzdz=0
fz=-fxxz-fyyz

z=z(x,y)
dz=zxdx+zydy
zxdx=dz-zydy
zxxz=1
xz=1/zx
similarly
yz=1/zy

fz=-fx/zx-fy/zy