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# 7.2: The Logarithmic Derivative

Difficulty Level: At Grade Created by: CK-12

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

Topic: Formal Differentiation

• Derive the Logarithmic Rule and the Generalized Logarithmic Rule for differentiating logarithmic functions.
• Prove that ln(x)=ln(a)loga(x)\begin{align*}\ln (x) =\ln (a) \cdot \log_a(x)\end{align*} by graphing f(x)=loga(x)\begin{align*}f(x)=\log_a(x)\end{align*} and g(x)=ln(x)\begin{align*}g(x) = \ln (x)\end{align*} 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 F6\begin{align*}F6\end{align*} :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 y=ln(x)\begin{align*}y = \ln (x)\end{align*}

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

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

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

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

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

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

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

Students should graph the functions y1=ln(x),y2=ln(2)log2(x),y3=ln(3)log3(x)\begin{align*}y1 = \ln (x), y2 = \ln (2) \cdot \log_2(x), y3 = \ln (3) \cdot \log_3(x)\end{align*}. 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 dydx=loga(e)x\begin{align*}\frac{dy}{dx} = \frac{\log_a (e)}{x}\end{align*}. Thus, if y=loga(x)\begin{align*}y = \log_a (x)\end{align*}, then dydx=1(xln(a))\begin{align*}\frac{dy}{dx} = \frac{1}{(x \ln (a))}\end{align*}.

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

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

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

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

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

• \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*}

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