#math/calculus
A composite output can change through several intermediate variables at once. The multivariable chain rule follows every dependency path, multiplies the rates along it, and adds the contributions. If
$
z=f(x,y),\qquad x=g(t),\qquad y=h(t),
$
then
$
\boxed{\frac{dz}{dt}=f_x\frac{dx}{dt}+f_y\frac{dy}{dt}}.
$
![[tikz-a2972432bd57.svg]]
%% tikz-source
```latex
\usetikzlibrary{arrows.meta,positioning}
\begin{document}
\begin{tikzpicture}[>=Stealth,node distance=1.3cm and 1.8cm,
box/.style={draw,rounded corners,minimum width=1.25cm,minimum height=.72cm,fill=blue!7},
every node/.style={font=\small}]
\node[box] (t) {$t$};
\node[box,above right=of t] (x) {$x=g(t)$};
\node[box,below right=of t] (y) {$y=h(t)$};
\node[box,right=3.0cm of t] (f) {$z=f(x,y)$};
\draw[->,thick] (t)--node[above left]{$x'$}(x);
\draw[->,thick] (t)--node[below left]{$y'$}(y);
\draw[->,thick] (x)--node[above]{$f_x$}(f);
\draw[->,thick] (y)--node[below]{$f_y$}(f);
\node[right=1.2cm of f,align=left]
{$\displaystyle \frac{dz}{dt}=f_x\frac{dx}{dt}+f_y\frac{dy}{dt}$\\sum over every path from $t$ to $z$};
\end{tikzpicture}
\end{document}
```
%%
For $u=f(x,y,z)$ with all three coordinates depending on $t$,
$
\frac{du}{dt}
=f_x\frac{dx}{dt}+f_y\frac{dy}{dt}+f_z\frac{dz}{dt}.
$
## Key example — a duck swimming through changing temperature
In the book, a duck follows the circular path
$
x=\cos t,
\qquad
y=\sin t,
$
through water whose temperature is
$
T(x,y)=x^2e^y-xy^3.
$
The temperature changes through both the $x$ and $y$ dependency paths. Since
$
T_x=2xe^y-y^3,
\qquad
T_y=x^2e^y-3xy^2,
$
and $x'=-\sin t$, $y'=\cos t$, the chain rule gives
$
\frac{dT}{dt}
=(2xe^y-y^3)(-\sin t)+(x^2e^y-3xy^2)\cos t.
$
Substituting the path produces
$
\frac{dT}{dt}
=-2\cos t\sin t\,e^{\sin t}+\sin^4t
+\cos^3t\,e^{\sin t}-3\cos^2t\sin^2t.
$
Directly substituting $x(t),y(t)$ into $T$ and differentiating gives the same result. The two terms in the chain rule are necessary because the duck's temperature changes through two simultaneous coordinate motions. [[Calculus III Ch 14 - Ch 16.pdf#page=86|Textbook Example 1, PDF p. 86]]
The rule is a sum, rather than one product, because the output changes through several intermediate variables. [[15D2 - General Chain Rule and Jacobians|Jacobian multiplication]] packages the same idea for vector-valued functions.
> [!ML/AI]
> For one neuron, let $a=wx+b$, $\hat y=\sigma(a)$, and $L=\tfrac12(\hat y-y)^2$. The chain rule gives $\partial L/\partial w=(\hat y-y)\sigma'(a)x$; backpropagation repeats exactly this calculation through every layer.