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The Product Rule

This original textbook reading covers part of the AP topic. Follow the AP course outline for complete topic instruction and practice.

This reading includes verified answer corrections and explicitly stated clarifications.

The Product Rule

We can use the product rule in order to differentiate functions in the form of F(x) = f(x)•g(x). The formula for the product rule, written in various ways to help you remember, is:

$\begin{aligned} F^{'} (x) & = g (x) \cdot f^{'} (x) + f (x) \cdot g^{'} (x) \\ F^{'} & = g \cdot f^{'} + f \cdot g^{'} \\ F^{'} (x) & = f^{'} (x) \cdot g (x) + f (x) \cdot g^{'} (x) \end{aligned}$

These formulas give us the same method of finding the derivative, just different ways of writing it. Let’s practice through example, though.

Example Problem 4.1.1

Find the slope of the tangent line at $x = - 2 of$ $F (x) = e^{x} \cdot x^{2}$

Given the function $F (x) = e^{x} \cdot x^{2}$ , for the sake of using the product rule we can say $f (x) = e^{x}$ and $g (x) = x^{2}$ , giving us $f^{'} (x) = e^{x}$ and $g^{'} (x) = 2 x$ . Now that we know these values, let’s plug them into our formula.

$\begin{aligned} F^{'} (x) & = f^{'} (x) \cdot g (x) + f (x) \cdot g^{'} (x) \\ = e^{x} \cdot x^{2} + e^{x} \cdot 2 x \end{aligned}$
This is the final derivative of the function. Plugging in x=-2 for this function gives us
$\begin{aligned} F^{'} (x) & = e^{- 2} \cdot (2)^{2} + e^{- 2} \cdot 2 (- 2) \\ = 0 \end{aligned}$

Interactive 4.1.1 Example Problem 4.1.1

Magnification 1×

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The Derivative Function

The blue function is the derivative function we calculated using the product rule. Note that it crosses y=0 (the x-axis) whenever the slope of a tangent line to the original function equals 0, namely at points x=-2 and x=0

The Original Function

The graph of the original function, y=ex•x2, is shown here. Note that it has a tangent line with a slope of 0 at x=-2 and x=0. Our point of interest is x=-2

Neat, huh? Let’s get some more practice in since this is a difficult formula to use. To show that I am not making up this rule, refer to interactive 4.1.1, which compares the derivative function to the original function.

There is a more advanced version of the product rule for three functions, which is written as:

For any function $F (x) = f (x) \cdot g (x) \cdot h (x)$ ,

$\begin{aligned} F^{'} (x) & = f^{'} (x) \cdot g (x) \cdot h (x) + f (x) \cdot g^{'} (x) \cdot h (x) \\ + f (x) \cdot g (x) \cdot h^{'} (x) \end{aligned}$

Example Problem 4.1.2

Find the derivative function of $F (x) = \frac{x^{3}}{2} \cdot \sin x \cdot \cos x$

For this case, let’s make a table:

$\begin{aligned} f (x) & = \frac{x^{3}}{2} & f^{'} (x) = \frac{3}{2} x^{2} \\ g (x) & = \sin x & g^{'} (x) = \cos x \\ h (x) & = \cos x & h^{'} (x) = - \sin x \end{aligned}$

Now let’s plug these values into our formula
$\begin{aligned} F^{'} (x) & = f^{'} (x) \cdot g (x) \cdot h (x) + f (x) \cdot g^{'} (x) \cdot h (x) \\ + f (x) \cdot g (x) \cdot h^{'} (x) \end{aligned}$

$= \frac{3}{2} x^{2} \sin x \cos x + \frac{x^{3}}{2} \cos x \cos x + \frac{x^{3}}{2} \sin x (- \sin x)$

$= \frac{3}{2} x^{2} \sin x \cos x + \frac{x^{3}}{2} \cos^{2} x - \frac{x^{3}}{2} \sin^{2} x$

$= \frac{3}{2} x^{2} \sin x \cos x + \frac{x^{3}}{2} (\cos^{2} x - \sin^{2} x)$

$= \frac{3}{4} x^{2} \cdot 2 \sin x \cos x + \frac{x^{3}}{2} (\cos (2 x))$

$= \frac{3}{4} x^{2} \cdot \sin (2 x) + \frac{x^{3}}{2} \cos (2 x)$

For the last two steps, refer to the double angle formulas

This version of the book will not show you the proof of how the product rule is derived because its application is difficult enough without the theory. If you are interested, however, there are plenty of proofs using differentials and composed rectangles online to look at during your leisure time.

