AP Calculus AB: The Product Rule

Finding the derivative of a function is a core skill in calculus, but what happens when the function is built from two simpler functions multiplied together? You cannot simply take the derivative of each part and multiply the results. The Product Rule provides the essential and correct procedure for differentiating products, a tool you will use constantly in both pure and applied mathematics, from physics to engineering economics.

What the Product Rule States

Formally, if you have a function $h(x)$ that is the product of two differentiable functions, $f(x)$ and $g(x)$, such that $h(x)=f(x)\cdot g(x)$, then the derivative of $h$ is given by:

$$h^{\prime }(x)=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$$

In Leibniz notation, if $y=u\cdot v$ where $u$ and $v$ are functions of $x$, the rule is often memorized as:

$$\frac{d}{dx}(uv)=\frac{du}{dx}\cdot v+u\cdot \frac{dv}{dx}$$

A helpful verbalization is: “The derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second.” The order within each term matters; the first function stays as-is while you differentiate the second, and vice versa. This rule is not just a formula but a logical consequence of the limit definition of the derivative, a relationship you can verify by exploring the limit of the difference quotient for $f(x)g(x)$.

Applying the Rule to Basic Functions

The best way to internalize the Product Rule is through deliberate practice. Start with polynomials, where you can check your work by first multiplying the functions and then using the Power Rule. The Product Rule, however, is far more powerful because it works even when expansion is impractical or impossible.

Example 1: Polynomials Find the derivative of $h(x)=(x^2+3x)(4x-5)$.

1. Identify your functions: Let $f(x)=x^2+3x$ and $g(x)=4x-5$.
2. Compute individual derivatives: $f^{\prime }(x)=2x+3$ and $g^{\prime }(x)=4$.
3. Apply the Product Rule formula: $h^{\prime }(x)=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$.
4. Substitute: $h^{\prime }(x)=(2x+3)(4x-5)+(x^2+3x)(4)$.
5. Simplify: $h^{\prime }(x)=(8x^2-10x+12x-15)+(4x^2+12x)=12x^2+14x-15$.

You can verify this by first expanding $h(x)$ to $4x^3+7x^2-15x$ and taking the derivative to get $12x^2+14x-15$.

Handling Trigonometric and Exponential Functions

The Product Rule’s true utility shines when differentiating products of functions from different families, such as a polynomial and a trigonometric function, or an exponential and a logarithmic function. The process remains identical: identify your first and second function, find their derivatives, and assemble the sum of the two required products.

Example 2: Polynomial and Trig Function Find the derivative of $y=x^3\sin (x)$.

1. Let $f(x)=x^3$ and $g(x)=\sin (x)$.
2. Then $f^{\prime }(x)=3x^2$ and $g^{\prime }(x)=\cos (x)$.
3. Apply the rule: $\frac{dy}{dx}=(3x^2)(\sin (x))+(x^3)(\cos (x))$.
4. The simplified result is $\frac{dy}{dx}=3x^2\sin (x)+x^3\cos (x)$.

Example 3: Exponential and General Function Find $f^{\prime }(x)$ if $f(x)=e^x\sqrt{x}$. First, rewrite $\sqrt{x}$ as $x^{1/2}$. Let $u=e^x$ and $v=x^{1/2}$. Then $u^{\prime }=e^x$ and $v^{\prime }=\frac{1}{2}{x}^{-1/2}$. Applying the rule: $$f^{\prime }(x)=(e^x)(x^{1/2})+(e^x)\left(\frac{1}{2}{x}^{-1/2}\right)=e^x\sqrt{x}+\frac{e^x}{2\sqrt{x}}$$ You can factor this result if needed: $f^{\prime }(x)=e^x\sqrt{x}\left(1+\frac{1}{2x}\right)$.

Common Pitfalls

1. Forgetting the Rule and Multiplying Derivatives: The most frequent critical error is assuming $(f\cdot g{)}^{\prime }=f^{\prime }\cdot g^{\prime }$. This is false. Always recall the "first times derivative of second plus..." structure. A quick mental check with a simple product like $x\cdot x$ (whose derivative you know is $2x$) can remind you that the Product Rule yields $(1)(x)+(x)(1)=2x$, while multiplying the derivatives $(1)(1)$ gives the wrong answer, 1.

1. Misapplying the Rule to Non-Products: Do not use the Product Rule on a composition of functions like $\sin (x^2)$; that requires the Chain Rule. Similarly, do not use it on a sum like $(x+\sin (x))$; the derivative of a sum is simply the sum of the derivatives. Only use the Product Rule when you see an explicit multiplication of two function expressions.

1. Algebraic Errors in Simplification: After applying the formula, you are often left with a sum of two terms that need to be combined. Carefully handle negative signs, especially when the second function involves a subtraction. Factoring a common factor from the result is an excellent way to simplify and present a clean final answer, as shown in Example 3.

1. Inconsistent Identification of $f$ and $g$: While the rule is symmetric mathematically ($f^{\prime }g+f{g}^{\prime }$ is the same as $g^{\prime }f+g{f}^{\prime }$), consistently labeling your chosen first and second functions helps avoid confusion, especially in more complex problems. Stick with your initial choice throughout the calculation.

Summary

• The Product Rule is mandatory for finding the derivative of a function defined as the product of two simpler functions: if $h(x)=f(x)g(x)$, then $h^{\prime }(x)=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$.
• The rule is applied systematically: differentiate the first function, multiply by the second; differentiate the second function, multiply by the first; then add the two results.
• It is widely applicable to products of any differentiable functions, including combinations of polynomials, trigonometric functions (e.g., $\sin (x)$, $\cos (x)$), exponential functions ($e^x$), and logarithmic functions.
• A common mistake is to incorrectly take the derivative of a product by simply multiplying the individual derivatives; the Product Rule's plus sign is non-negotiable.
• Mastery comes from practiced application followed by careful algebraic simplification of the resulting expression.