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Fundamental Properties of Differentiation

Michael Green

Michael Green

6 min read

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Study Guide Overview

This study guide covers derivatives of exponential and logarithmic functions. It includes differentiation of eˣ, eᵏˣ, aeᵏˣ, aˣ, aᵏˣ, ln x, a ln x, and ln kx. The guide provides worked examples, practice questions, and a glossary of key terms like the chain rule and constant multiple rule. Key takeaways emphasize correct application of these rules and properties of logarithms.

Derivative of Exponential and Logarithmic Functions

Table of Contents

  1. Derivative of the Exponential Function
  2. Derivative of a Number Raised to the Power of x
  3. Derivative of the Natural Logarithmic Function
  4. Practice Questions
  5. Glossary
  6. Summary and Key Takeaways

Derivative of the Exponential Function

Differentiating exe^x

The function f(x)=exf(x) = e^x is unique because its rate of change is equal to itself.

f′(x)=exf'(x) = e^x

Differentiating ekxe^{kx}

For the function g(x)=ekxg(x) = e^{kx}, its rate of change is proportional to itself.

g′(x)=kekxg'(x) = k e^{kx}

This result is due to the chain rule.

Differentiating aekxa e^{kx}

For a function with a constant multiple of the exponential, the differentiation follows a similar approach.

h(x)=aekxh(x) = a e^{kx} h′(x)=akekxh'(x) = a k e^{kx}

Worked Example 1

Differentiate the following functions

(a) f(x)=ex+e2x−4e7xf(x) = e^x + e^{2x} - 4e^{7x}

**Answer:**
  1. exe^x differentiates to exe^x
  2. ekxe^{kx} differentiates to kekxk e^{kx}
  3. aekxa e^{kx} differentiates to akekxa k e^{kx}

f′(x)=ex+2e2x−4⋅7e7xf'(x) = e^x + 2e^{2x} - 4 \cdot 7e^{7x} Simplify: f′(x)=ex+2e2x−28e7xf'(x) = e^x + 2e^{2x} - 28e^{7x}

(b) g(x)=3x+32xg(x) = 3^x + 3^{2x}

**Answer:**
  1. axa^x differentiates to axln⁡aa^x \ln a
  2. akxa^{kx} differentiates to akxkln⁡aa^{kx} k \ln a

g′(x)=3xln⁡3+32x⋅2⋅ln⁡3g'(x) = 3^x \ln 3 + 3^{2x} \cdot 2 \cdot \ln 3 Simplify: g′(x)=3xln⁡3+(32x)2ln⁡3g'(x) = 3^x \ln 3 + (3^{2x}) 2 \ln 3 This can also be factorized: g′(x)=3xln⁡3(1+2⋅3x)g'(x) = 3^x \ln 3 (1 + 2 \cdot 3^x)

Derivative of a Number Raised to the Power of x

Differentiating axa^x

For a constant raised to the power of xx:

f(x)=axf(x) = a^x f′(x)=axln⁡af'(x) = a^x \ln a

Differentiating akxa^{kx}

If the power is a multiple of xx:

g(x)=akxg(x) = a^{kx} g′(x)=akxkln⁡ag'(x) = a^{kx} k \ln a

This result is also derived using logarithms and the differentiation result for ekxe^{kx}:

  1. Rewrite y=akxy = a^{kx} as y=eln⁡(akx)=ekxln⁡ay = e^{\ln (a^{kx})} = e^{kx \ln a}
  2. Since g(x)=ekxg(x) = e^{kx} differentiates to g′(x)=kekxg'(x) = k e^{kx}:
  3. dydx=kln⁡a⋅ekxln⁡a\frac{dy}{dx} = k \ln a \cdot e^{kx \ln a}
  4. Simplify to: dydx=kln⁡a⋅akx\frac{dy}{dx} = k \ln a \cdot a^{kx}

Worked Example 2

Differentiate the following functions

(a) f(x)=ln⁡(2x5)+ln⁡(2x)f(x) = \ln(2x^5) + \ln(2x)

**Answer:**

Rewrite both logarithms using the laws of logarithms: f(x)=ln⁡2+ln⁡x5+ln⁡2+ln⁡xf(x) = \ln 2 + \ln x^5 + \ln 2 + \ln x Simplify: f(x)=2ln⁡2+6ln⁡xf(x) = 2 \ln 2 + 6 \ln x

