Section 4.2: The Chain Rule

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Before startingwith the chain rule, here is a quick quiz on using the "Calculation Thought Experiment (CTE)" discussed in the preceding tutorial. Press the "summary" button on the sidebar for a quick summary (approximately halfway down the summary page). For a more extensive review, press this button to bring up a new window with the pertinent material from the preceding tutorial.

The expression

3x1 x2ex

4

is written as:

	a difference			a quotient
	a power			none of the above

The expression

(2x5)¹(x²+3x¹)^1.5

is written as:

	a quotient			a product
	a power			none of the above

Question
Now that we are done with the preliminaries, what is the chain rule?

Answer
Here is an example: We know that the derivative of x³ is 3x². What, then, would you say is the deriviative of something more complicated raised top the third power, for instance (2x + x^1.4)³ ?

Question
Is it not just 3(2 + 1.4x^0.4)² ?

Answer

I'm afraid not. To find the correct answer, we use the chain rule.

The Chain Rule If u is a differentiable function of x, and f is a differentiable function of u, then: d dx [f(u)] = f'(u) du dx

Example Taking f(x) = x3, we get d dx u3 = 3u2 du dx In words: The derivative of a quantity cubed is 3 times that (original) quantity squared, times the derivative of the quantity. This is sometimes referred to as an example of the generalied power rule.

The following is similar to Example 1 on p. 267 of Calculus Applied to the Real World, or p. 965 of Finite Mathematics and Calculus Applied to the Real World.

d dx

(3x24)3

= ?

	18x(3x24)2			3(6x)2
	3(6x4)(3x24)2			3(3x24)2

Here is a table showing the chain rule and some examples.

Original Rule	Generalized Rule (Chain Rule)
d dx f(x) = g(x)	d dx f(u) = g(u) du dx	General form of Chain Rule
d dx xn = nx n1	d dx un = nun1 du dx	Generalized Power Rule
d dx 4x1/2 = 2x3/2	d dx 4u1/2 = 2u3/2 du dx	Example
d dx ex = ex	d dx eu = eu du dx	(See next tutorial for derivatives of exponential functions)
d dx sin x = cos x	d dx sin u = cos u du dx	(Press herefor text on trig functions.)

d dx

3x 1 x2 x1

5

= ?

	5 3x 1 x2 x1 4 3 2x + x2
	5 3x 1 x2 x1 4
	5 3(x2x1)(3x1)(2x+x2) (x2x1)2 4
	5 3x 1 x2 x1 4 3(x2x1)(3x1)(2x+x2) (x2x1)2

d dx

[(x2 1)3(3x + 4)1]

= ?

	6x(x2 1)2(3x + 4)2
	3(x2 1)2(3x+4)1 (x2 1)3(3x+4)2
	6x(x2 1)2(3x+4)1 3(x2 1)3(3x+4)2
	6x(3x+4)1 (x2 1)(3)2

Now try some of the exercises on pp. 277-280 of Calculus Applied to the Real World, or pp. 775-778 of Finite Mathematics and Calculus Applied to the Real World.