The Product Rule
Why this matters
Imagine you're running a small online business selling custom phone cases. Your total daily revenue is the number of cases you sell multiplied by the price of each case. Revenue = (Cases Sold) × (Price).
Now, what if both of those things are changing? Maybe you're running a promotion, so the price is slowly decreasing, but that's causing the number of cases you sell each day to increase. How is your total revenue changing at any given moment?
It's not as simple as just multiplying the change in sales by the change in price. That's the exact puzzle the product rule solves. It helps us find the rate of change for a "product" of two changing quantities. We'll break down this essential rule step-by-step.
Concept overview
flowchart TD
A[Start: Find derivative of h(x) = f(x) * g(x)] --> B{Is it a product?};
B -- Yes --> C[Identify f(x) and g(x)];
C --> D[Find f'(x)];
C --> E[Find g'(x)];
D --> F[Assemble the Rule];
E --> F[Assemble the Rule];
F[h'(x) = f'(x)g(x) + f(x)g'(x)] --> G[Substitute functions and derivatives];
G --> H[Simplify the final expression];
H --> I[End];
B -- No --> J[Use a different rule, e.g., Power Rule, Chain Rule];
Core explanation
Alright, let's dive into one of the most fundamental rules of differentiation: the product rule.
First, let's get the biggest misconception out of the way.
If you have a function h(x) that is the product of two other functions, say f(x) and g(x), so h(x) = f(x)g(x), you might think you can just take the derivative of each part and multiply them together.
This is the most common mistake students make.
d/dx [f(x)g(x)] is NOT f'(x)g'(x).
It feels like it should be that simple, but it doesn't work. Think back to our phone case business. The change in revenue isn't just the change in price times the change in sales. It's more nuanced, and the product rule gives us the correct, complete picture.
The Product Rule Formula
Here is the actual formula. If f and g are differentiable functions, then the derivative of their product is:
d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)
Let's break that down. I like to think of it in plain English:
"The derivative of a product is the derivative of the first function times the second, plus the first function times the derivative of the second."
Some people find this little jingle helpful: "Lefty-D-Righty plus Righty-D-Lefty," where D means "derivative of." Or, using u and v for the functions, (uv)' = u'v + uv'. Whatever helps you remember the pattern!
An Analogy: The Growing Rectangle
Imagine a rectangle whose width w(t) and height h(t) are both growing over time. The area of the rectangle is A(t) = w(t) * h(t). How is the area changing?
The change in area comes from two places:
- The rectangle gets wider. This adds a new slice of area on the side. The size of this new slice is
(the change in width) * (the current height). This is ourw'(t)h(t)term. - The rectangle gets taller. This adds a new slice of area on the top. The size of this slice is
(the current width) * (the change in height). This is ourw(t)h'(t)term.
The total change in area is the sum of these two new slices: A'(t) = w'(t)h(t) + w(t)h'(t). This is the product rule in action! You can see that the change depends on both the rates of change (w' and h') and the current dimensions (w and h).
How to Apply the Product Rule: A Step-by-Step Guide
Let's apply this to a real calculus problem. Suppose we want to find the derivative of y = x² sin(x).
- 1Identify your two functionsOur function is a product of
x²andsin(x). Letf(x) = x²(our "first" function). Letg(x) = sin(x)(our "second" function). - 2Find the derivative of each function separatelyUsing our basic derivative rules:
f'(x) = 2xg'(x) = cos(x) - 3Plug these four pieces into the product rule formulaRemember the formula:
f'(x)g(x) + f(x)g'(x)Substitute everything in its place:
y' = (2x)(sin(x)) + (x²)(cos(x)) - 4Simplify (if necessary)In this case, it's already quite clean.
y' = 2x sin(x) + x² cos(x)
And that's it! You've successfully used the product rule. The key is being organized. Always identify your two functions, find their derivatives separately, and then carefully assemble them into the final formula. Don't try to do it all in your head at once. Write down the parts. It prevents simple errors and makes your work much easier to check.
Worked examples
Let's walk through a couple of examples together. The key is to be methodical and write out each step.
Polynomial and Exponential
Problem: Find the derivative of h(x) = (3x⁴ - 5x)eˣ.
Solution Walkthrough:
- 1Identify the two functions being multipliedThis function is a product of a polynomial part and an exponential part. Let
f(x) = 3x⁴ - 5x. This is our "first" function. Letg(x) = eˣ. This is our "second" function. - 2Find the derivatives of these individual functionsFor
f(x), we use the power rule on each term:f'(x) = 12x³ - 5For
g(x), we know the derivative ofeˣis just itself:g'(x) = eˣ - 3Apply the product rule formula
h'(x) = f'(x)g(x) + f(x)g'(x)Now, we substitute our four pieces into the formula. Be careful with parentheses!h'(x) = (12x³ - 5)(eˣ) + (3x⁴ - 5x)(eˣ)This is a perfectly correct answer for the derivative. On a free-response question on the AP exam, you could often stop here.
- 4Simplify by factoring (good practice)Notice that both terms have a common factor of
eˣ. Let's factor it out to get a cleaner-looking answer.h'(x) = eˣ [ (12x³ - 5) + (3x⁴ - 5x) ]h'(x) = eˣ (3x⁴ + 12x³ - 5x - 5)
Finding the Equation of a Tangent Line
Problem: Find the equation of the line tangent to the curve y = x³ ln(x) at the point where x = 1.
