Remember that this rule tells us how to take the derivative of the composite in terms of the two derivatives and . Basically, the derivative of the product is the product of the derivatives, but we have to be careful where to evaluate each derivative. In Newton’s notation we write .
So now let’s take an integral of each side:
So if we’ve got an integrand that involves one function acting on an expression , it may be worth our while to see if we also see as a factor in the integrand, because we might then be able to reduce to integrating itself. We’re intentionally glossing over questions of where and must exist, have their ranges, be integrable or differentiable, and so on.
But let’s look a little closer at what’s really going on here. We say that the function is taking the interval and sending it to the interval . Actually, might send some points outside this image interval. Consider, for example, on the interval . We’re saying this goes to the interval , but clearly . What’s going on here.
We have to use the sign convention for intervals to understand this. First, let’s break up the domain interval into regions where is nonincreasing and where it’s nondecreasing. In this example, is nonincreasing on and nondecreasing on . The images of these monotonous intervals are exactly what we expect: and — how boring (sorry).
But now when we use the sign convention we see that our image interval is along with and . The first two of these two cancel out! In fact, anything outside the interval must be traversed an even number of times in opposite directions that will cancel each other out when we’re thinking about integrals.
So in this sense we’re using to tell us how to get our from to as we integrate . And as we change our variables from to we have to multiply by . Why? Because the derivative of is what tells us how to translate displacements in the domain of into displacements in the range of . That is, we think of a tiny little sliver of a rectangle in the integral over as having width . We need to multiply this by to get the width of a corresponding rectangle in the integral over .
Next time we’ll try to put this intuition onto a firmer ground not usually seen in calculus classes, and not often in advanced calculus classes either, these days.