Let’s now develop “The Product Rule,” which we need to find the derivative of the product of two differentiable functions, $f\cdot g.$ For instance, maybe we need find the derivative of $e^x \sqrt{x},$ which is the product of $e^x$ and $\sqrt{x}.$ Or perhaps we’d like to find the derivative of $x \sin x,$ which is the product of $x$ and $\sin x.$

The goal here is to develop a simple rule to compute the derivative of the product of two such differentiable functions, $f \cdot g.$ We’ll also build a sense for why the rule has the form that it does. We assume from the outset that the two functions, *f* and *g*, are both differentiable, so $f’$ and $g’$ exist.

A naive approach that does *not* work

Let’s think through a super-quick example you can do in your head to illustrate a crucial point: Consider the function $x \cdot x,$ which is the product of $x$ and $x.$ Of course we can instead view this product as being the function $x^2,$ and so you immediately know from the Power Rule that the derivative is $\dfrac{d\left(x^2\right)}{dx} = 2x.$ But what if we view it as the product $x \cdot x$ and choose to compute its derivative using whatever approach comes immediately to mind for “the product rule”? If you were going to naively calculate this derivative of the function $x$ multiplied by the function $x,$ without knowing anything else, what would you try?

If you’re like most people who first encounter this question, you would find the derivative of each function individually and then simply multiply them together. We are going to spoil this naive thought immediately and say that this approach is incorrect: $\left[ f(x)\,g(x)\right]’ \ne f'(x)\, g'(x).$ Our simple example immediately illustrates that this approach does *not* work: applying this naive reasoning to $x \cdot x,$ we find

\begin{align*}

\dfrac{d\left( x^2 \right)}{dx} = \dfrac{d(x \cdot x)}{dx} &\overbrace{=}^? \dfrac{dx}{dx} \cdot \dfrac{dx}{dx} &&\text{[Does naive approach work??]} \\[8px]
&\overbrace{=}^? 1 \cdot 1 \\[8px]
&\overbrace{=}^? 1 \ne 2x \quad \xmark &&\color{red}{\left[\text{No!} \quad \dfrac{d(f\cdot g)}{dx} \ne \dfrac{df}{dx} \cdot \dfrac{dg}{dx} \right]}

\end{align*}

Do *not* simply multiply the two derivatives together

That is, since the naive approach does Instead, the derivative of the product of two functions is a little (only a little!) more complicated.

Developing the Product Rule

Let’s ground our work here in what the derivative of this product of functions *tells* us: We’re finding the *rate* at which the product $f(x)\,g(x)$ changes when we vary the input *x* by a little bit. (Remember, a derivative is *always* telling how much a function changes, or “reacts,” when we change its input a little bit. The function we’re currently considering happens to be a function which is the product of two other functions.) For instance, we can imagine increasing the input *x* by a small amount *h,* to $x+h.$ The animation to the right uses our usual area model of multiplication to illustrate the idea.

To develop the correct Product Rule, we return — as always — to the Definition of the Derivative, now applied to the product of two functions, $f(x)\,g(x)$:

\[[f(x)\,g(x)]’ = \lim_{h \to 0} \frac{f(x+h)\,g(x+h)\, -\, f(x)\,g(x)}{h}\]
You can see how the numerator of this expression represents the difference in the areas between the newly-grown rectangle, $f(x+h)\,g(x+h),$ and the original rectangle, $f(x)\,g(x).$

Looking at the expression, it’s not immediately clear what to do with the limit and hence how to proceed. But the area model will help, since we can use it to write this initial expression in a way that includes the the terms

\[\frac{f(x+h) – f(x)}{h} \quad \text{ and } \quad \frac{g(x+h) – g(x)}{h} \]
Once those terms are in place, when we take the limit as $h \to 0,$ those two expressions will be $f'(x)$ and $g'(x)$ respectively, as you’ll see!

As the derivation in the animation shows (and as summarized in images you can view and download a bit further down on the screen), the Product Rule for Derivatives is

In prime notation:

\begin{align*}

[f(x) \, g(x)]’ &= f'(x)\,g(x) + f(x) \,g'(x) \\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }}]\, + \,[{\small \text{ (the first) } \times \text{ (derivative of the second)}}] \end{align*}

And in Leibnitz notation:

\begin{align*}

\dfrac{d}{dx}(fg)&= \left(\dfrac{d}{dx}f \right)g + f\left(\dfrac{d}{dx}g \right)\\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }}]\, + \,[{\small \text{ (the first) } \times \text{ (derivative of the second)}}] \end{align*}

Crucially, using our area model we can make sense of this rule as shown in the following figure:

Open/Close images that summarize the Product Rule derivation

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If you’d like to see a different approach to deriving the Product Rule, open this box. We’ll rely on the “trick” used in this derivation on the next screen when we develop the Quotient Rule.

