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Difference between revisions of "Chain Rule"

 
m (Chain rule moved to Chain Rule: capitalized proper noun)
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Revision as of 17:55, 20 June 2006

Statement

Single variable Chain Rule:


Let each of $I \subset \mathbb{R}, J \subset \mathbb{R}$ be an open interval, and suppose $g:I \to J$ and $f:J \to \mathbb{R}$. Let $h:I \to \mathbb{R}$ such that $h(x) = f(g(x)) \forall x \in I$. If $x_0 \in I$, $g$ is differentiable at ${x_0}$, and ${f}$ is differentiable at $g(x_0),$ then ${h}$ is differentiable at ${x_0}$, and ${h'(x_0) = f'(g(x_0))\cdot g'(x_0)}$.


Multi-dimensional Chain Rule:


Let $g:\mathbb{R}^n \to \mathbb{R}^m$ and $f:\mathbb{R}^m \to \mathbb{R}^p$. (Here each of $n$, $m$, and ${p}$ is a positive integer.) Let ${h}: \mathbb{R}^n \to \mathbb{R}^p$ such that $h(x) = f(g(x)) \forall x \in \mathbb{R}^n$. Let $x_0 \in \mathbb{R}^n$. If $g$ is differentiable at ${x_0}$, and ${f}$ is differentiable at $g(x_0),$ then $h$ is differentiable at ${x_0}$ and $h'(x_0) = f'(g(x_0))\cdot g'(x_0)$. (Here, each of $h'(x_0)$,$f'(g(x_0))$, and $g'(x_0)$ is a matrix.)


Intuition

The single-variable Chain Rule is often explained by pointing out that


$\frac{f(g(x+\Delta x)) - f(g(x))}{\Delta x} = \frac{f(g(x+\Delta x)) - f(g(x))}{g(x+ \Delta x)-g(x)}\cdot \frac{g(x+ \Delta x)-g(x)}{\Delta x}$.


The first term on the right approaches $f'(g(x))$, and the second term on the right approaches $g'(x)$, as $\Delta x$ approaches $0$. This can be made into a rigorous proof. (But we do have to worry about the possibility that $g(x+\Delta x) - g(x)=0$, in which case we would be dividing by $0$.)


This explanation of the chain rule fails in the multi-dimensional case, because in the multi-dimensional case $\Delta x$ is a vector, as is $g(x+\Delta x) - g(x)$, and we can't divide by a vector.


However, there's another way to look at it.


Suppose a function $F$ is differentiable at $x$, and $\Delta x$ is "small". Question: How much does $F$ change when its input changes from $x$ to $x+ \Delta x$? (In other words, what is $F(x+ \Delta x) - F(x)$?) Answer: approximately $F'(x) \cdot \Delta x$. This is true in the multi-dimensional case as well as in the single-variable case.


Well, suppose that (as above) $h(x) = f(g(x))$, and $\Delta x$ is "small", and someone asks you how much $h$ changes when its input changes from $x$ to $x+ \Delta x$. That is the same as asking how much $f$ changes when its input changes from $g(x)$ to $g(x+ \Delta x)$. Which is the same as asking how much $f$ changes when its input changes from $g(x)$ to $g(x) + \Delta g$, where $\Delta g = g(x+ \Delta x) - g(x)$. And what is the answer to this question? The answer is: approximately, $f'(g(x)) \cdot \Delta g$.


But what is $\Delta g$ ? In other words, how much does $g$ change when its input changes from $x$ to $x+ \Delta x$? Answer: approximately $g'(x) \cdot \Delta x$.


Therefore, the amount that $h$ changes when its input changes from $x$ to $x+ \Delta x$ is approximately ${f'(g(x)) \cdot g'(x) \cdot \Delta x}$.


We know that $h'(x)$ is supposed to be a matrix $M$ (or number $M$, in the single-variable case) such that $h'(x) \cdot \Delta x$ is a good approximation to $h(x+ \Delta x) - h(x)$. Thus, it seems that $f'(g(x)) \cdot g'(x)$ is a good candidate for being the matrix (or number) that $h'(x)$ is supposed to be.


This can be made into a rigorous proof. The standard proof of the multi-dimensional chain rule can be thought of in this way.

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