nLab
chain rule

Statement

The chain rule is the statement that differentiation $d:\mathrm{Diff}\to \mathrm{Diff}$ is a functor on Diff:

given two smooth functions between smooth manifolds $f:X\to Y$ and $g:Y\to Z$ we have

$$d(g\circ f):TX\stackrel{df}{\to }TY\stackrel{dg}{\to }TZ.$$d(g \circ f) : T X \stackrel{d f}{\to} T Y \stackrel{d g}{\to} T Z \,.

If one thinks of a tangent vector $v\in T_{x}X$ to be an equivalence class of a smooth path ${\gamma }_v:[-ϵ,ϵ]\to X$, for some $ϵ>0$, with $\gamma (0)=x$, then the chain rule is the associativity of the composite

$$[-ϵ,ϵ]\stackrel{{\gamma }_v}{\to }X\stackrel{f}{\to }Y\stackrel{g}{\to }Z.$$[-\epsilon,\epsilon] \stackrel{\gamma_v}{\to} X \stackrel{f}{\to} Y \stackrel{g}{\to} Z \,.

Bracketed as $(g\circ f)\circ {\gamma }_v$ this represents $d(g\circ f)(v)$. Bracketed as $g\circ (f\circ {\gamma }_v)$ is represents $dg(df(v))$.

Alternatively, in a context of synthetic differential geometry where with $D$ being the infinitesimal interval we may identify $v$ with $v:D\to X$, the chain rule is the associativity of

$$D\stackrel{v}{\to }X\stackrel{f}{\to }Y\stackrel{g}{\to }Z.$$D \stackrel{v}{\to} X \stackrel{f}{\to} Y \stackrel{g}{\to} Z \,.

Examples

Let $X=Y=Z=ℝ$ the real line. Then the tangent bundle $TX$ is canonically identified with $ℝ\times ℝ$.

Given two functions, $f,g:ℝ\to ℝ$ their derivatives are traditionally regarded again as functions $f\prime ,g\prime :ℝ\to ℝ$, though strictly speaking we are to think of them as the maps

$$df,dg:ℝ\times ℝ=Tℝ\to Tℝ=ℝ\times ℝ$$d f, d g : \mathbb{R} \times \mathbb{R} = T \mathbb{R} \to T \mathbb{R} = \mathbb{R} \times \mathbb{R}

given by

$$df:(x,v)\mapsto (f(x),vf\prime (x))$$d f : (x,v) \mapsto (f(x), v f'(x))

and

$$dg:(x,v)\mapsto (g(x),vg\prime (x)).$$d g : (x,v) \mapsto (g(x), v g'(x)) \,.

The composite

$$f(g\circ f):ℝ\times ℝ=Tℝ\to Tℝ=ℝ\times ℝ$$f (g \circ f) : \mathbb{R} \times \mathbb{R} = T \mathbb{R} \to T \mathbb{R} = \mathbb{R} \times \mathbb{R}

is therefore the map

$$d(g\circ f):(x,v)\mapsto (f(x),vf\prime (x))\mapsto (g(f(x)),vf\prime (x)g\prime (f(x))).$$d(g \circ f) : (x,v) \mapsto (f(x), v f'(x)) \mapsto (g(f(x)), v f'(x) g'(f(x))) \,.

Therefore we have

$$(g\circ f)\prime (x)=f\prime (x)g\prime (f(x)).$$(g \circ f)'(x) = f'(x) g'(f(x)) \,.

This is the form in which the chain rule is taught to kids. It’s just a test to see if they understand what’s really going on. One of these tests that are never being graded.