Show/Hide an algebra-based derivation of the Product Rule

Another very common derivation starts again (of course) with the Definition of the Derivative as applied to the product $f(x)\,g(x):$

\[[f(x)\,g(x)]’=\lim_{h \to 0} \frac{f(x+h)\,g(x+h)\, -\, f(x)\,g(x)}{h}\] Instead of using the area model to rewrite the numerator as we did above, this approach employs an algebraic “trick” to allow us to factor and then see the terms $f'(x)$ and $g'(x).$ The trick is that, in the numerator, we will add 0 in the form of $f(x)\, g(x+h)\, -\, f(x)\, g(x+h)$, and then do some fancy factoring from there. (Students often ask how you’re supposed to know to use this “trick.” The answer is you’re not: essentially with this approach you already know the answer you’re after, and then figure out how to cleverly manipulate the terms to get there.)

\begin{align*}

[f(x)\,g(x)]’ &= \lim_{h \to 0} \frac{f(x+h)\,g(x+h)\, \overbrace{-\,f(x)\,g(x+h) + f(x)\,g(x+h)}^{= 0}\, -\, f(x)\,g(x)}{h} &&\text{[Now factor this]}\\[8px] &= \lim_{h \to 0} \frac{[f(x+h)-f(x)]\,g(x+h) + f(x)\,[g(x+h)-g(x)]}{h} \\[8px] &= \lim_{h \to 0}\frac{[f(x+h)-f(x)]\,g(x+h)}{h} + \lim_{h \to 0}\frac{f(x)\,[g(x+h)-g(x)]}{h} \\[8px] &= \overbrace{\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}}^{=f'(x)} \cdot \lim_{h \to 0}g(x+h) + \lim_{h \to 0}f(x) \cdot \overbrace{\lim_{h \to 0} \frac{g(x+h)-g(x)}{h}}^{=g'(x)} \\[8px] &= f'(x)\,g(x) + f(x)\,g'(x) \quad \cmark

\end{align*}

From the fourth to the fifth line, we used the fact that

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Most students find that the Product Rule becomes automatic and routine after you use it a fair number of times — and you will be using it *often* in the work to come. We’ll of course let you practice below, and you should expect similar problems on both your midterm and final exams.

Let’s first consider a few Examples to show how it works in practice.

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

*Solution.*

To apply the Product Rule, we first need to identify the two functions being multiplied, and then find the derivative of each:

\[h(x) = \overbrace{e^x}^{f(x)} \cdot \overbrace{\sqrt{x}}^{g(x)}\]
\begin{align*}

f(x) = e^x &\quad && g(x) = \sqrt{x} = x^{1/2} \\[8px]
f'(x) = e^x &\quad && g'(x) = \dfrac{1}{2\sqrt{x}}

\end{align*}

We can now apply the Product Rule:

\begin{align*}

h(x) &= f(x) \cdot g(x) \\[8px]
h'(x) &= f'(x)\cdot g(x) + f(x)\cdot g'(x) \\[8px]
h'(x) &= \overbrace{e^x}^{f'(x)} \cdot \overbrace{\sqrt{x}}^{g(x)} + \overbrace{e^x}^{f(x)} \cdot \overbrace{\frac{1}{2\sqrt{x}}}^{g'(x)} \\[8px]
&= e^x \sqrt{x} + \frac{e^x}{2\sqrt{x}} \quad \cmark

\end{align*}

That’s it. As long as you remember to find the derivative of each function separately (even if just in your head) and then make the correct substitutions in the Product Rule, this process is straightforward.

Consider $f(x) = (x^3 + x) \cdot (3x^3 + 9x – 6)$.

**(a)** Use the Product Rule to find $f'(x).$

**(b)** Instead, multiply out the two polynomials to make a single polynomial, and then find $f'(x)$ using only the Power Rule. Compare the result to that of (a).

*Solution.*

**(a)** First, we identify the two functions being multiplied, and take their derivatives. (Since the overall function is called $f(x)$, we’re going to call the two functions being multiplied $u(x)$ and $w(x),$ but you could call them whatever you want. Soon, you won’t call them anything special at all, and will just take these derivatives in your head as you go.)

\[f(x) = \overbrace{(x^3 + x)}^{u(x)} \cdot \overbrace{(3x^3 + 9x\,-\,6)}^{w(x)}\]
\begin{align*}

u(x) = x^3 + x &\quad && w(x) =3x^3 + 9x – 6 \\[8px]
u'(x) = 3x^2 + 1 &\quad && w'(x) = 9x^2 + 9

\end{align*}

Then applying the Product Rule, we get

\begin{align*}

f(x) &= u(x) \cdot w(x) \\[8px]
f'(x) &= u'(x)\cdot w(x) + u(x)\cdot w'(x) \\[8px]
&= \left[ (3x^2 + 1)\cdot (3x^3 + 9x \,-\, 6)\right] + \left[ (x^3 + x)\cdot(9x^2 + 9)\right] \\[8px]
&= \left[ 3x^2 (3x^3 + 9x \,-\, 6) + 1(3x^3 + 9x \,-\, 6)\right] + \left[ x^3 (9x^2 + 9) + x(9x^2 + 9)\right]\\[8px]
&= \left[9x^5 + 27x^3 -18x^2 + 3x^3 + 9x \,-\, 6\right] + \left[ 9x^5 + 9x^3 + 9x^3 + 9x \right]\\[8px]
&= 18x^5 + 48x^3 – 18x^2 + 18x \,-\, 6 \quad \cmark

