Differentiating Products of Functions Using the Product Rule

Introduction

Key Concepts

Understanding the Product Rule

The product rule is a formula used to find the derivative of the product of two functions. If you have two functions, \( u(x) \) and \( v(x) \), their product \( f(x) = u(x)v(x) \) can be differentiated using the product rule. Mathematically, the product rule is expressed as:

This means that the derivative of the product \( fg \) is equal to the derivative of \( f \) multiplied by \( g \) plus \( f \) multiplied by the derivative of \( g \).

Derivation of the Product Rule

The product rule can be derived using the limit definition of the derivative. Consider the derivative of \( f(x) = u(x)v(x) \): $$ f'(x) = \lim_{h \to 0} \frac{u(x+h)v(x+h) - u(x)v(x)}{h} $$ By adding and subtracting \( u(x+h)v(x) \) within the numerator, we get: $$ f'(x) = \lim_{h \to 0} \left[ u(x+h)\frac{v(x+h) - v(x)}{h} + v(x)\frac{u(x+h) - u(x)}{h} \right] $$ Taking the limit as \( h \to 0 \): $$ f'(x) = u(x) \cdot v'(x) + v(x) \cdot u'(x) $$ This confirms the product rule formula.

Applying the Product Rule: Basic Examples

Let's consider a simple example to apply the product rule.

Example 1: Differentiate \( f(x) = x^2 \sin(x) \).

Here, \( u(x) = x^2 \) and \( v(x) = \sin(x) \). First, find the derivatives of \( u \) and \( v \): $$ u'(x) = 2x \\ v'(x) = \cos(x) $$ Applying the product rule: $$ f'(x) = u'(x)v(x) + u(x)v'(x) = 2x \cdot \sin(x) + x^2 \cdot \cos(x) $$ Therefore, $$ f'(x) = 2x \sin(x) + x^2 \cos(x) $$

Complex Polynomials and Trigonometric Functions

Consider differentiating a function involving a polynomial and a trigonometric function:

Example 2: Differentiate \( f(x) = (3x^3 - 2x)(\cos(x)) \).

Let \( u(x) = 3x^3 - 2x \) and \( v(x) = \cos(x) \). Then, $$ u'(x) = 9x^2 - 2 \\ v'(x) = -\sin(x) $$ Applying the product rule: $$ f'(x) = u'(x)v(x) + u(x)v'(x) = (9x^2 - 2)\cos(x) + (3x^3 - 2x)(-\sin(x)) $$ Simplifying, $$ f'(x) = (9x^2 - 2)\cos(x) - (3x^3 - 2x)\sin(x) $$

Multiple Applications of the Product Rule

When differentiating products of more than two functions, the product rule can be extended by applying it iteratively. For three functions \( u(x) \), \( v(x) \), and \( w(x) \), the derivative of \( f(x) = u(x)v(x)w(x) \) is:

Example 3: Differentiate \( f(x) = x e^x \ln(x) \).

Here, \( u(x) = x \), \( v(x) = e^x \), and \( w(x) = \ln(x) \). Their derivatives are: $$ u'(x) = 1 \\ v'(x) = e^x \\ w'(x) = \frac{1}{x} $$ Applying the extended product rule: $$ f'(x) = 1 \cdot e^x \ln(x) + x \cdot e^x \cdot \ln(x) + x \cdot e^x \cdot \frac{1}{x} $$ Simplifying: $$ f'(x) = e^x \ln(x) + x e^x \ln(x) + e^x $$

Higher-Order Derivatives

While the product rule is primarily used for first-order derivatives, it can also be applied iteratively for higher-order derivatives. However, this often involves more complex calculations and the use of additional rules like the chain rule.

Example 4: Find the second derivative of \( f(x) = x^2 \sin(x) \).

First, find the first derivative using the product rule: $$ f'(x) = 2x \sin(x) + x^2 \cos(x) $$ Now, differentiate \( f'(x) \): $$ f''(x) = 2 \sin(x) + 2x \cos(x) + 2x \cos(x) - x^2 \sin(x) $$ Combining like terms: $$ f''(x) = 2 \sin(x) + 4x \cos(x) - x^2 \sin(x) $$

Practical Applications in Physics and Engineering

The product rule is not only a theoretical concept but also has practical applications in various fields such as physics and engineering. For instance, in physics, it is used to determine the rate of change of physical quantities that are products of different functions, like momentum (mass times velocity) when mass or velocity is changing.

