Question

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`.