﻿ ﻿THE CHAIN RULE; THE DERIVATIVE OF A COMPOSITE FUNCTION - Differentiation - Calculus AB and Calculus BC

## CHAPTER 3 Differentiation

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

Formula (3) says that 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 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: 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 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: 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: 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 SOLUTION: EXAMPLE 4 SOLUTION: EXAMPLE 5

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

SOLUTION: EXAMPLE 6

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

SOLUTION: Then, f (0) = 1(3 · 1 + 2 · 0) = 3.

EXAMPLE 7 SOLUTION: Note that neither f (v) nor f (v) exists where the denominator equals zero, namely, where 1 − 2v2 = 0 or where v equals EXAMPLE 8

If x ≠ 0, find f (x).

SOLUTION: EXAMPLE 9

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

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

EXAMPLE 10

If x = cos3 (1 − 3θ), find SOLUTION: EXAMPLE 11

If y = e(sin x) + 1, find SOLUTION: EXAMPLE 12

If y = (x + 1)ln2(x + 1), find SOLUTION: EXAMPLE 13

If g(x) = (1 + sin2 3x)4, find SOLUTION: Then EXAMPLE 14

If y = sin−1 x + find y .

SOLUTION: EXAMPLE 15

If u = ln find SOLUTION: .

EXAMPLE 16

If s = e−t(sin t − cos t), find s .

SOLUTION: EXAMPLE 17

Let y = 2u3 − 4u2 + 5u − 3 and u = x2x. Find SOLUTION: EXAMPLE 18

If y = sin (ax + b), with a and b constants, find SOLUTION: = [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 SOLUTION: We can use both formula (13), and the Chain Rule to get Alternatively, we can rewrite the given function using a property of logarithms: ln (kx) = ln k + ln x. Then 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. ﻿