Example 5: If the mass of an object varies with time, and its velocity is also a function of time, the derivative of momentum \( p(t) = m(t)v(t) \) with respect to time is given by: $$ p'(t) = m'(t)v(t) + m(t)v'(t) $$>

Common Mistakes and How to Avoid Them

When applying the product rule, students often make the following mistakes:

• Forgetting to Apply the Rule to All Terms: Ensure that you differentiate each function in the product while keeping the other functions unchanged.
• Misapplying the Derivatives: Carefully compute the derivatives of individual functions before applying the product rule.
• Incorrect Algebraic Manipulation: After applying the product rule, simplify the expression correctly to avoid errors in the final answer.

To avoid these mistakes, practice consistently and double-check each step when applying the product rule.

Illustrative Examples

Example 6: Differentiate \( f(x) = (2x^3 + x)(5x^2 - 4) \).

Let \( u(x) = 2x^3 + x \) and \( v(x) = 5x^2 - 4 \). Then, $$ u'(x) = 6x^2 + 1 \\ v'(x) = 10x $$ Applying the product rule: $$ f'(x) = u'(x)v(x) + u(x)v'(x) = (6x^2 + 1)(5x^2 - 4) + (2x^3 + x)(10x) $$ Expanding the terms: $$ f'(x) = (30x^4 - 24x^2 + 5x^2 - 4) + (20x^4 + 10x^2) $$ Combining like terms: $$ f'(x) = 50x^4 - 19x^2 - 4 $$

Visualization and Graphical Interpretation

Visualizing the product of two functions and their derivatives can provide deeper insight. Consider graphing both the original functions and their derivatives to understand how the product rule influences the slope of the resulting function. Tools like graphing calculators or software can aid in this visualization.

Practice Problems

To reinforce the understanding of the product rule, attempt the following practice problems:

1. Differentiate \( f(x) = (x^2 + 3x)(\ln(x)) \).
2. Find the derivative of \( f(x) = (e^x)(\tan(x)) \).
3. Differentiate \( f(x) = (4x^3 - x)(\sqrt{x}) \).
4. Find \( f'(x) \) if \( f(x) = (x^2)(x^3) \).
5. Differentiate \( f(x) = (3x + 2)(x^2 - 5) \).

Advanced Concepts

Higher-Order Product Rules

While the basic product rule handles the differentiation of two functions, higher-order product rules extend this concept to products of multiple functions. For instance, differentiating a product of three functions requires applying the product rule multiple times. This can become increasingly complex with more functions, necessitating a structured approach to avoid errors.

Example 7: Differentiate \( f(x) = u(x)v(x)w(x) \).

Using the extended product rule: $$ f'(x) = u'(x)v(x)w(x) + u(x)v'(x)w(x) + u(x)v(x)w'(x) $$>

Repeat this process for each differentiation step if more functions are involved.

Product Rule in Implicit Differentiation

Implicit differentiation involves differentiating equations where the dependent variable is not isolated. The product rule plays a crucial role in such scenarios.

Example 8: Differentiate implicitly \( x^2y + y^3 = 6 \).

Applying the product rule to \( x^2y \): $$ \frac{d}{dx}(x^2y) = 2xy + x^2y' $$ Differentiating \( y^3 \): $$ \frac{d}{dx}(y^3) = 3y^2y' $$ Differentiating the constant 6: $$ \frac{d}{dx}(6) = 0 $$ Combining these: $$ 2xy + x^2y' + 3y^2y' = 0 $$ Solving for \( y' \): $$ y'(x^2 + 3y^2) = -2xy \\ y' = -\frac{2xy}{x^2 + 3y^2} $$>

Logarithmic Differentiation

Logarithmic differentiation is a technique that applies the natural logarithm to both sides of an equation to simplify the differentiation process, especially useful for products of multiple functions.

Example 9: Differentiate \( f(x) = (x^2 + 1)^5 (3x - 2)^4 \).

Taking the natural logarithm: $$ \ln(f(x)) = 5\ln(x^2 + 1) + 4\ln(3x - 2) $$ Differentiating both sides: $$ \frac{f'(x)}{f(x)} = 5 \cdot \frac{2x}{x^2 + 1} + 4 \cdot \frac{3}{3x - 2} $$ Multiplying both sides by \( f(x) \): $$ f'(x) = f(x) \left( \frac{10x}{x^2 + 1} + \frac{12}{3x - 2} \right ) $$>

Substituting back \( f(x) \): $$ f'(x) = (x^2 + 1)^5 (3x - 2)^4 \left( \frac{10x}{x^2 + 1} + \frac{12}{3x - 2} \right ) $$>

Applications in Optimization Problems

The product rule is instrumental in solving optimization problems where two variables are dependent and their product needs to be maximized or minimized under certain constraints.

