Introduction

Welcome to another article with another rule! In this article, you will be going over the product rule. The product rule is an extremely vital rule that you must know to succeed in AP Calculus AB/BC, as it tells you how to differentiate a product of two functions. In other words, you get to differentiate even more stuff!

The Product Rule

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

This is the CORRECT way to differentiate the product of two functions. Notice that this rule is a more general rule. $f(x)$ and $g(x)$ are both functions which really can be anything. 

It may be easier to remember this, where the functions are named with one letter.

$\frac{d}{dx}[fg]=f’g+g’f$

To be clear, $f$ and $g$ both represent the two functions. This is simply a shorter version you can remember more easily. 

Differentiating Using the Product Rule

Calculate the derivative of $e^x\ln x$ with respect to $x$.

We can simply plug in the functions, but it would be easier if we could organize these functions. Whenever you differentiate using the power rule, you are required to have both functions and their individual derivatives, so $f$, $g$, $f’$, and $g’$. We can list them out for starters.

$f(x)=e^x$

$g(x)=\ln x$

It is okay for you to set $g(x)=e^x$, it doesn’t matter, and you will get the same answer because of all these commutative properties of addition and multiplication (ie $f’g+g’f=g’f+f’g$). Our next step is to find $f’$ and $g’$, so we will differentiate these functions like we’ve done in the past.

$f’(x)=e^x$

$g’(x)=\frac{1}{x}$

Now that we have all our information, we can simply plug it into our formula, and we should get the following. 

$f’g+g’f=e^x\ln x+\frac{1}{x}\cdot e^x=e^x\ln x+\frac{e^x}{x}$

You could leave it like this or factor, but what matters is that you display the final answer correctly with the correct work.

Derivative at a Point

$h(x)=\sin (2x)$. Solve for $h’\left(\frac{\pi }{2}\right)$

Notice that this is one function here, not two functions being multiplied with each other. However, we can turn this into a product of two functions using a trigonometric identity.

$\sin (2x)=2\sin (x)\cos (x)$

We can now rewrite the function as $h(x)=2\sin (x)\cos (x)$, and do the same thing we did before, which is to solve for all the required information and plug it all into the rule.

$f(x)=2\sin (x)$

$g(x)=\cos (x)$

$f’(x)=2\cos (x)$ $g’(x)=-\sin (x)$

$f’g+g’f=2\cos (x)\cdot \cos (x)+(-\sin (x)\cdot 2\sin (x))=2{\cos }^2(x)-2{\sin }^2(x)=2\cos (2x)$

For the last step, this is simply a trigonometric identity. You won’t use trigonometric identities a ton in AP Calculus AB/BC, but it’s a good way to simplify possibly complex functions. 

Now, we aren’t done yet. We have solved for $h’(x)$, but not for $h’\left(\frac{\pi }{2}\right)$ yet, so let’s just quickly do that.

$h’\left(\frac{\pi }{2}\right)=2{\cos }^2\left(\frac{\pi }{2}\right)-2{\sin }^2\left(\frac{\pi }{2}\right)=2(0{)}^2-2(1{)}^2=-2$

The derivative of $h(x)$ at $x=\frac{\pi }{2}$ is $-2$.

Conceptual Understanding of the Product Rule

Example 3:

Given $f(3)=5$, $f’(3)=\sqrt{8}$, $g(3)=\sqrt{2}$, $g’(3)=-1$, If $j(x)$ = $f(x)\cdot g(x)$, calculate $j(3)$ and $j’(3)$.

Don’t try to overcomplicate this, as simply being organized and listing out what is required will allow you to easily solve this. We need the individual functions and their derivatives, but we have the functions at a point. However, do not fret because the point at which we want to find the value and derivative of $j(x)$ is the same point that is given to us in the individual functions. Everything is at $x=3$. For now, let’s just solve for $j(3)$ only, no derivative.

$j(3)=f(3)\cdot g(3)=5\cdot \sqrt{2}=5\sqrt{2}$

We can use the product rule, and let’s just list out all the functions for now.

$f(x)=f(x)$

$g(x)=g(x)$

$f’(x)=f’(x)$

$g’(x)=g’(x)$

Looks a little bit confusing, but hopefully you get why each one is what it is. If you don’t, look back at the question. Using the rule one more time, we get the following.

$f’g+g’f=f’\cdot g+g\cdot f’$

We have solved for $j’(x)$, and now let’s just plug in $x=3$.

$j^{\prime }(3)=f’(3)g(3)+g(3)f’(x)=(\sqrt{8})(\sqrt{2})+(-1)(5)=4-5=-1$.

Practice

You know the drill!