Writing out chain rule for the following function

Question

$\frac{dh}{dx}$, where $h(x) = f(x, u(x), v(x))$.

First of all, this function doesn't even make sense to me. It's a function of one variable, with domain $\mathbb{R}$ and range $\mathbb{R}$. How can it have 3 different inputs? That's like saying, e.g. $f(x,x,x) = x^2$ which is nonsense to me. I understand that $u(x)$ and $v(x)$ are different functions of single variables and I know you can compose functions, but how can you simultaneously compose two different functions of one variable into another function of one variable? Like you can do $f \circ g$ and $f \circ h$, but how can you simultaneously perform $f \circ g$ and $f \circ h$ at the same exactly time like it seems to be the case here?

Answer 1

Maybe an example will help:

\begin{align*} f(a,b,c) &= a + b + c \\ u(x) &= x^2 \\ v(x) &= x^3 \\ h(x) &= f(x, u(x), v(x)) = f(x, x^2, x^3) = x + x^2 + x^3 \end{align*}

The function $f$ has three inputs, but the function $h$, which is the one you're differentiating, has only one.

Answer 2

What's really happening: you take the derivative of a real-valued function $f$, defined on subset $G$ of $R^3$ along a path $c$ in $R^3$ defined on an interval $J$ of $R$, provided that $c(J)\subset G$. Then $f\circ c$ is a real-valued function defined on $J$. You're just take a journey in $f$ along $c$ and you want to compute the velocity at any time $t$ of your journey, namely $(f\circ c)'(t)$. Well, now that makes sense, doesn't it? Applying the chain rule we'll arrive in $$(f\circ c)'(t)=\langle\nabla f\bigl(c(t)\bigr),c'(t)\rangle.$$