The Product Rule

How do you differentiate $x^2 \sin x$ or $xe^x$? You can’t expand them, and the derivative of a product is not the product of the derivatives. The product rule gives the right answer, and it’s one of the rules you’ll use most for the rest of calculus.

Key ideas

The product rule

If $f$ and $g$ are differentiable, then

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

In words: derivative of the first times the second, plus the first times the derivative of the second. Some students remember it as "$(fg)' = f'g + fg'$".

The order of the two terms doesn’t matter, because they are added.

Why it works: an area picture

Think of $u = f(x)$ and $v = g(x)$ as the sides of a rectangle with area $uv$. When $x$ changes a little, $u$ grows by $\Delta u$ and $v$ grows by $\Delta v$.

The extra area is $v\,\Delta u + u\,\Delta v + \Delta u\,\Delta v$. The tiny corner becomes negligible.

Divide the extra area by $\Delta x$ and let $\Delta x \to 0$. The two strips become $v\,\dfrac{du}{dx} + u\,\dfrac{dv}{dx}$, and the corner piece $\Delta u \cdot \dfrac{\Delta v}{\Delta x}$ goes to $0$ (since $\Delta u \to 0$). That’s the product rule.

Three factors

For a product of three functions, take turns differentiating one factor at a time:

$$(fgh)' = f'gh + fg'h + fgh'$$

Products from a table

AP questions often give values instead of formulas. If $h(x) = f(x)g(x)$, then $h'(a) = f'(a)g(a) + f(a)g'(a)$, so you just need the four numbers $f(a)$, $f'(a)$, $g(a)$, and $g'(a)$.

Worked examples

Example 1: A power times a trig function

Differentiate $y = x^2 \sin x$.

Solution. Let $f = x^2$ (so $f' = 2x$) and $g = \sin x$ (so $g' = \cos x$):

$$\frac{dy}{dx} = 2x\sin x + x^2\cos x$$

Example 2: Two polynomials, checked by expanding

Find $f'(x)$ for $f(x) = (3x^2 - 1)(x^3 + 2x)$.

Solution. With the product rule:

$$\begin{aligned} f'(x) &= (6x)(x^3 + 2x) + (3x^2 - 1)(3x^2 + 2) \\ &= 6x^4 + 12x^2 + 9x^4 + 6x^2 - 3x^2 - 2 \\ &= 15x^4 + 15x^2 - 2 \end{aligned}$$

Check by expanding first: $f(x) = 3x^5 + 6x^3 - x^3 - 2x = 3x^5 + 5x^3 - 2x$, so $f'(x) = 15x^4 + 15x^2 - 2$. Same answer.

Example 3: From a table

Let $h(x) = f(x)g(x)$, where $f(3) = 2$, $f'(3) = -1$, $g(3) = 5$, and $g'(3) = 4$. Find $h'(3)$.

Solution.

$$h'(3) = f'(3)g(3) + f(3)g'(3) = (-1)(5) + (2)(4) = -5 + 8 = 3$$

Example 4: Tangent line and horizontal tangent

Let $y = xe^x$.

• (a) Find the tangent line at $x = 1$.
• (b) Where is the tangent line horizontal?

Solution. By the product rule:

$$y' = (1)e^x + x e^x = e^x(1 + x)$$

(a) At $x = 1$: the point is $(1, e)$ and the slope is $e(2) = 2e$.

$$y - e = 2e(x - 1) \quad\text{or}\quad y = 2ex - e$$

(b) $e^x(1 + x) = 0$. Since $e^x$ is never $0$, we need $1 + x = 0$, so $x = -1$. The point is $\left(-1, -\dfrac{1}{e}\right)$.

Common mistakes

Multiplying the derivatives. $\dfrac{d}{dx}\big[x^2 \sin x\big]$ is not $2x\cos x$. Always two terms: $f'g + fg'$.

Forgetting a term. Write the structure first, ”( )( ) + ( )( )”, then fill in the blanks. This also helps with three factors, which need three terms.

