<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Skip Navigation

7.2: The Logarithmic Derivative

Difficulty Level: At Grade Created by: CK-12
Turn In

This activity is intended to supplement Calculus, Chapter 6, Lesson 3.

ID: 9092

Time required: 45 minutes

Activity Overview

Students will determine the derivative of the function \begin{align*}y = \ln (x)\end{align*}y=ln(x) and work with the derivative of both \begin{align*}y = \ln (u)\end{align*}y=ln(u) and \begin{align*}y = \log_a(u)\end{align*}y=loga(u). In the process, the students will show that \begin{align*} \lim_{h \to 0} \frac{ \ln (a + h) - \ln (a)}{h} = \frac{1}{a}\end{align*}limh0ln(a+h)ln(a)h=1a.

Topic: Formal Differentiation

  • Derive the Logarithmic Rule and the Generalized Logarithmic Rule for differentiating logarithmic functions.
  • Prove that \begin{align*}\ln (x) =\ln (a) \cdot \log_a(x)\end{align*}ln(x)=ln(a)loga(x) by graphing \begin{align*}f(x)=\log_a(x)\end{align*}f(x)=loga(x) and \begin{align*}g(x) = \ln (x)\end{align*}g(x)=ln(x) for some a and deduce the Generalized Rule for Logarithmic differentiation.
  • Apply the rules for differentiating exponential and logarithmic functions.

Teacher Preparation and Notes

  • This investigation derives the definition of the logarithmic derivative. The students should be familiar with keystrokes for the limit command, the derivative command, entering the both the natural logarithmic and the general logarithmic functions, drawing a graph, and setting up and displaying a table.
  • Before starting this activity, students should go to the home screen and select \begin{align*}F6\end{align*}F6 :Clean Up > 2:NewProb, then press \begin{align*}\div\end{align*}÷. This will clear any stored variables, turn off any functions and plots, and clear the drawing and home screens.
  • This activity is designed to be student-centered with the teacher acting as a facilitator while students work cooperatively.

Associated Materials

Problem 1 – The Derivative for \begin{align*}y = \ln (x)\end{align*}y=ln(x)

In this problem, students are asked to use the limit command (\begin{align*}F3\end{align*}F3 :Calc > 3:limit() to find the values. Remind the students to be very careful of their parentheses.

The students should get \begin{align*}\frac{1}{x}\end{align*}1x for the answer to the last problem.

In this portion, the students are asked to use the derivative command (\begin{align*}F3\end{align*}F3 :Calc > 1:d( differentiate) to find the derivative.

Problem 2 – The Derivative of \begin{align*}y = \log_a(x)\end{align*}y=loga(x)

Students should notice that both functions have a value of zero for \begin{align*}\ln (1)\end{align*}ln(1) and \begin{align*}\log_2(1)\end{align*}log2(1) but that \begin{align*}\log_2(x)\end{align*}log2(x) is a multiple of \begin{align*} \ln (x)\end{align*}ln(x). They both have about the same shape but \begin{align*}\ln (x)\end{align*}ln(x) is smaller (both positively and negatively) than \begin{align*}\log_2(x)\end{align*}log2(x).

When \begin{align*}\log_4(x)\end{align*}log4(x) is compared to \begin{align*}\ln(x)\end{align*}ln(x), they also intercept at 1 but \begin{align*}\ln (x)\end{align*}ln(x) is larger.

Have the students notice that \begin{align*}\ln (4)\end{align*}ln(4) is listed as \begin{align*}2(\ln (2))\end{align*}2(ln(2)) instead of \begin{align*}\ln (4)\end{align*}ln(4). The answer to the last problem is \begin{align*}\ln (a)\end{align*}ln(a).

Students should graph the functions \begin{align*}y1 = \ln (x), y2 = \ln (2) \cdot \log_2(x), y3 = \ln (3) \cdot \log_3(x)\end{align*}y1=ln(x),y2=ln(2)log2(x),y3=ln(3)log3(x). They should notice that all three functions yield the same graph. Students can check by graphing each one individually.

Students are asked to find the Generalized Logarithmic Rule for Differentiation. They should find \begin{align*}\frac{dy}{dx} = \frac{\log_a (e)}{x}\end{align*}dydx=loga(e)x. Thus, if \begin{align*}y = \log_a (x)\end{align*}y=loga(x), then \begin{align*}\frac{dy}{dx} = \frac{1}{(x \ln (a))}\end{align*}dydx=1(xln(a)).

Problem 3 – Derivative of Exponential and Logarithmic Functions Using the Chain Rule

Students are asked to identify \begin{align*}u(x)\end{align*}u(x) and \begin{align*}a\end{align*}a for each function and then find the derivative by hand or using the Derivative command to find the derivative.

Recall: \begin{align*}y = a^u \rightarrow \frac{dy}{dx} = a^u \ \frac{du}{dx}\end{align*}y=audydx=au dudx where \begin{align*}u\end{align*}u depends on \begin{align*}x\end{align*}x.

  • \begin{align*}y = \log_a(u) \rightarrow \frac{dy}{dx} = \frac{1}{(u \ln (a))} \cdot \frac{du}{dx}\end{align*}y=loga(u)dydx=1(uln(a))dudx or \begin{align*}\frac{dy}{dx} = \frac{ \frac{du}{dx}}{u(\ln (a))}\end{align*}dydx=dudxu(ln(a))

\begin{align*}f(x) = 5^{(x^2)}, u(x) = x^2, a = 5 \end{align*}f(x)=5(x2),u(x)=x2,a=5

  • \begin{align*}f'(x) = 2 \cdot \ln (5) \cdot x \cdot 5^{(x^2)}\end{align*}

\begin{align*}g(x) = e^{(x^3 + 2)}, u(x) = x^3 + 2, a = e \end{align*}

  • \begin{align*}g'(x) = 3 \cdot x^2 \cdot e^{(x^3 + 2)}\end{align*}

\begin{align*}h(x) = \log_3(x^4 + 7), u(x) = x^4 + 7, a = 3\end{align*}

  • \begin{align*}h'(x) = \frac{4 \cdot x^3 \log_3(e)}{x^4 + 7}\end{align*}

\begin{align*}j(x) = \ln \left (\sqrt{x^6 + 2}\right ), u(x) = \sqrt{x^6 + 2}, a = e\end{align*}

  • \begin{align*}j'(x) = \frac{3 \cdot x^5}{x^6 + 2}\end{align*}

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

Image Attributions

Show Hide Details
Date Created:
Feb 23, 2012
Last Modified:
Nov 04, 2014
Files can only be attached to the latest version of section
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original