Composition and the Chain Rule

By the end you should be able to: Apply the chain rule to compositions of functions and extend it to multivariable functions whose inputs depend on a common parameter.

Composition

Given $f(x) = x^2$ and $g(x) = \sin x$ , feeding one into the other gives

(f \circ g)(x) = f(g(x)) = \sin^2 x \qquad (g \circ f)(x) = g(f(x)) = \sin(x^2)

Order matters. These are different functions; try $x = 1$ and compare.

The chain rule, one variable

\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)

Or in Leibniz form, which is easier to remember and to generalise:

\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}

What it says. A small nudge in $x$ changes $u$ by $g'(x)$ times as much; that change in $u$ changes $y$ by $f'(u)$ times as much. Multiply the two amplification factors.

For $y = \sin(x^2)$ : put $u = x^2$ , so $y = \sin u$ .

\frac{dy}{dx} = \cos(u) \cdot 2x = 2x\cos(x^2)

The multivariable chain rule

Now let $z = f(x,y)$ where both $x$ and $y$ depend on a parameter $t$ . As $t$ moves, both inputs move, and each drags $z$ with it:

\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}

Sum the contributions, one per input. Each term is "how much $z$ responds to that input" times "how fast that input is moving".

Note the mixture of $\partial$ and $d$ : partial for $z$ against its own inputs, ordinary for anything depending on $t$ alone.

A worked instance with a check

$z = x^2 + y^2$ , with $x = \cos t$ and $y = \sin t$ .

\frac{\partial z}{\partial x} = 2x, \quad \frac{\partial z}{\partial y} = 2y, \quad \frac{dx}{dt} = -\sin t, \quad \frac{dy}{dt} = \cos t

\frac{dz}{dt} = 2x(-\sin t) + 2y(\cos t) = -2\cos t\sin t + 2\sin t\cos t = 0

Check it directly. Substituting first, $z = \cos^2 t + \sin^2 t = 1$ , a constant — so $dz/dt = 0$ . The two routes agree.

The geometry: the point $(\cos t, \sin t)$ travels around the unit circle, which is itself a level curve of $x^2+y^2$ . Moving along a contour, height never changes. The chain rule and the picture say the same thing.

More inputs

The pattern extends. For $w = f(x,y,z)$ with all three depending on $t$ :

\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}

One term per input, always.

Why it matters later

Backpropagation in a neural network is the chain rule applied through many layers. The gradient of a loss with respect to an early weight is a product of the sensitivities at every layer in between — exactly this rule, at scale.