# Product Rule: Definition, Examples

Contents:

## What is the Product Rule?

The product rule is used to differentiate many functions where one function is multiplied by another. The formal definition of the rule is:

(f * g)′ = f′ * g + f * g′.
While this looks tricky, you’re just multiplying the derivative of each function by the other function. Recognizing the functions that you can differentiate using the product rule in calculus can be tricky. Working through a few examples will help you recognize when to use the product rule and when to use other rules, like the chain rule.

## Product Rule Example 1: y = x3 ln x

The derivative of x3 is 3x2, but when x3 is multiplied by another function—in this case a natural log (ln x), the process gets a little more complicated.

Step 1: Name the first function “f” and the second function “g.” Go in order (i.e. call the first function “f” and the second “g”).

• f = x3
• g = ln x

Step 2: Rewrite the equation using the new function names f and g you started using in Step 1:
Multiply f by the derivative of g, then add the derivative of f multiplied by g. You don’t need to actually differentiate at this point: just rewrite the equation.
y’= x3 D (ln x) + D (x3) ln x

Step 3: Take the derivative of the two functions in the equation you wrote in Step 2. Leave the two other functions in the sequence alone.
y′= x3 (1/x) + (3x2 ln x).

Step 4: Use algebra to simplify the result. This step is optional, but it keeps things neat and tidy.
y′= x2 + 3x2 ln x.

That’s it! If you differentiate y=x3 ln 3, the answer is y’= x2 + 3x2 ln x.

## Product Rule Example 2: y = (x3 + 7x – 7)(5x + 2)

Step 1: Label the first function “f” and the second function “g”.

• f = (x3 + 7x – 7)
• g = (5x + 3)

Step 2: Rewrite the functions: multiply the first function f by the derivative of the second function g and then write the derivative of the first function f multiplied by the second function, g. The tick marks mean “derivative” but we’ll use “D” instead.
y′ = (x3 + 7x – 7) D(5x + 3) + D(x3 + 7x – 7)(5x + 3)

Step 3: Take the derivative of the two functions identified in the equation you wrote in Step 2.
y′ = (x3 + 7x – 7) (5) + (3x2 + 7)(5x + 3)

Step 4: Use algebra to multiply out and neaten up your answer:
y′ = 5x3 + 35x – 35 + 15x3 + 9x2 + 35x + 21 = 20x3 + 9x2 + 70x – 14

That’s it!

## Product Rule Example 3: y = x-3 (17 + 3x-3)

Example problem: Differentiate y = x-3(17 + 3x-3) using the product rule.

Step 1: Name the functions so that the first function is “f” and the second function is “g.” In this example, we have:

• f = x-3 and
• g = (17 + 3x-3)

Step 2: Rewrite the equation: Multiply f by the derivative of g, added to the derivative of f multiplied by g.
f′ = x-3 D (17 + 3x-3) + D(x-3) (17 + 3x-3).

Step 3: Take the two derivatives of the equation from Step 2:
f′ = x-3 (-9x-4) + (-3x-4) (17 + 3x-3).

Step 4: Use algebra to expand and simplify the equation:
f′ = -9x-7-51x-4-9x-7 = -18x-7-51x-4.

That’s it!

## Product Rule Example 4: y = 6x3/2 cot x.

Step 1: Label the first function “f” and the second function “g”.
f = 6x3/2
g = cot x

Step 2: Rewrite the functions: multiply the first function f by the derivative of the second function g and then write the derivative of the first function f multiplied by the second function, g. The tick(‘) in the formal definition means “derivative” but we’ll use “D” instead.
y′ = (6x3/2)* D (cot x) + D(6x3/2)* cot x

Step 3: Take the derivative of the two functions from Step 2.
y′ = (6x3/2)* (– csc2 x) + (6(3/2)x1/2)* cot x

Step 4: Use algebra to multiply out and neaten up your answer:
f′ = 6x3/2 – csc2 x + 9x1/2cot x
= 3x1/2(2x – csc2 x + 3 cot x)

That’s it!

Tip: Don’t be tempted to skip steps, especially when multiplying out algebraically. Although you might think you’re in calculus (and therefore know it all when it comes to algebra!), common mistakes usually happen in differentiation not by the actual differentiating process itself, but when you try and multiply out “in your head” instead of being careful to multiply out piece-wise.

## Product Derivative Theorem

The product derivative theorem states that if two functions f and g are differentiable at some point x = a, then f * g is also differentiable at z. In other words, if two different functions have derivatives at a point, then their product inherits the differentiable property (Swann & Johnson, 2014).

The theorem also tells us that, for some point a:
D(f * g) = f(a)Dg + g(a)Df
In words, that’s: the derivative of f * g at some point a is equal to:

• The function f’s value at a, multiplied by the derivative of g at that point, plus
• The function g’s value at a, multiplied by the derivative of f at that point.

## References

Swann, H. & Johnson, J. (2014). Prof. E. McSquared’s Calculus Primer. Expanded Intergalactic Version! Dover Publications.