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Fundamental Properties of Differentiation

Michael Green

Michael Green

4 min read

Next Topic - The Quotient Rule

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Study Guide Overview

This study guide covers the product rule for finding the derivative of two functions multiplied together. It defines the rule, provides worked examples with functions like e2x(x5+3x)e^{2x}(x^5 + 3x)e2x(x5+3x) and ln(3x)∗sin(2x)ln(3x) * sin(2x)ln(3x)∗sin(2x), and offers practice questions. The guide also emphasizes the difference between the product rule and the chain rule for composite functions, and includes a glossary of terms like product rule and composite function.

#Derivatives of Products

#Table of Contents

  1. Introduction
  2. Product Rule
  3. Exam Tip
  4. Worked Examples
    • Example (a)
    • Example (b)
  5. Practice Questions
  6. Glossary
  7. Summary and Key Takeaways

#Introduction

Understanding how to differentiate the product of two functions is crucial in calculus. This guide will cover the product rule, provide clear examples, and include practice questions to solidify your understanding.

#Product Rule

The product rule is used to find the derivative of the product of two functions.

#Definition

If f(x)=g(x)⋅h(x)f(x) = g(x) \cdot h(x)f(x)=g(x)⋅h(x), then the derivative f′(x)f'(x)f′(x) is given by:

f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x)f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x)

Alternatively, if y=u⋅vy = u \cdot vy=u⋅v, then:

dydx=dudx⋅v+u⋅dvdx\frac{dy}{dx} = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx}dxdy​=dxdu​⋅v+u⋅dxdv​

In a more concise form:

y′=u′v+uv′y' = u'v + uv'y′=u′v+uv′

#Exam Tip

Exam Tip

Don't confuse the product of two functions with a composite function.

  • The product of two functions like f(x)⋅g(x)f(x) \cdot g(x)f(x)⋅g(x) is two functions multiplied together.
  • A composite function like f(g(x))f(g(x))f(g(x)) is a function of a function.
  • For how to differentiate composite functions, see the "Chain rule" study guide.

#Worked Examples

#Example (a)

Find the derivative of the function f(x)=e2x(x5+3x)f(x) = e^{2x} (x^5 + 3x)f(x)=e2x(x5+3x).

Solution:

  1. Assign uuu and vvv to each function: u=e2xu = e^{2x}u=e2x v=x5+3xv = x^5 + 3xv=x5+3x

  2. Find the derivatives of uuu and vvv: u′=2e2xu' = 2e^{2x}u′=2e2x v′=5x4+3v' = 5x^4 + 3v′=5x4+3

  3. Apply the product rule, y′=u′v+uv′y' = u'v + uv'y′=u′v+uv′: y′=2e2x(x5+3x)+e2x(5x4+3)y' = 2e^{2x} (x^5 + 3x) + e^{2x} (5x^4 + 3)y′=2e2x(x5+3x)+e2x(5x4+3)

  4. Write the final answer in the required format: f′(x)=2e2x(x5+3x)+e2x(5x4+3)f'(x) = 2e^{2x} (x^5 + 3x) + e^{2x} (5x^4 + 3)f′(x)=2e2x(x5+3x)+e2x(5x4+3)

  5. Factorize the answer: f′(x)=e2x(2x5+5x4+6x+3)f'(x) = e^{2x} (2x^5 + 5x^4 + 6x + 3)f′(x)=e2x(2x5+5x4+6x+3)

#Example (b)

Find the derivative of the function g(x)=ln⁡(3x)⋅sin⁡(2x)g(x) = \ln (3x) \cdot \sin (2x)g(x)=ln(3x)⋅sin(2x).

Solution:

  1. Assign uuu and vvv to each function: u=ln⁡(3x)u = \ln (3x)u=ln(3x) v=sin⁡(2x)v = \sin (2x)v=sin(2x)

  2. Find the derivatives of uuu and vvv: u′=1xu' = \frac{1}{x}u′=x1​ v′=2cos⁡(2x)v' = 2 \cos (2x)v′=2cos(2x)

  3. Apply the product rule, y′=u′v+uv′y' = u'v + uv'y′=u′v+uv′: y′=1x⋅sin⁡(2x)+ln⁡(3x)⋅2cos⁡(2x)y' = \frac{1}{x} \cdot \sin (2x) + \ln (3x) \cdot 2 \cos (2x)y′=x1​⋅sin(2x)+ln(3x)⋅2cos(2x)

  4. Simplify: g′(x)=sin⁡(2x)x+2ln⁡(3x)cos⁡(2x)g'(x) = \frac{\sin (2x)}{x} + 2 \ln (3x) \cos (2x)g′(x)=xsin(2x)​+2ln(3x)cos(2x)

#Practice Questions

Practice Question
  1. Find the derivative of h(x)=x3⋅cos⁡(x)h(x) = x^3 \cdot \cos(x)h(x)=x3⋅cos(x).
Practice Question
  1. Differentiate p(x)=x⋅exp(x) = \sqrt{x} \cdot e^xp(x)=x​⋅ex.
Practice Question
  1. Calculate the derivative of q(x)=tan⁡(x)⋅ln⁡(x)q(x) = \tan(x) \cdot \ln(x)q(x)=tan(x)⋅ln(x).

#Glossary

  • Product Rule: A rule used to find the derivative of the product of two functions.
  • Composite Function: A function formed by applying one function to the result of another.

#Summary and Key Takeaways

  • The product rule is essential for differentiating the product of two functions.
  • Always remember to distinguish between products and composite functions.
  • Practice using the product rule with various functions to master the concept.
Key Concept

The product rule is expressed as f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x)f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x).


By mastering the product rule, you will be well-equipped to tackle complex differentiation problems involving multiple functions.

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Question 1 of 8

What is the correct formula for the product rule when differentiating f(x)=g(x)⋅h(x)f(x) = g(x) \cdot h(x)f(x)=g(x)⋅h(x)? 🤔

f′(x)=g′(x)+h′(x)f'(x) = g'(x) + h'(x)f′(x)=g′(x)+h′(x)

f′(x)=g(x)⋅h′(x)−g′(x)⋅h(x)f'(x) = g(x) \cdot h'(x) - g'(x) \cdot h(x)f′(x)=g(x)⋅h′(x)−g′(x)⋅h(x)

f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x)f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x)

f′(x)=g(x)⋅h(x)f'(x) = g(x) \cdot h(x)f′(x)=g(x)⋅h(x)