Calculus AB and Calculus BC

CHAPTER 3 Differentiation

C. THE CHAIN RULE; THE DERIVATIVE OF A COMPOSITE FUNCTION

Formula (3) says that

Image

This formula is an application of the Chain Rule. For example, if we use formula (3) to find the derivative of (x2x + 2)4, we get

Image

In this last equation, if we let y = (x2x + 2)4 and let u = x2x + 2, then y = u4. The preceding derivative now suggests one form of the Chain Rule:

Image

as before. Formula (3) the previous page gives the general case where y = un and u is a differentiable function of x.

Now suppose we think of y as the composite function f (g(x)), where y = f (u) and u = g(x) are differentiable functions. Then

Chain rule

Image

as we obtained above. The Chain Rule tells us how to differentiate the composite function: “Find the derivative of the ‘outside’ function first, then multiply by the derivative of the ‘inside’ one.”

For example:

Image

Many of the formulas listed above in §B and most of the illustrative examples that follow use the Chain Rule. Often the chain rule is used more than once in finding a derivative.

Note that the algebraic simplifications that follow are included only for completeness.

EXAMPLE 1

If y = 4x3 − 5x + 7, find y (1) and y (1).

SOLUTION: Image

Then y (1) = 12 · 12 − 5 = 7 and y (1) = 24 · 1 = 24.

EXAMPLE 2

If f (x) = (3x + 2)5, find f (x).

SOLUTION: f (x) = 5(3x + 2)4 · 3 = 15(3x + 2)4.

EXAMPLE 3

Image

SOLUTION: Image

EXAMPLE 4

Image

SOLUTION: Image

EXAMPLE 5

If s(t) = (t2 + 1)(1 − t)2, find s (t).

SOLUTION: Image

EXAMPLE 6

If f (t) = e2t sin 3t, find f (0).

SOLUTION: Image

Then, f (0) = 1(3 · 1 + 2 · 0) = 3.

EXAMPLE 7

Image

SOLUTION: Image

Note that neither f (v) nor f (v) exists where the denominator equals zero, namely, where 1 − 2v2 = 0 or where v equals Image

EXAMPLE 8

If Image x ≠ 0, find f (x).

SOLUTION: Image

EXAMPLE 9

If y = tan (2x2 + 1), find y .

SOLUTION: y = 4x sec2 (2x2 + 1).

EXAMPLE 10

If x = cos3 (1 − 3θ), find Image

SOLUTION: Image

EXAMPLE 11

If y = e(sin x) + 1, find Image

SOLUTION: Image

EXAMPLE 12

If y = (x + 1)ln2(x + 1), find Image

SOLUTION: Image

EXAMPLE 13

If g(x) = (1 + sin2 3x)4, find Image

SOLUTION: Image

Then Image

EXAMPLE 14

If y = sin−1 x + Image find y .

SOLUTION: Image

EXAMPLE 15

If u = ln Image find Image

SOLUTION: Image.

EXAMPLE 16

If s = e−t(sin t − cos t), find s .

SOLUTION: Image

EXAMPLE 17

Let y = 2u3 − 4u2 + 5u − 3 and u = x2x. Find Image

SOLUTION: Image

EXAMPLE 18

If y = sin (ax + b), with a and b constants, find Image

SOLUTION: Image = [cos(ax + b)] · a = a cos(ax + b).

EXAMPLE 19

If f (x) = aekx (with a and k constants), find f and f .

SOLUTION: f (x) = kaekx and f = k2 aekx.

EXAMPLE 20

If y = ln (kx), where k is a constant, find Image

SOLUTION: We can use both formula (13), and the Chain Rule to get

Image

Alternatively, we can rewrite the given function using a property of logarithms: ln (kx) = ln k + ln x. Then

Image

EXAMPLE 21

Given f (u) = u2u and u = g(x) = x3 − 5 and F(x) = f (g(x)), evaluate F (2).

SOLUTION: F (2) = f (g(2))g (2) = f (3) · (12) = 5 · 12 = 60.

Now, since g (x) = 3x2, g (2) = 12, and since f (u) = 2u − 1, f (3) = 5. Of course, we get exactly the same answer as follows.

Image