Multi-variable chain rule with multi-variable functions as arguments What is the chain rule of a multi-variable function with arguments that are also multi-variable functions?
Suppose $x$, $y$, $z$ are independent variables. I mean changing $x$ won't change $y$ and $z$.
Is the general form of multi-variable chain rule similar to the following?
$$\frac{\partial w(u(x, y), v(y, z), t(y, z))}{\partial x} = {\frac{\partial w}{\partial u}}\cdot{\frac{\partial u}{\partial x}} + {\frac{\partial w}{\partial v}}\cdot{\frac{\partial v}{\partial x}} $$
Thanks.

That's nearly right, but you left off the third term that accounts for $t$. So you should have
$$\frac{\partial w}{\partial x} = \frac{\partial w}{\partial u}\cdot\frac{\partial u}{\partial x} + \frac{\partial w}{\partial v}\cdot\frac{\partial v}{\partial x} + \frac{\partial w}{\partial t}\cdot\frac{\partial t}{\partial x}$$
where $w$ is a function of $u,v,t$ and $u,v,t$ are functions of $x,y,z$.
One mnemonic device for remembering this is to think of it as summing over all products of "fractions" that will partially cancel and each give $\partial w$ in the numerator and $\partial x$ in the denominator. But be warned that that's not really what's happening, these aren't fractions and they aren't cancelling. Nonetheless, this helps you to remember it.

Addendum: Of course, if $x$ doesn't actually appear in the formula for $u$, $v$, or $t$, then that partial is zero. What I wrote is the most general formula, not just the formula for your specific case. In other words, even if $v=v(y,z)$ , you can still think of it as $v(x,y,z)$; but then ${\partial v}/{\partial x} = 0$ , and similarly for the others. Not sure if you really wrote what you meant, but my answer is applicable to the general case.
