The product rule is a very useful tool to use in finding the derivative of a function that is simply **the product of two simpler functions**. There are a few different ways you might see the product rule written. I would recommend picking whichever one is easiest for you to remember and understand so that you can work with it from memory. Although you may only need to remember one of these, you should be able to recognize the other representations as the product rule when you see them. All of the following are different ways of writing the product rule:

Take any two functions *f(x)* and *g(x)*.

$$\frac{d}{dx} \Big[f(x)g(x) \Big]=f(x) \frac{d}{dx} \Big[g(x) \Big]+g(x) \frac{d}{dx} \Big[f(x) \Big]$$

$$\big(f \cdot g \big)’=f \cdot g’+g \cdot f’$$

$$\big(f(x) \cdot g(x) \big)’=f(x) \cdot g'(x)+g(x) \cdot f'(x)$$

Of course, when you add two things together the order doesn’t matter. In other words \(a+b=b+a\). The same is true for multiplication, \(a \cdot b=b \cdot a\). As a result we can reorder the product rule equations shown above, so

$$\big(f \cdot g \big)’=f’ \cdot g+g’ \cdot f.$$

## How to remember the product rule

As long as you multiply the first function with the derivative of the of the second and multiply the second function with the derivative of the first, then add the two together, it will work out. The phrase I like to think about when I need to remember the product rule is:

“The derivative of the first times the second plus the derivative of the second times the first.”

I like that phrasing personally, but I recommend you come up with some trick for remembering the product rule because it comes up frequently.

## How to start with the product rule

The next thing that is important to discuss is how to decide which function will be \(f(x)\) and which function will be \(g(x)\) when you realize you need to use the product rule. As long as you can reduce the full function to the product of two smaller functions, the product rule can be applied and it doesn’t really matter which part of the product you call \(f\) and which one you call \(g\).

Let’s consider the following example:

## Example 1

Find the derivative of \(h(x) = e^x \sqrt{x}\).

#### Recognizing when to use product rule

This is perhaps one of the more obvious cases where we can see that the product rule can be used here. This function is simply the product of the two simpler functions: \(e^x\) and \(\sqrt{x}\).

#### How to begin

The first thing we need to do is decide which of those two functions should be \(f\) and which will be \(g\). As I mentioned before, this decision doesn’t matter as long as you **stick with it through the whole problem** (you will have issues if you switch the two half way through the problem).

I will say \(f(x)=e^x\) and \(g(x)=\sqrt{x}\). Once you name your \(f\) and \(g\) functions, the next thing I would recommend doing is figure out the derivative of each.

#### Dealing with each piece of the formula

Remember, the product rule formula requires the use of \(f’\) and \(g’\), so it’s easiest to figure those out now so we can just plug them in.

First we’ll find \(f'(x)\). Luckily this one is easy. The derivative of \(e^x\) is also \(e^x\). Or $$\frac{d}{dx}e^{x}=e^{x}$$ Therefore we have $$f'(x)=e^{x}.$$

Now we’ll find \(g'(x)\). This one is a bit more tricky, but keep in mind, there is another way to write \(\sqrt{x}\) that makes finding the derivative seem a lot easier. Always remember, \(\sqrt{x}\) can also be written as \(x^{\frac{1}{2}}\). $$\sqrt{x}=x^{1/2}.$$ Now that we have it written as \(x\) to some power, we can find this derivative using the power rule.

Just move the power down in front and lower the power of \(x\) by \(1\). This means $$\frac{d}{dx}x^{\frac{1}{2}}=\frac{1}{2}x^{-\frac{1}{2}}.$$ Therefore, we know

$$g'(x)=\frac{1}{2}x^{-\frac{1}{2}}.$$

#### Putting it all together

Now we have figured out \(f’\) and \(g’\), so we have all the pieces we need to plug into the product rule formula. To summarize, so far we have:

$$f(x)=e^{x}$$

$$g(x)=\sqrt{x}=x^{\frac{1}{2}}$$

$$f'(x)=e^{x}$$

$$g'(x)=\frac{1}{2}x^{-\frac{1}{2}}$$

When you are using the product rule, I would recommend listing out \(f\), \(g\), \(f’\), and \(g’\) before trying to plug anything into the power rule formula. **It is best to list these out all in one place, like I have done above so it’s easy to refer back to all of the pieces you will need to use.**

Now we just need to plug all of these into our product rule formula. I will use \(h'(x)=f'(x)\cdot g(x)+g'(x)\cdot f(x)\), but as I said before, any of the formulas shown previously will work. Use the one with notation that you are most comfortable with.

$$h'(x)=\big(e^{x}\big)\Big(x^{\frac{1}{2}}\Big)+\bigg(\frac{1}{2}x^{-\frac{1}{2}}\bigg)\big(e^{x}\big)$$

This is a perfectly acceptable answer for \(h'(x)\), but we can also simplify this by factoring out an \(e^{x}\). Doing this tells us that:

$$h'(x)=\big(e^{x}\cdot \sqrt{x}\big)’=e^{x}\bigg(x^{\frac{1}{2}}+\frac{1}{2}x^{-\frac{1}{2}}\bigg)$$

## Product Rule with 3 Functions

My favorite way to think about using the produce rule with 3 terms is the way it’s explained in The Calculus Lifesaver by Adrian Banner. This book explains everything you need to know in calculus in a very intuitive way that makes a lot of sense when you’re trying to learn calculus and it is very affordable priced under $20 on Amazon. I highly recommend checking that book out.

In that book, Adrian Banner describes how you can actually apply the product rule to find the derivative of a function that is the product of any number of smaller functions.

When it is being applies to 3 functions, the formula is fairly simple and follows the same pattern as using the product rule on only 2 terms.

Let’s say we want to find the derivative of $$f(x)=(x^3+1)(x^4-2x+6)(x^2-x)$$

First we start with assigning each term to a new variable. So let’s say:

$$u=x^3+1$$ $$v=x^4-2x+6$$ $$w=x^2-x$$

Then we can apply these three terms to the product rule formula for 3 terms.

$$f'(x)= \frac{du}{dx}vw + u \frac{dv}{dx} w + uv \frac{dw}{dx}$$

So you can see that we will need to find the derivative of each of our 3 terms individually before we plug them into the product rule formula. All 3 of these derivatives can be found using the power rule.

$$\frac{du}{dx}= 3x^2$$ $$\frac{dv}{dx}= 4x^3-2$$ $$\frac{dw}{dx}= 2x-1$$

And now we can simply plug all of these pieces into the formula to get the derivative.

$$f'(x)= (3x^2)(x^4-2x+6)(x^2-x)$$ $$\ + (x^3+1) (4x^3-2) (x^2-x)$$ $$\ + (x^3+1)(x^4-2x+6) (2x-1)$$

And that’s all there is to it. If you want to get some more practice with derivatives and other rules and shortcuts you should check out the derivatives page. There you will find a list of lesson and full problems I have worked through to help you make sense of derivatives. You can also see more product rule practice problems here. If you can’t find what you’re looking for there then let me know with an email to **jakesmathlessons@gmail.com**. Send me your questions and I’ll do my best to answer them and may even post a full lesson to address your question.

** Links on this page may be affiliate links meaning I would get a small commission from your purchase at no additional cost to you. **