# nLab chain rule

### Context

#### Differential geometry

differential geometry

synthetic differential geometry

# Contents

## 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

$d\left(g\circ f\right):TX\stackrel{df}{\to }TY\stackrel{dg}{\to }TZ\phantom{\rule{thinmathspace}{0ex}}.$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 ${\gamma }_{v}:\left[-ϵ,ϵ\right]\to X$, for some $ϵ>0$, with $\gamma \left(0\right)=x$, then the chain rule is the associativity of the composite

$\left[-ϵ,ϵ\right]\stackrel{{\gamma }_{v}}{\to }X\stackrel{f}{\to }Y\stackrel{g}{\to }Z\phantom{\rule{thinmathspace}{0ex}}.$[-\epsilon,\epsilon] \stackrel{\gamma_v}{\to} X \stackrel{f}{\to} Y \stackrel{g}{\to} Z \,.

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

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

$D\stackrel{v}{\to }X\stackrel{f}{\to }Y\stackrel{g}{\to }Z\phantom{\rule{thinmathspace}{0ex}}.$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 $ℝ×ℝ$.

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

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

given by

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

and

$dg:\left(x,v\right)↦\left(g\left(x\right),vg\prime \left(x\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$d g : (x,v) \mapsto (g(x), v g'(x)) \,.

The composite

$f\left(g\circ f\right):ℝ×ℝ=Tℝ\to Tℝ=ℝ×ℝ$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

$d\left(g\circ f\right):\left(x,v\right)↦\left(f\left(x\right),vf\prime \left(x\right)\right)↦\left(g\left(f\left(x\right)\right),vf\prime \left(x\right)g\prime \left(f\left(x\right)\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$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

$\left(g\circ f\right)\prime \left(x\right)=f\prime \left(x\right)g\prime \left(f\left(x\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$(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.

Revised on December 1, 2011 01:12:19 by Zoran Škoda (161.53.130.104)