Example Problem 4.1.3

Find y’’ if $y = \cos x \sqrt{x}$

Let’s challenge ourselves with the first part of this problem and do it without making any tables, and doing the derivatives in our heads. When solving problems for exams, it really helps writing down the first derivative of each “product” function. This is just an exercise to increase our proficiency with the product rule.
Here, we have f(x) = cos(x) and g(x) = $\sqrt{x}$ . So, $y^{'} = \frac{\cos x}{2 \sqrt{x}} - \sin x \sqrt{x}$
A good shortcut to remember when taking the derivative of $\sqrt{x}$ is that it is equal to $\frac{1}{2 \sqrt{x}}$ . You can just as easily prove this shortcut using the power rule, since $\sqrt{x} = x^{\frac{1}{2}}$ .

Here we may want to make a table.

$\begin{aligned} f_{1} (x) & = \cos x & f_{1}^{'} (x) = - \sin (x) \\ g_{1} (x) & = x^{- \frac{1}{2}} & g_{1}^{'} (x) = - \frac{1}{2} x^{- \frac{3}{2}} \\ f_{2} (x) & = - \sin x & f_{2}^{'} (x) = - \cos x \\ g_{2} (x) & = \sqrt{x} & g_{2}^{'} (x) = \frac{1}{2} x^{- \frac{1}{2}} \end{aligned}$

$\begin{aligned} y^{''} & = \frac{1}{2} (- \sin (x) \cdot x^{- \frac{1}{2}} - \frac{1}{2} x^{- \frac{3}{2}} \cdot \cos x) \\ + (- \cos x \sqrt{x} + \frac{1}{2} x^{- \frac{1}{2}} (- \sin x)) \end{aligned}$
Now we add like terms and distribute.

$y^{''} = - \sin x \cdot x^{- \frac{1}{2}} - \frac{1}{4} \cos x \cdot x^{- \frac{3}{2}} - \cos x \sqrt{x}$

$y^{''} = - \sin x \cdot x^{- \frac{1}{2}} - \cos x (\frac{1}{4} \cdot x^{- \frac{3}{2}} + \sqrt{x})$

This is our final answer, that was so easy!

Physical Meaning of the Second Derivative: Let’s shift our gear into Physics for a few minutes. For any position function showing an object’s displacement changing with time, d(t), we can take that function’s derivative and get its velocity, v(t). We can also take its derivative again to get the change in velocity with respect to time, or the acceleration, a(t). Here are how the units for each of these terms works out, assuming distance is measured in meters, and time in seconds (these units can be replaced with km and hours, or miles and hours).

$d (t) = meters (m), a measurement of displacement$

$\overset{\cdot}{d} = v (t) = \frac{m}{s} a measurement of velocity$

$\overset{\cdot\cdot}{d} = \overset{\cdot}{v} = a (t) = \frac{m}{s^{2}} a measurement of acceleration$

Notice how increasing the order of derivatives adds another “independent variable” term to the denominator of the units. This is why the second derivative is often expressed as $\frac{d^{2} y}{d x^{2}}$ .

Pop Quiz 4.1.1 Product rule and second derivatives

Question 1 of 5 · 0 correct

If the displacement of a ball flying vertically through the air can be modeled by the equation $y = - 4.9 t^{2} + 20 t$ , what is the initial velocity of the ball (assuming units of meters and seconds)? What is the acceleration?

Glossary and deeper reading · 2 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Product rule
The product rule differentiates a product of two differentiable functions. It gives (fg)′=f′g+fg′, changing one factor at a time and adding the results. For example, the derivative of x sin x is sin x+x cos x. The product of the two derivatives is generally not the derivative of the product.
Read more: 2.8 · Product rule →
Derivative
A derivative gives the instantaneous rate of change of a function with respect to its input. At a point, it is the finite limit of the difference quotient when that limit exists. For example, the derivative of x² is 2x, so the tangent slope at x=3 is 6. A derivative function collects those rates across the inputs where the function is differentiable.
Read more: 2.2 · Derivative →

Interactive 4.1.1 Example Problem 4.1.1

The Derivative Function

The Original Function

Pop Quiz 4.1.1 Product rule and second derivatives

Product rule

Derivative

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