The Chain Rule - CALCULUS - MATHEMATICS IN HISTORY - Mathematics for the liberal arts

Mathematics for the liberal arts (2013)

Part I. MATHEMATICS IN HISTORY

Chapter 4. CALCULUS

4.7 The Chain Rule

The sum rule allows us to take the derivative of f(x) + g(x). The product rule tells us how to differentiate f(x)g(x), and the quotient rule lets us take the derivative of images. But there is another way we can combine functions, one that we haven’t yet discussed.

Compositions of Functions

Can we take the derivative of a composition of functions? If h(x) = g(f(x)), is there a way to say what the derivative will be? For example, if f(x) = images and g(x) = x2 then

images

Is there a way to compute the derivative of h, a way that lets us use what we know about f and what we know about g? It turns out that there is. It’s called the chain rule.

The chain rule: If h(x) = g(f(x)), then h′(x) = g′(f(x))f′(x).

images EXAMPLE 4.19

Find the derivative of h(x) = (x2)3.

Solution: By properties of exponents, h(x) = x6, and the power rule directly tells us that h′(x) = 6x5. We can also use the chain rule to arrive at this. For the purposes of the chain rule, the “inside” function of the composition is f(x) = x2. The “outside” function is g(x) = x3.

By the power rule, f′(x) = 2x and g′(x) = 3x2, and according to the chain rule,

images

images

images EXAMPLE 4.20

Let’s apply the chain rule to h(x) = images. Here f(x) = images and g(x) = x2. The derivative of g is easy: g′(x) = 2x. For the derivative of f, we use the quotient rule:

images

According to the chain rule, the derivative of h is

images

images EXAMPLE 4.21

In Example 4.15 we showed that the derivative of g(x) = images is g′(x) = images. Find the derivative of h(x) = images.

Solution: In this case, the inside of the composition is f(x) = 1 + x2, and the outside is g(x) = images. By the chain rule,

images

images

Roots and Fractional Exponents

If fractional exponents are a dim memory for you, remember that a square root can be written as an exponent of 1/2. This is not crazy. Just as you multiply images by itself to get x, when you multiply x1/2 · x1/2 you get x1 by adding exponents.

In Example 4.15 we used the product rule to find the derivative of images, that is, the derivative of the function y = x1/2. Knowing the chain rule, we can find the derivative of any root.

Let f(x) = images = x1/n, where n is a positive integer (1, 2, 3, …). If we take the nth power of both sides, we learn that (f(x))n = x. Now, x is something we know the derivative of, and its derivative is 1. The left side of the equality is a composition, however, and we can apply the chain rule.

For the purposes of the chain rule, the outside function is g(x) = xn. The inside function is f(x). Taking the derivative, we get

images

It’s really f′ that we are interested in, and solving for f′, we get

images

If we use fractional exponents for the root, we can make this a bit simpler:

images

Thus, the chain rule tells us that the derivative of f(x) = x1/n is images. This is exactly as we might have guessed from the power rule.

images EXAMPLE 4.22

If f(x) = images = x1/2, then images. This agrees with our conclusion in Example 4.15.

images EXAMPLE 4.23

Find the derivative of h(x) = x2/3.

Solution: We can write images. This is a composition of two functions. The outside function is g(x) = x1/3, and the inside function is f(x) = x2. By the chain rule,

images

Notice that this agrees with the pattern of the power rule. The derivative of h(x) x2/3 is images.

It turns out that the power rule works for any fractional exponent. Although we don’t prove it here, the power rule works for all exponents, even irrational ones.

The power rule: If f(x) = xr, where r is any real number, then f′(x) = rxr–1.

images EXAMPLE 4.24

Find the tangent to images when x = 1.

Solution: To find a tangent line, we need two pieces of information, the point of tangency and the slope of the line. When x = 1, we have y = f(1) = images = 1, so the point of tangency is (x, y) = (1, 1).

The derivative of the function tells us the slope of the tangent, and the derivative of images is simply images. When x = 1, the slope is f′(1) = images.

Putting these together using the point-slope form of a line, we obtain the tangent line

images

If you prefer the y = mx + b form of a line, you can simplify to get

images

A plot on a calculator or computer, like Figure 4.11, can help us verify that this answer is correct and we have made no mistakes.

images

Figure 4.11 Tangent to the power function images.

images

EXERCISES

4.38 Find the derivative of each function:

a) images

b) images

c) images

d) h(s) = s3/4

e) y(t) = tπ

f) y(x) = (3x2 – 5x + 4)2/3

4.39 For each function in the previous exercise:

a) Find the value of function when the independent variable is equal to 1.

b) Find the slope of the tangent line when the independent variable is 1.

c) Find the equation of the tangent line to the function at the point where the independent variable is 1. If you have access to a computer or calculator, plot the function and tangent line together to check your answer.