Product and Quotient Rule

(This topic is also in Section 4.3 in Applied Calculus 5e and Section or Section 11.3 of Finite Mathematics and Applied Calculus 5e)

Here's a little quiz to warm up. (Go to the tutorial on derivatives of powers if you want to first review derivatives of powers of x.)

Q Your friend tells you that . This claim is:

The above example suggests the following:

Q So how do we deal with products and quotients? Must we always somehow simplify a given expression into exponent form in order to use the power rule?
A No. Luckily, there are two convenient rules for handling products and quotients:

Product and Quotient Rules

Product Rule

    d

    dx
    		 [f(x)g(x)] = f'(x) g(x) + f(x)g'(x)

Product Rule In Words:

The derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second.

Quick Examples:

    \frac{d}{dx}[ (3x^3 + 7x^2)(4x^2-x+4)] = Deriv of First × Second + First × Deriv of Second
    = (9x^2 + 14x)(4x^2-x+4) + (3x^3 + 7x^2)(8x - 1)

    Note  In some circumstances, it is helpful to simplfy the answer. Use your best judgement as the situation arises. Below are some for you to try. You need not simplify the answers for these.
    \frac{d}{dx}[ ] =
       
    \frac{d}{dx}[ ] =
         
    \frac{d}{dx}[ ] =
         

Quotient Rule

    \frac{d}{dx}\Bigleft[\frac{f(x)}{g(x)}\Bigright] = \frac{f'(x)\.g(x) - f(x)\.g'(x)}{g(x)^2}

Quotient Rule In Words:

The derivative of a quotient is the derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.

Quick Examples:

    \frac{d}{dx}\Bigleft[\frac{3x^3 + 7x^2}{4x^2-x+4}\Bigright] =
    Deriv of Top × Bottom - Top × Deriv of Bottom

    (Bottom)2
    = \frac{(9x^2 + 14x)(4x^2-x+4) - (3x^3 + 7x^2)(8x - 1)}{(4x^2-x+4)^2}
    \frac{d}{dx}  =
          ↑ Unsimplified
    \frac{d}{dx}  =
          ↑ Simplified!
    \frac{d}{dx}  =
            ↑ Unsimplified
    \frac{d}{dx}  =
            ↑ Unsimplified

 
Q Where do these rules come from?
A You can find a proof of the product rule in Section 4.1 of Applied Calculus, or Section 11.1 of Finite Mathematics and Applied Calculus. Press here for a proof of the quotient rule.

Q This is all very well if what you're given is an obvious product or quotient. But we all know that instructors are fond of "in-between" expressions, such as

Which rule do we use for that ??

A To deal with things like that -- or any mathematical expression whatsoever, we use the following little secret desribed in Applied Calculus and Finite Mathematics and Applied Calculus, called the Calculation Thought Experiment:

Calculation Thought Experiment (CTE)

The calculation thought experiment is a technique to determine whether to treat an algebraic expression as a product, quotient, sum, or difference:

    Given an expression, consider the steps you would use in computing its value. If the last operation is multiplication, treat the expression as a product; if the last operation is division, treat the expression as a quotient, and so on.

Using the Calculation Thought Experiment (CTE) to Differentiate a Function

    If the CTE says, for instance, that the expression is a sum of two smaller expressions, then apply the rule for sums as a first step. This will leave you having to differentiate simpler expressions, and you can use the CTE on these, and so on...

Examples

1. (3x2-4)(2x+1) can be computed by first calculating the expressions in parentheses and then multiplying. Since the last step is multiplication, we can treat the expression as a product.

2. (2x-1)/x can be computed by first calculating the numerator and denominator, and then dividing one by the other. Since the last step is division, we can treat the expression as a quotient.

3. x2 + (4x-1)(x+2) can be computed by first calculating x2, then calculating the product (4x-1)(x+2), and finally adding the two answers. Thus, we can treat the expression as a sum.

4. (3x2-1)5 can be computed by first calculating the expression in parentheses, and then raising the answer to the fifth power. Thus, we can treat the expression as a power.

5. The expression is written as

6. The expression is written as

7. The expression is written as

8. The expression is written as

Using the Calculation Thought Experiment (CTE)

Let us use the CTE to find the derivative of

To use this method, pretend you were calculating, one step at a time, the value of this funtion for, say, x = 5. (You don't need to actually do the calculation.) One way of doing the calculation would be to use the following procedure:

Since the last operation is multiplication, the CTE tells us that the given expression is a product and so we must use the product rule.

Thus,

Remember that the expressions "d/dx" are short-hand for "the derivative of ..." In other words, we haven't done the work yet; the line above is just telling us what we need to do: take two derivatives. (If we wanted, we could take a coffee break and come back to it later to do the work.)

To finish the calculation, we must compute the magenta- and blue-colored derivatives one-at-a-time and plug them in to the expression above:

The first (magenta) derivative is easy:

To calculate the second (blue) derivative, we need the quotient rule (use the CTE on the expression if you don't believe this...).

Now substitute these derivatives into formula (I) to obtain the answer:

Whew ! Now you do one:
(Similar to an example of Applied Calculus, or Finite Mathematics and Applied Calculus. )

Q The Calculation Thought Experiment (CTE) tells us that

Q Thus, a valid first step in the calculation of the derivative of

is to write down:

For the next question, you need to enter an algebraic expression using proper graphing calculator format as above (spaces are ignored).

Q Finally, the derivative of equals

 

Now try some of the exercises on Section 4.3 of Applied Calculus, or Section 11.3 of Finite Mathematics and Applied Calculus.

Top of Page

Last Updated: October, 2009
Copyright © 2008 Stefan Waner