## 4.1 Product and Quotient Rule

(This topic is also in Section 4.1 in Applied Calculus and Section or Section 11.1 of Finite Mathematics and Applied Calculus)

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

 Q The statement ddx x2x5 = 2x5x4 is:

 wrong, because the correct answer is -3x4 wrong, because the correct answer is 13x2 wrong, because the correct answer is 03x2 = 0 wrong, because the correct answer is ln |x3|

The above example suggests the following:

The derivative of a quotient is not the quotient of the derivatives.

The derivative of a product is not the product of the derivatives.

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

 ddx [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.

Quotient Rule

 ddx f(x)g(x) = 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..

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

Using the Product Rule
(There are examples similar to the following in Section 4.1 of Applied Calculus, or Section 11.1 Finite Mathematics and Applied Calculus. )

Let us find the derivative of

 f(x) = (4x3-x4) (11x - x ).

Before we start, first recognize that f(x) is a product of two factors:

 (4x3-x4)

and

 (11x - x ).

Therefore, the product rule applies. Before using the rule, let us first rewrite the function in exponent form:

f(x) = (4x3-x4) (11x-x0.5 ).

Now we can apply the product rule. The rule says that:

 ddx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x).

Therefore,

 ddx (4x3-x4) (11x-x0.5 )
 = (12x2-4x3)(11x-x0.5 ) + (4x3-x4)(11-0.5x0.5 ). (deriv. of first)(second left alone) + (first left alone) (deriv. of second)

In some circumstances, it is helpful to simplfy the answer. Use your best judgement as the situation arises. Now here are some for you to try. You need not simply the answers for these.

Note Use proper graphing calculator format to input your answers (spaces are ignored). For example, input

(3x-2 + 2/x)(x + 1)     as     (3*x^(-2) + 2/x)*(x+1)

Q
 ddx (x2 - x)(5 + x-0.5)
 =
Q
 ddx x2 - 1x2 (5 + x-0.5)
 =

Using the Quotient Rule

Next, let us calculate the derivative of

 f(x) = 3x2-2xx3 + x .

First, we recognize that f(x) is a quotient (one expression divided by another) and so we need to use the quotient rule:

 ddx f(x)g(x) = f'(x)g(x)-f(x)g'(x)[g(x)]2
 ddx 3x2-2xx3 + x = (6x-2)(x3 + x) -(3x2-2x) (3 x2 + 1)(x3 + x)2

That is the answer, although it is sometimes useful to simplify the numerator. In the question below, you will need to manipulate the answer a little in order to match the correct choice.

 Q The derivative of f(x) = x2 + 3x + 2x-1 is ?

 3x2+ 4x-1x2-2x + 1 x2-2x-5x2-2x + 1 x2-2x-5x2-1 2x + 31

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" creatures, such as

 (3x+1) x2+ 4x2+x .

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

Using the Calculation Thought Experiment (CTE)

Let us use the CTE to find the derivative of

 (3x+1) x2+ 4x2+x .

To use this method, pretend you were calculating, one step at a time, the value of 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:

1. Calculate the quotient (x2+4)/(x2 + x) = (25+4)/(25+5) = 29/30.
2. Calculate (3x+1) = 15+1 = 16
3. Multiply the answers from steps 1 and 2.

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

 f(x) = (3x+1) x2+ 4x2+x .

Thus,

 f'(x) = ddx (3x+1) x2+ 4x2+x + (3x+1) ddx x2+ 4x2+x . . . . (I) (deriv. of first) (second left alone) + (first left alone) (deriv. of second)

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. (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:

 ddx (3x+1) = 3

To calculate the second (blue) derivative, we need the quotient rule (use the CTE on the expression if you don't believe this...)
 ddx x2+ 4x2+x = (2x)(x2+x)-(x2+4)(2x+1)(x2+x)2 = x2-8x-4(x2+x)2

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

 f'(x) = 3 x2+ 4x2+x + (3x+1) x2-8x-4(x2+x)2

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

 4x2 + (x-1)(4x+1)3x Select one a product a quotient a sum a difference a power none of the above

Q A valid first step in the calculation of the derivative of

 4x2 + (x-1)(4x+1)3x

is to write down:

 8x + (1)(4)(3)
 ddx (4x2) + ddx (x-1)(4x+1)3x
 8x + ddx (x-1)(4x+1)3x
 8x + (1)(4)(3x) - (x-1)(4x + 1)(3)9x2

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

(3x-2 + 2/x)(x + 1)     as     (3*x^-2 + 2/x)*(x+1)  Q Finally, the derivative of 4x2 + (x-1)(4x+1)3x simplifies to

Now try some of the exercises on Section 4.1 of Applied Calculus, or Section 11.1 of Finite Mathematics and Applied Calculus.

Last Updated: March, 2007
Copyright © 1999, 2003, 2006, 2007 Stefan Waner