Solution Walkthrough:
-
Recognize what you need. To find the equation of a line, we need a point and a slope.
- PointWe're given
x = 1. To find the y-coordinate, plug it into the original function:y = (1)³ ln(1) = 1 * 0 = 0. So, our point is(1, 0). - SlopeThe slope of the tangent line is the value of the derivative at that point. We need to find
y'first, using the product rule.
- Point
-
Apply the Product Rule to find
y'. Our function isy = x³ ln(x).f(x) = x³=>f'(x) = 3x²g(x) = ln(x)=>g'(x) = 1/x
Now, assemble the rule:
y' = f'(x)g(x) + f(x)g'(x)y' = (3x²)(ln(x)) + (x³)(1/x) -
Simplify the derivative.
y' = 3x² ln(x) + x² -
Calculate the slope at
x = 1. Plugx = 1into our expression fory'.m = y'(1) = 3(1)² ln(1) + (1)²m = 3(1)(0) + 1m = 0 + 1 = 1So, the slope of the tangent line atx=1is 1. -
Write the equation of the line. Using point-slope form
y - y₁ = m(x - x₁), with our point(1, 0)and slopem = 1:y - 0 = 1(x - 1)y = x - 1
This example shows how the product rule is a tool used in larger, multi-step problems—a very common AP exam scenario.
Try it yourself
Ready to try on your own? Don't worry about getting it perfect on the first try. The goal is to practice the process.
Problem 1: Find the derivative, dy/dx, for the function y = x⁴ tan(x).
Problem 2: At what x-values does the graph of f(x) = x eˣ have a horizontal tangent line?
In simple terms, the product rule is a special formula you use in calculus to find the rate of change of a function that is made by multiplying two other functions together.
- FUN-3.B: Calculate derivatives of products and quotients of differentiable functions.
- FUN-3.B.1
- Derivatives of products of differentiable functions can be found using the product rule.
flowchart TD
A[Start: Find derivative of h(x) = f(x) * g(x)] --> B{Is it a product?};
B -- Yes --> C[Identify f(x) and g(x)];
C --> D[Find f'(x)];
C --> E[Find g'(x)];
D --> F[Assemble the Rule];
E --> F[Assemble the Rule];
F[h'(x) = f'(x)g(x) + f(x)g'(x)] --> G[Substitute functions and derivatives];
G --> H[Simplify the final expression];
H --> I[End];
B -- No --> J[Use a different rule, e.g., Power Rule, Chain Rule];
Read what Saavi narrates
Hi everyone, it's Saavi from Shrutam. Let's talk about a super important tool for your calculus toolkit: the product rule.
Imagine you're selling phone cases. Your revenue is the number of cases sold, times the price. But what if both of those things are changing? Maybe a sale increases the number you sell, but lowers the price. How is your total revenue changing? That's what the product rule helps us figure out.
The big idea is this: when you have a function that's made by multiplying two other functions, you can't just take the derivative of each piece and multiply them. That's the number one mistake I see.
Instead, we have a specific formula: the derivative of the first function times the second, PLUS the first function times the derivative of the second.
Let's try one. Say we have the function h of x equals, open parenthesis, 3x to the fourth minus 5x, close parenthesis, times e to the x.
First, we identify our two functions. The first function is 3x to the fourth minus 5x. The second is e to the x.
Next, we find their derivatives. The derivative of the first function is 12x cubed minus 5. The derivative of e to the x is just... e to the x.
Now we assemble the rule. It's the derivative of the first, which is 12x cubed minus 5... times the original second function, e to the x.
Then we add... the original first function, 3x to the fourth minus 5x... times the derivative of the second, which is e to the x.
So our full derivative is... open parenthesis, 12x cubed minus 5, close parenthesis, times e to the x... PLUS... open parenthesis, 3x to the fourth minus 5x, close parenthesis, times e to the x.
See how we methodically put the pieces together? The most common mistake is just multiplying the two derivatives together. Please, don't fall into that trap! Always use the full formula: f prime g, plus f g prime.
You've got this. Just be patient, write out your steps, and you'll master the product rule in no time. Keep practicing!
This ignores the interaction between the two functions. The correct formula accounts for how each function's change affects the other.
Memorize and use the full product rule: `f'(x)g(x) + f(x)g'(x)`.
The product rule has two parts that are added together. You're only accounting for half of the change.
Say the rule to yourself as you write it: "Derivative of the first times the second, *plus* the first times the derivative of the second."
The product rule is useless if the building blocks (`f'`, `g'`) are incorrect.
Double-check your basic derivative rules (Power, Trig, Exponential, etc.) before you even start assembling the product rule.
The `x²+1` needs to be treated as a single group. Without parentheses, `x²+1(1)` becomes `x²+1`, which is correct here, but for `(x²+1)(2x)` it would be `x²+2x`, which is wrong.
Always wrap your functions and their derivatives in parentheses when you substitute them into the formula: `h'(x) = (2x)(x-2) + (x²+1)(1)`. This prevents algebraic errors.