How do you estimate the instantaneous rate of change at the point for $x=5$ for $f(x) = ln(x)$ ?

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How do you estimate the instantaneous rate of change at the point for $x=5$ for $f(x) = ln(x)$ ?

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Answer 1 · 2015-04-03T18:36:55

Since $f'(x)=1/x$ , the answer is just $f'(5)=\frac{1}{5}=0.2$ .

When $x=5$ (so that $y=f(5)=ln(5)\approx 1.609$ ), small changes in $x$ produce changes in $y$ that are about 5 times as small. For example, if $\Delta x=0.1$ (if $x$ changes from 5 to 5.1), then $\Delta y=f(5.1)-f(5)=ln(5.1)-ln(5)\approx 0.0198$ , which is about 5 times smaller than $\Delta x=0.1$ .

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How do you estimate the instantaneous rate of change at the point for $x=5$ for $f(x) = ln(x)$ ?

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How do you estimate the instantaneous rate of change at the point for x=5x=5 for f(x) = ln(x)f(x) = ln(x)?

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How do you estimate the instantaneous rate of change at the point for $x=5$ for $f(x) = ln(x)$ ?