## 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.

## Examples

## 1. y = x^{3} ln x (Video)

Watch the video or read the steps below:

## Product Rule Example 1: y = x^{3} ln x

**The derivative of x ^{3} is 3x^{2}**, but when x

^{3}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 = x
^{3} - 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’= x^{3} D (ln x) + D (x^{3}) 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′= x^{3} (1/x) + (3x^{2} ln x).

Step 4: **Use algebra **to simplify the result. This step is optional, but it keeps things neat and tidy.

y′= x^{2} + 3x^{2} ln x.

*That’s it!* If you differentiate y=x^{3} ln 3, the answer is y’= x^{2} + 3x^{2} ln x.

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## Product Rule Example 2: y = (x^{3} + 7x – 7)(5x + 2)

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

- f = (x
^{3}+ 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′ = (x^{3} + 7x – 7) D(5x + 3) + D(x^{3} + 7x – 7)(5x + 3)

Step 3: **Take the derivative** of the two functions identified in the equation you wrote in Step 2.

y′ = (x^{3} + 7x – 7) (5) + (3x^{2} + 7)(5x + 3)

Step 4: Use algebra to multiply out and neaten up your answer:

y′ = 5x^{3} + 35x – 35 + 15x^{3} + 9x^{2} + 35x + 21 = 20x^{3} + 9x^{2} + 70x – 14

That’s it!

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## 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!

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## Product Rule Example 4: y = 6x^{3/2} cot x.

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

f = 6x^{3/2}

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′ = (6x^{3/2})* D (cot x) + D(6x^{3/2})* cot x

Step 3: Take the derivative of the two functions from Step 2.

y′ = (6x^{3/2})* (â€“ csc^{2} x) + (6^{(3/2)}x^{1/2})* cot x

Step 4: Use algebra to multiply out and neaten up your answer:

f′ = 6x^{3/2} â€“ csc^{2} x + 9x^{1/2}cot x

= 3x^{1/2}(2x â€“ csc^{2} x + 3 cot x)

That’s it!

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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.