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7.2: The Logarithmic Derivative

Difficulty Level: At Grade Created by: CK-12
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This activity is intended to supplement Calculus, Chapter 6, Lesson 3.

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

If \begin{align*}(x, y)\end{align*}(x,y) is a point on \begin{align*}y = f(x)\end{align*}y=f(x) and \begin{align*}y = g(x)\end{align*}y=g(x) is the inverse of \begin{align*}f(x)\end{align*}f(x), then \begin{align*}(y, x)\end{align*}(y,x) is a point on \begin{align*}g(x)\end{align*}g(x). We know that \begin{align*}e^0 = 1\end{align*}e0=1 and \begin{align*}\ln(1) = 0\end{align*}ln(1)=0, so \begin{align*}(0, 1)\end{align*}(0,1) is a point on \begin{align*}y = e^x\end{align*}y=ex and \begin{align*}(1, 0)\end{align*}(1,0) is a point on \begin{align*}y = \ln(x)\end{align*}y=ln(x). We could do this for several points and keep getting the same inverse results.

Thus, if \begin{align*}y = e^x\end{align*}y=ex, then \begin{align*}x = e^y\end{align*}x=ey will be equivalent to \begin{align*}y = \ln(x)\end{align*}y=ln(x) because they are inverses of one another. Now we can take the implicit derivative with respect to \begin{align*}x\end{align*}x of \begin{align*}x = e^y\end{align*}x=ey.

\begin{align*}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}\end{align*}x=ey1=dydxey1=dydxx1x=dydx

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

  • Find \begin{align*}\lim_{h \to 0} \frac{\ln(2+h)- \ln(2)}{h}\end{align*}limh0ln(2+h)ln(2)h.
  • Do the same with \begin{align*}\lim_{h \to 0} \frac{\ln(3+h)- \ln(3)}{h}\end{align*}limh0ln(3+h)ln(3)h.
  • What is \begin{align*}\lim_{h \to 0} \frac{\ln(x+h)- \ln(x)}{h} ?\end{align*}limh0ln(x+h)ln(x)h?

Use the derivative command to find the derivative of the logarithmic function \begin{align*}f(x) = \ln(x)\end{align*}f(x)=ln(x).

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

What happens if our logarithm has a base other than \begin{align*}e\end{align*}e? We need to know how to take the derivative of the function \begin{align*}y = \log_a(x)\end{align*}y=loga(x).

First we want to compare \begin{align*}y1 = \ln(x)\end{align*}y1=ln(x) and \begin{align*}y2 = \log_2(x)\end{align*}y2=log2(x).

To enter \begin{align*}\log_2(x)\end{align*}log2(x), use the alpha keys to spell out log.

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

  • Graph both functions (\begin{align*}\ln(x)\end{align*}ln(x) and \begin{align*}\log_2(x)\end{align*}log2(x)) on the same set of axes. Sketch your graph to the right. What do you notice?

  • Do the same steps with \begin{align*}y1 = \ln(x)\end{align*}y1=ln(x) and \begin{align*}y3 = \log_4(x)\end{align*}y3=log4(x). What do you notice?

  • Simplify the following ratios.

\begin{align*}\frac{\ln(x)}{\log_2(x)} && \frac{\ln(x)}{\log_4(x)} && \frac{\ln(x)}{\log_a(x)}\end{align*}ln(x)log2(x)ln(x)log4(x)ln(x)loga(x)

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

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

What happens when we take the derivative of \begin{align*}y = \log_a(x)\end{align*}. Use the derivative command to find the derivatives of the functions below.

\begin{align*}f(x) = \log_2(x) && g(x) = \log_3(x) && h(x) = \log_a(x)\end{align*}

  • Do you notice a pattern?

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

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

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

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

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

  • Suppose that \begin{align*}y = \log_a(u)\end{align*}, where \begin{align*}u\end{align*} depends on \begin{align*}x\end{align*}. Using the chain rule, take the derivative of this function.

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

Identify \begin{align*}u(x)\end{align*} and \begin{align*}a\end{align*} for each function before you find the derivative.

  • \begin{align*}f(x) = 5^{(x^2)} \quad u(x)=\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \quad a =\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

\begin{align*}f'(x) = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

  • \begin{align*}g(x) = e^{(x^3+2)} \quad u(x)=\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \quad a = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

\begin{align*}g'(x) = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

  • \begin{align*}h(x) = \log_3(x^4 + 7) \quad u(x)=\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \quad a =\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

\begin{align*}h'(x) =\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

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

\begin{align*}j'(x) =\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

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Date Created:
Feb 23, 2012
Last Modified:
Nov 04, 2014
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