<meta http-equiv="refresh" content="1; url=/nojavascript/"> The Logarithmic Derivative | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Texas Instruments Calculus Student Edition Go to the latest version.

# 7.2: The Logarithmic Derivative

Created by: CK-12

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

## Problem 1 – The Derivative of $y = \ln(x)$

If $(x, y)$ is a point on $y = f(x)$ and $y = g(x)$ is the inverse of $f(x)$, then $(y, x)$ is a point on $g(x)$. We know that $e^0 = 1$ and $\ln(1) = 0$, so $(0, 1)$ is a point on $y = e^x$ and $(1, 0)$ is a point on $y = \ln(x)$. We could do this for several points and keep getting the same inverse results.

Thus, if $y = e^x$, then $x = e^y$ will be equivalent to $y = \ln(x)$ because they are inverses of one another. Now we can take the implicit derivative with respect to $x$ of $x = e^y$.

$x=e^y \rightarrow 1 = \frac{dy}{dx} \cdot e^y \rightarrow 1 = \frac{dy}{dx} \cdot x \rightarrow \frac{1}{x} = \frac{dy}{dx}$

Use the limit command to test this formula. Be careful with your parentheses.

• Find $\lim_{h \to 0} \frac{\ln(2+h)- \ln(2)}{h}$.
• Do the same with $\lim_{h \to 0} \frac{\ln(3+h)- \ln(3)}{h}$.
• What is $\lim_{h \to 0} \frac{\ln(x+h)- \ln(x)}{h} ?$

Use the derivative command to find the derivative of the logarithmic function $f(x) = \ln(x)$.

## Problem 2 – The Derivative of $y = \log_a(x)$

What happens if our logarithm has a base other than $e$? We need to know how to take the derivative of the function $y = \log_a(x)$.

First we want to compare $y1 = \ln(x)$ and $y2 = \log_2(x)$.

To enter $\log_2(x)$, use the alpha keys to spell out log.

Within the parentheses, enter the expression, then the base.

• Graph both functions ($\ln(x)$ and $\log_2(x)$) on the same set of axes. Sketch your graph to the right. What do you notice?

• Do the same steps with $y1 = \ln(x)$ and $y3 = \log_4(x)$. What do you notice?

• Simplify the following ratios.

$\frac{\ln(x)}{\log_2(x)} && \frac{\ln(x)}{\log_4(x)} && \frac{\ln(x)}{\log_a(x)}$

Sometimes the ratio $\frac{\ln(x)}{\log_a(x)}$ is written as $\ln(x) = \ln(a) \cdot \log_a(x)$. We can rewrite this ratio as $\log_a(x) = \frac{\ln(x)}{\ln(a)}$ and call it an identity.

• Graph the following functions on the same set of axes: $y1 = \ln(x), y2 = \ln(2) \cdot \log_2(x), y3 = \ln(3) \cdot \log_3(x)$. What was the result?

What happens when we take the derivative of $y = \log_a(x)$. Use the derivative command to find the derivatives of the functions below.

$f(x) = \log_2(x) && g(x) = \log_3(x) && h(x) = \log_a(x)$

• Do you notice a pattern?

What does $\log_2(e)$ equal? If we use the formula from earlier in this class, we get $\log_2(e) = \frac{\ln(e)}{\ln(2)} = \frac{1}{\ln(2)}$.

Therefore, the general result is $y = \log_a(x) \rightarrow \frac{dy}{dx} = \frac{1}{(x \ln(a))}$.

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

Now we want to take the derivative of more complicated functions:

Recall: $y = a^u \rightarrow \frac{dy}{dx} = a^u \ \frac{du}{dx}$ where $u$ depends on $x$.

• Suppose that $y = \log_a(u)$, where $u$ depends on $x$. Using the chain rule, take the derivative of this function.

Find the derivative of the following functions with the chain rule.

Identify $u(x)$ and $a$ for each function before you find the derivative.

• $f(x) = 5^{(x^2)} \quad u(x)=\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \quad a =\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$

$f'(x) = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$

• $g(x) = e^{(x^3+2)} \quad u(x)=\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \quad a = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$

$g'(x) = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$

• $h(x) = \log_3(x^4 + 7) \quad u(x)=\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \quad a =\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$

$h'(x) =\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$

• $j(x) = \ln \left (\sqrt{x^6 + 2} \right ) \quad u(x)=\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \quad a = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$

$j'(x) =\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$

Feb 23, 2012

Nov 04, 2014