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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.

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{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}

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Feb 23, 2012

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Nov 04, 2014
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TI.MAT.ENG.SE.1.Calculus.7.2

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