\end{align*}

**(b)** We first multiply out the polynomials:

\begin{align*}

f(x) &= (x^3+x)(3x^3+9x-6) \\[8px]
&= x^3(3x^3+9x-6) + x(3x^3+9x-6) \\[8px]
&= 3x^6 + 9x^4 – 6x^3 + 3x^4 + 9x^2 \,-\, 6x \\[8px]
&= 3x^6 + 12x^4 -6x^3 + 9x^2 \,-\, 6x

\end{align*}

Then, using the Power Rule

\[f'(x) = 18x^5 + 48x^3 – 18x^2 + 18x \,-\, 6 \quad \cmark \]
This is happily the same result as we found in (a).

You can see that we can use using either method to find $f'(x)$ here. With experience, you’ll see how in some cases using the Product Rule is easier, while in other cases simplifying the multiplication and then taking the derivative term-by-term is easier.

Time to practice! This new rule may take a little getting used to, so please practice until you feel more comfortable with it. We’ll be using it a *lot* going forward, so the more you can make it part of your working toolkit now, the easier things will be later.

Show/Hide Statement of the Product Rule

In prime notation:

\begin{align*}

[f(x) \, g(x)]’ &= f'(x)\,g(x) + f(x) \,g'(x) \\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }}]\, + \,[{\small \text{ (the first) } \times \text{ (derivative of the second)}}] \end{align*}

And in Leibnitz notation:

\begin{align*}

\dfrac{d}{dx}(fg)&= \left(\dfrac{d}{dx}f \right)g + f\left(\dfrac{d}{dx}g \right)\\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }}]\, + \,[{\small \text{ (the first) } \times \text{ (derivative of the second)}}] \end{align*}

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Let’s consider some problems where we need to find the derivative of the product of 3 functions. The Product Rule extends as you might expect:

\begin{align*}

\dfrac{d}{dx}(fgh)&= \left(\dfrac{d}{dx}f \right)gh + f\left(\dfrac{d}{dx}g \right)h + + fg\left(\dfrac{d}{dx}h \right)\\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }\times \text{ (the third) }}] \\

&\qquad + [{\small \text{ (the first) } \times \text{ (derivative of the second)}\times \text{ (the third) }}] \\

&\qquad \qquad + [{\small \text{ (the first) } \times\text{ (the second) } \times \text{ (derivative of the third)}}] \end{align*}

The Product Rule for 4 or more functions extends similarly.

Show/Hide Development of Product Rule for 3 Functions

We developed above the rule for finding the derivative of the product of 2 functions $f(x)$ and $j(x)$:

\[\dfrac{d}{dx}(fj)= \left(\dfrac{d}{dx}f \right)j + f\left(\dfrac{d}{dx}j \right)\] Let’s say now that the function

\[\dfrac{d}{dx}j = \dfrac{d}{dx}(gh)= \left(\dfrac{d}{dx}g \right)h + g\left(\dfrac{d}{dx}h \right)\] Substituting the preceding equation into the one above gives us the result we’re after, for the derivative of the product of three functions:

\begin{align*}

\dfrac{d}{dx}(fj) &= \left(\dfrac{d}{dx}f \right)j + f\left(\dfrac{d}{dx}j \right) \\[8px] \dfrac{d}{dx}(fj) = \dfrac{d}{dx}(fgh) &= \left(\dfrac{d}{dx}f \right)j + f\left[\left(\dfrac{d}{dx}g \right)h + g\left(\dfrac{d}{dx}h \right) \right] \\[8px] &= \left(\dfrac{d}{dx}f \right)gh + f\left(\dfrac{d}{dx}g \right)h + + fg\left(\dfrac{d}{dx}h \right)\\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }\times \text{ (the third) }}] \\

&\qquad + [{\small \text{ (the first) } \times \text{ (derivative of the second)}\times \text{ (the third) }}] \\

&\qquad \qquad + [{\small \text{ (the first) } \times\text{ (the second) } \times \text{ (derivative of the third)}}] \end{align*}

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- The Product Rule for finding the derivative of the product of two functions is

In prime notation:

\begin{align*}

[f(x) \, g(x)]’ &= f'(x)\,g(x) + f(x) \,g'(x) \\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }}]\, + \,[{\small \text{ (the first) } \times \text{ (derivative of the second)}}] \end{align*}

And in Leibnitz notation:

\begin{align*}

\dfrac{d}{dx}(fg)&= \left(\dfrac{d}{dx}f \right)g + f\left(\dfrac{d}{dx}g \right)\\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }}]\, + \,[{\small \text{ (the first) } \times \text{ (derivative of the second)}}] \end{align*}

On the next screen, we’ll develop the Quotient Rule, so we can easily find the derivative of the quotient of two functions.

For now, what do you think about the Product Rule? Weird? Difficult to use? Easy? Please post on the Forum and let the Community know your thoughts . . . and also ask any questions that you have, or help answer someone else’s!