The theorem states that if , then wherever those expressions make sense.
For example, if , , and , then .
Here are some more precise statements for the single-variable and multi-variable cases.
Single variable Chain Rule
Let each of be an open interval, and suppose and . Let such that . If , is differentiable at , and is differentiable at then is differentiable at , and .
Multi-dimensional Chain Rule
Let and . (Here each of , , and is a positive integer.) Let such that . Let . If is differentiable at , and is differentiable at then is differentiable at and . (Here, each of ,, and is a matrix.)
The single-variable Chain Rule is often explained by pointing out that
The first term on the right approaches , and the second term on the right approaches , as approaches . This can be made into a rigorous proof. (But we do have to worry about the possibility that , in which case we would be dividing by .)
This explanation of the chain rule fails in the multi-dimensional case, because in the multi-dimensional case is a vector, as is , and we can't divide by a vector.
There's another way to look at it:
Suppose a function is differentiable at , and is "small". Question: How much does change when its input changes from to ? (In other words, what is ?) Answer: approximately . This is true in the multi-dimensional case as well as in the single-variable case.
Suppose that (as above) , and is "small", and someone asks you how much changes when its input changes from to . That is the same as asking how much changes when its input changes from to , which is the same as asking how much changes when its input changes from to , where . And what is the answer to this question? The answer is: approximately, .
We must determine how much does change when its input changes from to ? Answer: approximately .
Therefore, the amount that changes when its input changes from to is approximately .
We know that is supposed to be a matrix (or number, in the single-variable case) such that is a good approximation to . Thus, it seems that is a good candidate for being the matrix (or number) that 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.
The following is a proof of the multi-variable Chain Rule. It's a "rigorized" version of the intuitive argument given above.
This proof uses the following fact: Assume , and . Then is differentiable at if and only if there exists an by matrix such that the "error" function has the property that approaches as approaches . (In fact, this can be taken as a definition of the statement " is differentiable at .") If such a matrix exists, then it is unique, and it is called . Intuitively, the fact that approaches as approaches just means that is approximated well by .
Let and . (Here each of , , and is a positive integer.) Let such that . Let , and suppose that is differentiable at and is differentiable at .
In the intuitive argument, we stated that if is "small", then , where . In this proof, we'll fix that statement up and make it rigorous. What we can say is, if , then , where is a function which has the property that .
In the intuitive argument, we said that . In this proof, we'll make that rigorous by saying , where has the property that .
Putting these together, we find that , where I have taken that messy error term and called it .
Now, we need to show that as , in order to prove that is differentiable at and that .
In order to finish off the proof, it's needed to look at and "play around with it", so to speak. The conclusion can be reached by the following fact: If is an by matrix, then there exists a number such that for all .
by the triangle inequality.
We'll call the first term on the right here the "first error term" and the second term on the right the "second error term." If we can show that the "first error term" and the "second error term" each approach as , then we'll be done.
which approaches as . So the "first error term" approaches . That's good. ( is the -norm of the matrix .)
Consider the "second error term", . On top we have the norm of with a certain (slightly complicated) input. We know that is supposed to be small, as long as its input is small. In fact, we know more than that. If you take , and divide it by the norm of its input, then that quotient is also supposed to be small, as long as the input of is small. This suggests an idea: divide by the norm of the input of , and look at what we get. But to make up for the fact that we are dividing by the norm of the input of , we will also have to multiply by the norm of the input of .
The first term on the right should approach , and the second term on the right hopefully at least remains bounded, as .
This idea is promising, but there is a problem with it. When we divide by the norm of the input of , we may be dividing by . The following argument can resolve this anomaly.
We introduce a function such that is equal to if , and is if . Then for all , and as .
Certainly as . Also, since , we know that as . So as , which means that as .
This remains bounded as .
We have shown that the "second error term" is a product of one term that approaches and another term that remains bounded as . Therefore, the "second error term" approaches as .