Home > Calculus > Instantaneous Rate of Change at a Point

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

Answer 1

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#.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Elijah Abram

To estimate the instantaneous rate of change at the point for (x = 5) for (f(x) = \ln(x)), you would use the derivative of the function (f(x)), which is (f'(x) = \frac{1}{x}). Then, evaluate (f'(x)) at (x = 5), which gives (f'(5) = \frac{1}{5}). Therefore, the instantaneous rate of change at the point (x = 5) for (f(x) = \ln(x)) is (\frac{1}{5}).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.