Example 10: A rectangle has a perimeter of 20 meters. Express the area of the rectangle as a function of its length and find the dimensions that maximize the area.

Let the length be \( l \) and width be \( w \). Given: $$ 2l + 2w = 20 \\ w = 10 - l $$ The area \( A \) is: $$ A(l) = l \cdot w = l(10 - l) = 10l - l^2 $$>

To find the maximum area, differentiate \( A(l) \) with respect to \( l \): $$ A'(l) = 10 - 2l $$ Set \( A'(l) = 0 \): $$ 10 - 2l = 0 \\ l = 5 $$ Thus, \( w = 10 - 5 = 5 \). The rectangle is a square with sides of 5 meters, maximizing the area.

Interdisciplinary Connections

The product rule extends beyond pure mathematics, finding applications in fields like physics, economics, and engineering.

• Physics: In kinematics, the product rule is used to differentiate position functions that are products of time-dependent variables.
• Economics: Calculating elasticity of demand involves differentiating products of price and quantity functions.
• Engineering: In control systems, the product rule helps in determining the derivatives of transfer functions.

Understanding the product rule enhances problem-solving skills across these disciplines, showcasing the versatility of calculus in real-world applications.

Integration with Other Differentiation Rules

The product rule often works in tandem with other differentiation rules such as the chain rule and quotient rule to solve more complex differentiation problems.

Example 11: Differentiate \( f(x) = \sin(x) \cdot e^{x^2} \).

Applying the product rule: $$ f'(x) = \cos(x) \cdot e^{x^2} + \sin(x) \cdot \frac{d}{dx}(e^{x^2}) $$>

Using the chain rule for \( \frac{d}{dx}(e^{x^2}) \): $$ \frac{d}{dx}(e^{x^2}) = e^{x^2} \cdot 2x $$>

Thus, $$ f'(x) = \cos(x) e^{x^2} + \sin(x) \cdot 2x e^{x^2} \\ f'(x) = e^{x^2} (\cos(x) + 2x \sin(x)) $$>

Partial Product Rule for Multivariable Functions

In multivariable calculus, the product rule extends to functions of several variables, involving partial derivatives.

Example 12: Differentiate \( f(x, y) = x^2 y^3 \) with respect to \( x \).

Applying the product rule: $$ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2) \cdot y^3 + x^2 \cdot \frac{\partial}{\partial x}(y^3) $$>

Since \( y \) is treated as a constant with respect to \( x \): $$ \frac{\partial f}{\partial x} = 2x y^3 + 0 = 2x y^3 $$>

Generalization to n Functions

For the product of \( n \) differentiable functions, the derivative is the sum of all possible products where each term's derivative is taken once while keeping others constant. Formally, for \( f(x) = u_1(x)u_2(x)...u_n(x) \): $$ f'(x) = \sum_{i=1}^{n} \left( u'_i(x) \prod_{j=1, j \neq i}^{n} u_j(x) \right ) $$>

This generalization underscores the importance of the product rule in handling complex differentiation scenarios involving multiple functions.

Alternative Proofs and Theoretical Insights

Beyond the standard derivation, alternative proofs of the product rule can provide deeper theoretical insights. One such method involves using the concept of infinitesimals or leveraging linear approximations.

Understanding these alternative approaches can enhance a student's conceptual grasp and appreciation of the underlying mathematical structures.

Applications in Differential Equations

The product rule is frequently utilized in solving differential equations, especially linear differential equations where solutions are expressed as products of functions.

Example 13: Solve the differential equation \( \frac{dy}{dx} + P(x)y = Q(x) \), assuming it can be expressed in a product form.

Using an integrating factor \( \mu(x) = e^{\int P(x) dx} \), the product rule aids in rewriting and solving the equation.

Non-Elementary Functions

When dealing with non-elementary functions or those not easily expressed in closed form, the product rule still applies but may require numerical methods or approximations for differentiation.

Understanding the limitations and extensions of the product rule in such contexts is crucial for advanced studies and applications.

Comparison Table

Summary and Key Takeaways