The chain rule is one of the most useful tools in differential calculus. Equipped with your knowledge of specific derivatives, and the power, product and quotient rules, the chain rule will allow you to find the derivative of any function.
The chain rule is a bit tricky to learn at first, but once you get the hang of it, it's really easy to apply, even to the most stubborn of functions. Knowing it will also allow you to forget about memorizing the quotient rule because any denominator h(x) can be expressed as a multiple, [h(x)]-1.
Each of the functions in the box on the right is a composition of functions, g(h(x)). For example, in the first, h(x) = x2 - 4 and the outer function is g(x) = sin(x): g(h(x)) = sin(x2 - 4).
To derive the chain rule, consider a function y(t) that is actually a composition of functions, y(x) and x(t). It might be something like y(t) = cos(3t2), where y(x) = cos(x) and x(t) = 3t2. In terms of the changes in y and x, we have:
where the Δx terms cancel to give the overall change in y with respect to t. We can use the same cancellation in terms of differentials, but a word about that is in order:
While Leibniz, who invented the df/dx notation, never intended it to be used as such, we can safely consider df/dx to be simply a ratio of differentials, making the analogous cancellation possible:
Thus the derivative of a composition of functions can be expanded as a product of derivatives, the first the derivative of the outer function with respect to the full inner function (e.g. by treating it like a single "placeholder" variable), the second just the derivative of the inner function with respect to its independent variable.
It's best to learn the chain rule by working through examples. In the first two examples below, we'll substitute a "dummy variable" for the inner function, but you'll find that in no time you wont need to do that.
The outer function is f(x) = sin(x) and the inner function is g(x) = x2. Let's let a = x2 and differentiate the outer function as f(a) = sin(a):
f'(a) = cos(a).
Then we differentiate the inner function, g'(x) = 2x, and multiply it by f(a)':
f' = 2x·cos(a)
I didn't write a variable behind f' above because at the moment I've got a weird hybrid function of x and a. The last step is to re-substitute x2 for a:
f'(x) = 2x·cos(x2).
The outer function is f(x) = x3 and the inner function is g(x) = x2- 5. Let's let a = x2 - 5 and differentiate the outer function as f(a) = a3:
f'(a) = 3a2.
Then we differentiate the inner function, g'(x) = x2- 5 with respect to x, and multiply it by f':
f' = 2x·3a2
The last step is to re-substitute (x2 - 5) for a:
f'(x) = 6x·(x2 - 5)2.
Now we'll do without substitution of the placeholder variable and use the chain rule directly. The first step is to take the derivative of the whole outer function, treating the inner function (x2 + x) as a single unit.
½(x2 + x)-1/2 ← not finished, just pausing!
Then multiply by the derivative of the inner function, 2x + 1
f'(x) = (2x + 1)1/2(x2 + x)-1/2
The outer function is x-1 and the inner is x2 - 4. The derivative is f'(x) = -1(x2 - 4)-2(2x), which can be simplified a bit, but I'll leave that to you.
There are a bunch of video examples of chain rule derivatives here.
Find the derivatives of these compound functions. You can download solutions below (try not to peek until you've tried them).
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