The Product Rule - AP Calculus BC Study Guide

The Core Idea: The Product Rule

When working with derivatives, we often encounter functions that are formed by multiplying two other functions together. A common misconception is to assume that the derivative of such a product is simply the product of the individual derivatives. This is incorrect. The Product Rule provides the correct method for differentiating the product of two differentiable functions.

The rule establishes a specific structure for this calculation: it is a sum of two terms. The first term is the first function multiplied by the derivative of the second function. The second term is the second function multiplied by the derivative of the first function. Mastering this rule is essential for accurately finding the rate of change of functions that are constructed through multiplication.

Key Formulas

The Product Rule provides the formula for finding the derivative of a function $h(x)$ that is the product of two other differentiable functions, $f(x)$ and $g(x)$.

• If $h(x)=f(x)g(x)$, then its derivative $h^{\prime }(x)$ is given by:

  $$h^{\prime }(x)=\frac{d}{dx}[f(x)g(x)]=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$$

In words, the formula is: The derivative of a product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function.

Understanding the Structure

The most critical aspect of the Product Rule is correctly identifying the two distinct functions being multiplied. For a function like $h(x)=x^3\cos (x)$, you must recognize that $f(x)=x^3$ and $g(x)=\cos (x)$. The rule is not applicable to a function like $\sin (5x)$, which is a composition of functions, not a product.

The order of the two terms in the sum does not matter due to the commutative property of addition. That is, $f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$ is identical to $f(x)g^{\prime }(x)+f^{\prime }(x)g(x)$. While both are correct, consistently using one form helps prevent errors. The structure $(derivativeoffirst)(second)+(first)(derivativeofsecond)$ is a reliable way to remember and apply it. It is crucial to avoid the common mistake of simply multiplying the derivatives, $f^{\prime }(x)g^{\prime }(x)$, which is not the correct derivative of the product.

Core Concepts & Rules

• Purpose: The Product Rule is used to find the derivative of a function that is the product of two other differentiable functions.

• The Formula: The derivative of $f(x)g(x)$ is $f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$.

• Conceptual Breakdown: The derivative is calculated as a sum of two products:

  1. The derivative of the first function times the original second function.

  2. The original first function times the derivative of the second function.

• Prerequisite: Both functions involved in the product, $f(x)$ and $g(x)$, must be differentiable for the Product Rule to apply.

Step-by-Step Example 1: Basic Application

Problem: Find the derivative of the function $h(x)=x^2{e}^x$.

Step 1: Identify the two functions.

The function $h(x)$ is a product of two simpler functions. Let:

• $f(x)=x^2$

• $g(x)=e^x$

Step 2: Find the derivatives of the individual functions.

Using the Power Rule for $f(x)$ and the rule for the natural exponential function for $g(x)$:

• $f^{\prime }(x)=\frac{d}{dx}(x^2)=2x$

• $g^{\prime }(x)=\frac{d}{dx}(e^x)=e^x$

Step 3: Apply the Product Rule formula.

The formula is $h^{\prime }(x)=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$. Substitute the functions and their derivatives from the previous steps.

• $h^{\prime }(x)=(2x)(e^x)+(x^2)(e^x)$

Step 4: Simplify the result (if possible).

In this case, we can factor out the common term $x{e}^x$.

• $h^{\prime }(x)=x{e}^x(2+x)$

This is the derivative of $h(x)$.

Step-by-Step Example 2: Exam-Style Application

Problem: The differentiable functions $f$ and $g$ have the values shown in the table below. If $P(x)=f(x)g(x)$, what is the value of $P^{\prime }(1)$?

Step 1: Write down the general formula for the derivative of $P(x)$.

Since $P(x)$ is the product of $f(x)$ and $g(x)$, we use the Product Rule.

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

Step 2: Evaluate the derivative at the specific point, $x=1$.

Substitute $x=1$ into the derivative formula.

• $P^{\prime }(1)=f^{\prime }(1)g(1)+f(1)g^{\prime }(1)$

Step 3: Find the required values from the table.

Look at the row for $x=1$ in the table to find the values of $f(1)$, $g(1)$, $f^{\prime }(1)$, and $g^{\prime }(1)$.

• $f(1)=3$

• $g(1)=-2$

• $f^{\prime }(1)=5$

• $g^{\prime }(1)=4$

Step 4: Substitute the values into the expression for $P^{\prime }(1)$ and compute the result.

• $P^{\prime }(1)=(5)(-2)+(3)(4)$

• $P^{\prime }(1)=-10+12$

• $P^{\prime }(1)=2$

The value of $P^{\prime }(1)$ is 2.

Using Your Calculator

The Product Rule is a purely analytical tool; a calculator cannot symbolically apply the formula $f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$. However, a graphing calculator is extremely useful for checking your answer at a specific point.

To check the result of Example 1 ($h(x)=x^2{e}^x$) at $x=3$:

1. Calculate the analytical derivative's value:

   Our derivative was $h^{\prime }(x)=x{e}^x(2+x)$.

   Evaluate this at $x=3$: $h^{\prime }(3)=3e^3(2+3)=15e^3\approx 301.283$.

2. Use the calculator's numerical derivative function:

   On a TI-84 style calculator, you can use the nDeriv function.

   • Press MATH and select 8: nDeriv(.

   • Enter the expression as $nDeriv(X^2\ast e^X,X,3)$.

   • This command calculates the numerical derivative of $x^2{e}^x$ with respect to $x$, evaluated at $x=3$.

3. Compare the results:

   The calculator will return a value very close to $301.283$. If the value from your hand-calculated derivative matches the calculator's numerical derivative, you can be confident your application of the Product Rule was correct.

AP Exam Quick Hit

Common Question Types

• Direct Differentiation: You will be given a function defined as a product and asked to find its derivative.

  • Example: Find $dy/dx$ if $y=\sqrt{x}\sin (x)$.

• Finding the Derivative at a Point (from a table): You will be given a table of function and derivative values for $f$ and $g$ at specific points and asked to find the derivative of their product at one of those points.

  • Example: Using the table from Example 2 above, find the derivative of $f(x)g(x)$ at $x=4$.

• Finding the Derivative at a Point (from graphs): You will be given the graphs of two functions, $f$ and $g$, and asked to find the derivative of their product $h(x)=f(x)g(x)$ at a point $x=a$.

  • Example: To find $h^{\prime }(a)$, you must find $f(a)$ and $g(a)$ (the y-values on the graphs) and $f^{\prime }(a)$ and $g^{\prime }(a)$ (the slopes of the tangent lines to the graphs at $x=a$).

Common Mistakes

• The "Fake" Product Rule: The most frequent error is to assume $\frac{d}{dx}[f(x)g(x)]=f^{\prime }(x)g^{\prime }(x)$. This is incorrect. Always use the full formula: $f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$.

• Mixing up $f$ and $f^{\prime }$: When substituting into the formula, it is easy to accidentally use $f(x)$ where $f^{\prime }(x)$ is required, or vice versa. A systematic approach of writing down $f$, $g$, $f^{\prime }$, and $g^{\prime }$ before substituting can prevent this.

• Forgetting the Product Rule Entirely: When a problem is complex, students sometimes forget that a product exists within it. For example, when finding the derivative of $y=3x^2\ln (x)$, it is easy to focus on the $\ln (x)$ and forget that $3x^2$ is a function being multiplied by it, requiring the Product Rule.

• Algebraic Errors After Differentiation: The Product Rule application might be correct, but subsequent simplification (factoring, distributing, combining terms) can introduce errors. Be careful with your algebra after the calculus is done.