Using the product rule when a constant is involved. For $5x^3$, you can use the product rule, but the constant multiple rule is much faster: $15x^2$. Save the product rule for when both factors contain $x$.

Mixing up values in a table problem. In $f'(3)g(3) + f(3)g'(3)$, each term has exactly one derivative. Circle the four numbers you need before you substitute.

Over-simplifying and making errors. On the AP exam, an unsimplified correct derivative is fine. Simplify only when you need to, such as when solving $y' = 0$ (where factoring, like $e^x(1 + x)$, is the key step).

Practice

1. (Warm-up) Differentiate $y = x^3 e^x$.

Solution

$$\frac{dy}{dx} = 3x^2 e^x + x^3 e^x = x^2 e^x(3 + x)$$

2. (Warm-up) Find $f'(x)$ for $f(x) = x\cos x$.

Solution

$$f'(x) = (1)\cos x + x(-\sin x) = \cos x - x\sin x$$

3. (Warm-up) Let $p(x) = f(x)g(x)$, where $f(1) = 4$, $f'(1) = 2$, $g(1) = -3$, and $g'(1) = 5$. Find $p'(1)$.

Solution

$$p'(1) = f'(1)g(1) + f(1)g'(1) = (2)(-3) + (4)(5) = -6 + 20 = 14$$

4. (Core) Find $\dfrac{dy}{dx}$ for $y = \sqrt{x}\,\ln x$.

Solution

$\dfrac{d}{dx}\big[\sqrt{x}\big] = \dfrac{1}{2\sqrt{x}}$ and $\dfrac{d}{dx}[\ln x] = \dfrac{1}{x}$:

$$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\ln x + \sqrt{x}\cdot\frac{1}{x} = \frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}}$$

(Using $\dfrac{\sqrt{x}}{x} = \dfrac{1}{\sqrt{x}}$.)

5. (Core) Find $\dfrac{dy}{dx}$ for $y = \sin x \cos x$.

Solution

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

6. (Core) Find the equation of the tangent line to $f(x) = (x^2 + 1)(x - 3)$ at $x = 2$.

Solution

Point: $f(2) = (5)(-1) = -5$.

Slope: $f'(x) = 2x(x - 3) + (x^2 + 1)(1)$, so $f'(2) = 4(-1) + 5 = 1$.

$$y - (-5) = 1(x - 2) \quad\text{or}\quad y = x - 7$$

7. (Core) Find the points where the graph of $y = x^2 e^x$ has a horizontal tangent.

Solution

$$y' = 2xe^x + x^2 e^x = xe^x(2 + x)$$

$e^x \ne 0$, so $y' = 0$ when $x = 0$ or $x = -2$.

The points are $(0, 0)$ and $\left(-2, \dfrac{4}{e^2}\right)$.

8. (Challenge) Differentiate $y = xe^x \sin x$.

Solution

Use the three-factor rule with $f = x$, $g = e^x$, $h = \sin x$:

$$\frac{dy}{dx} = (1)e^x\sin x + x e^x \sin x + x e^x \cos x = e^x\big(\sin x + x\sin x + x\cos x\big)$$

9. (Challenge) A bakery’s weekly revenue is $R(t) = P(t)\,N(t)$, where $P(t)$ is the price of a loaf in dollars and $N(t)$ is the number of loaves sold per week, $t$ weeks after opening. At $t = 4$: $P = 12$, $P' = -0.50$ dollars per week, $N = 300$, and $N' = 20$ loaves per week, per week (sales are growing by $20$ loaves a week, every week). Find $R'(4)$ and explain what it means.

Solution

$$R'(4) = P'(4)N(4) + P(4)N'(4) = (-0.50)(300) + (12)(20) = -150 + 240 = 90$$

At $t = 4$ weeks, the bakery’s weekly revenue is increasing at $90 per week. The price drop alone would lower revenue by $150 per week, but the growth in sales adds $240 per week.