Function of Function Rule

When the two variables $x,y$ of $W(x,y)$ are themselves each functions of a single variable $t$ (with the same domain), the composite $W(x(t),y(t))$ ultimately depends only on $t$ — so it should be treatable as an ordinary one-variable function, and its derivative $\dfrac{dW}{dt}$ should be computable. This is not a coincidence:

Theorem 8.2 (Function of Function Rule). Suppose $W(x,y)$ has partial derivatives $\dfrac{\partial W}{\partial x},\dfrac{\partial W}{\partial y}$. If $x,y$ are both differentiable functions of a single variable $t$, then $W$ is a differentiable function of $t$, and

The tree diagram is the standard visual aid: $W$ branches down to $x$ and $y$ (labelled by $\partial W/\partial x$ and $\partial W/\partial y$), and each of $x,y$ branches further down to $t$ (labelled $dx/dt$ and $dy/dt$); multiplying along each full branch and summing over the branches reproduces (16) exactly.

Verifying the theorem directly. For $F(x,y)=x^2-2y^2+2xy$ with $x(t)=\cos t,\,y(t)=\sin t$: substituting first gives $F=\cos^2t-2\sin^2t+2\cos t\sin t$, a function of $t$ alone, whose derivative (by direct differentiation, using $\dfrac{d}{dt}(2\cos t\sin t)=2(\cos^2t-\sin^2t)$) works out to $-6\cos t\sin t+2(\cos^2t-\sin^2t)$. Computing instead via (16): $F_x=2x+2y,\,F_y=-4y+2x$, $dx/dt=-\sin t,\,dy/dt=\cos t$, so

which simplifies to the same $-6\cos t\sin t+2(\cos^2t-\sin^2t)$ — confirming (16). Both routes always agree, so computing both is a genuine self-check; whichever route is algebraically shorter for a given problem is the one to use in practice.

Two-parameter version. Sometimes $x=x(s,t)$ and $y=y(s,t)$ both depend on two parameters $s,t\in\mathbb R$, making $W$ ultimately a function of $s,t$ as well:

Theorem 8.3 (Chain Rule, two parameters). If $W(x,y)$ has partial derivatives, and $x=x(s,t),\,y=y(s,t)$ both have partial derivatives with respect to $s$ and $t$, then

$\frac{\partial W}{\partial s} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial W}{\partial y}\frac{\partial y}{\partial s}, \qquad \frac{\partial W}{\partial t} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial W}{\partial y}\frac{\partial y}{\partial t}. \qquad(17,\,18)$ …