﻿ ﻿THE CHAIN RULE - Differentiating - AP CALCULUS AB & BC REVIEW - Master AP Calculus AB & BC

## Master AP Calculus AB & BC

Part II. AP CALCULUS AB & BC REVIEW

CHAPTER 4. Differentiating

THE CHAIN RULE

The Power Rule, although wonderful in some instances, falls short in others. For example, is solved easily using the Power Rule, and the answer is 3x2. However, if cos v is cubed instead of just x, the Power Rule fails! The derivative of cos3x is not 3(cos x)2. Another rule is necessary, and it is called the Chain (rhymes with pain) Rule.

The Chain Rule applies to all composite functions (expressions in which one function is plugged into another function). For example, (cos x)3 is the cosine function plugged into the cubed function. Mathematically, you can write To make things clearer, you can refer to cos x as the “inner function” and x3 as the “outer function.”

NOTE. It is no coincidence that chain rhymes with pain. You will use the Chain Rule often for the remainder of this course, so be sure to understand it, or the pain will be relentless.

Example 8: Each of the following is a composition of functions. For each, identify which is the “inner” and which is the “outer” function.

(a) √csc x

The inner function is csc x, and the outer function is √x. Note that the outer function always acts upon the inner function; in other words, the inner function is always “plugged into” the outer function.

(b) sec 3x

Because 3x is plugged into secant, 3x is the inner function, and sec x is the outer.

(c) ln (x2 + 4)

The inner function is x2 + 4, and the outer function is In x. Again, you are finding the natural log of x2 + 4, so x2 + 4 is being plugged into lnx.

(d) 3sin x

The inner function is sin x, and the outer function is 3x. Here, the exponential function is being raised to the sin x power, so sin x is being plugged into 3x. With this terminology, the Chain Rule is much easier to translate and understand.

The Chain Rule: The derivative of f(g(x)), with respect to x, is f'(g(x)) ∙ g'(x).

Translation: In order to find the derivative of a composite function, take the derivative of the outer function, leaving the inner function alone; then, multiply by the derivative of the inner function.

TIP. When applying the Chain Rule, use the mantra, “Take the derivative of the outside, leaving the inside alone, then multiply by the derivative of the inside.” It becomes automatic if you say it enough.

Now it is possible to find the derivative of f(x) = cos3x. Remember, cos3x is the same as (cos x)3. The outer function is x3, so we use the power rule to take the derivative, leaving the inner function, cos x, alone:

f’(x) = 3 (cos x)2

But, we already said that wasn’t right! That’s because we weren’t finished. Now, multiply that by the derivative of the inner function: Example 9: Find d/dx for each of the following:

(a) √csc x

This expression can be rewritten (esc x)1/2. You already know the inner and outer functions from Example 8. Thus, the derivative is as follows: (b) sec 3x

TIP. Don’t forget to leave the inner function alone when you begin the Chain Rule! For example, d/dx(tan 4x) ≠ sec2(4). Instead, the derivative is sec2(4x) ∙ 4.

The derivative of sec x is sec x tan x, so the derivative of sec 3x is as follows:

sec 3x tan 3x ∙ (3), or

3sec 3x tan 3x

since 3 is the derivative of the inner function 3x.

(c) ln (x2 + 4)

The derivative of In x is 1/x, so this derivative will be times the derivative of (x2 + 4): (d) 3sin x

Remember, the derivative of 3x is 3x ∙ (ln 3), so the derivative of 3sin x is as follows: EXERCISE 4

Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.

DO NOT USE A GRAPHING CALCULATOR FOR ANY OF THESE PROBLEMS.

In questions 1 through 6, find d/dx. 7. Given f is a continuous and differentiable function such that f and f' contain the values given by the below table: (a) Evaluate m’(2) if m(x) = (f(x))3.

(b) Evaluate g'(3) if g(x) = arctan (f(x)).

(c) Evaluate k'(0) if k(x) = f(ex).

8. Let h be a continuous and differentiable function defined on [0,2π]. Some function values of h and h' are given by the chart below: If p(x) = sin2(h(2x)), evaluate p'(π/2)

1. 2. Rewrite the expression as The derivative will be NOTE. Problem 6 uses the logarithmic property loga xn = n ∙ loga x.

3.  In this problem, the x is plugged into the arcsin x function, which is then plugged into the csc x function. The Chain Rule is applied twice.

4. When you derive 4u, you get The u is the inner function and is left alone at first. Once the outer function is differentiated, you multiply by the derivative of u, u'.

5. The Chain Rule is applied twice in this problem. Begin with the outermost function (leaving the rest alone) and work your way inside: NOTE. It is not necessary to write |ex| in the arcsecant derivative formula because ex has no negative range elements.

6. Using logarithmic properties, rewrite the expression as Because ex and ln x are inverse functions, they cancel each other out, and you get (arcsec ex)6. The Chain Rule will be applied three times when you differentiate:  8. The Chain Rule will have to be applied three times in this example. To make it easier to visualize, we have underlined the “inner” functions to be left alone as we find the derivative. Notice how the 2x is left alone until the very end. Because you can write: ﻿