Since 2ln⁡22 \ln 2 is a constant, it differentiates to zero; ln⁡x\ln x differentiates to 1x\frac{1}{x}: f′(x)=6⋅1xf'(x) = 6 \cdot \frac{1}{x} f′(x)=6xf'(x) = \frac{6}{x}

Derivative of the Natural Logarithmic Function

Differentiating ln⁡x\ln x

For a natural logarithm:

f(x)=ln⁡xf(x) = \ln x f′(x)=1xf'(x) = \frac{1}{x}

Differentiating aln⁡xa \ln x

If there is a constant multiple of the logarithm:

h(x)=aln⁡xh(x) = a \ln x h′(x)=a(1x)=axh'(x) = a \left( \frac{1}{x} \right) = \frac{a}{x}

Differentiating ln⁡kx\ln kx

If there is a constant multiple of xx inside the logarithm:

  1. Use the laws of logarithms to rewrite: g(x)=ln⁡kx=ln⁡k+ln⁡xg(x) = \ln kx = \ln k + \ln x
  2. ln⁡k\ln k is a constant, hence its derivative is zero:
  3. Therefore: g′(x)=1xg'(x) = \frac{1}{x}
Exam Tip

Don't forget that the derivative of ln⁡kx\ln kx is 1x\frac{1}{x}, the same as the derivative for ln⁡x\ln x. Differentiating ln⁡kx\ln kx as kx\frac{k}{x} is a common mistake on exams!

Worked Example 3

Differentiate the following function:

f(x)=ln⁡(2x5)+ln⁡(2x)f(x) = \ln(2x^5) + \ln(2x)

**Answer:**

Rewrite both logarithms using the laws of logarithms: f(x)=ln⁡2+ln⁡x5+ln⁡2+ln⁡xf(x) = \ln 2 + \ln x^5 + \ln 2 + \ln x Simplify: f(x)=2ln⁡2+6ln⁡xf(x) = 2 \ln 2 + 6 \ln x

Since 2ln⁡22 \ln 2 is a constant, it differentiates to zero; ln⁡x\ln x differentiates to 1x\frac{1}{x}: f′(x)=6⋅1xf'(x) = 6 \cdot \frac{1}{x} f′(x)=6xf'(x) = \frac{6}{x}

Practice Questions

Practice Question
  1. Differentiate: y=e3x+5e2x−7exy = e^{3x} + 5e^{2x} - 7e^{x}
Practice Question
  1. Differentiate: y=4x+42xy = 4^x + 4^{2x}
Practice Question
  1. Differentiate: y=2ln⁡(5x)y = 2 \ln(5x)

Glossary

  • Exponential Function: A function of the form exe^x or ekxe^{kx} where ee is the base of natural logarithms.
  • Natural Logarithm: The logarithm to the base ee, denoted as ln⁡x\ln x.
  • Chain Rule: A rule for differentiating compositions of functions.
  • Constant Multiple Rule: A rule stating that the derivative of a constant times a function is the constant times the derivative of the function.

Summary and Key Takeaways

Summary

  • The derivative of exe^x is exe^x.
  • The derivative of ekxe^{kx} is kekxk e^{kx} using the chain rule.
  • The derivative of aekxa e^{kx} is akekxa k e^{kx}.
  • The derivative of axa^x is axln⁡aa^x \ln a.
  • The derivative of akxa^{kx} is akxkln⁡aa^{kx} k \ln a.
  • The derivative of ln⁡x\ln x is 1x\frac{1}{x}.
  • The derivative of aln⁡xa \ln x is ax\frac{a}{x}.

Key Takeaways

  • Always apply the chain rule when differentiating composite functions.
  • Remember the properties of logarithms to simplify differentiation.
  • Be cautious of common mistakes such as incorrectly applying the derivative rules for logarithmic functions.

By following these structured notes, you will gain a comprehensive understanding of differentiating exponential and logarithmic functions, which are crucial topics in calculus and essential for success in your exams.

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Question 1 of 12

What is the derivative of f(x)=exf(x) = e^x?

exe^x

1

xex−1xe^{x-1}

0