# How to find instantaneous rate of change for f(x) = ln(x) at x=5?

Jan 3, 2017

$\frac{1}{5}$

#### Explanation:

The instantaneous rate of change of the function $f$ at $x = a$ is expressible through $f ' \left(a\right)$, since this is the slope (rate of change) of the tangent line at that point.

So, for this question, we must know that $\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$; that is, the derivative of $\ln \left(x\right)$ is $\frac{1}{x}$. This is a very well known fact and can also be shown through another very well known derivative: $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$.

Anyway, we see that $f \left(x\right) = \ln \left(x\right)$, so the derivative is $f ' \left(x\right) = \frac{1}{x}$.

The instantaneous rate of change at $x = 5$ is $f ' \left(5\right)$, and $f ' \left(5\right) = \frac{1}{5}$.

This means that when the point $\left(5 , \ln \left(5\right)\right)$, which lies on the graph of $\ln \left(x\right)$, is included in a line with a slope of $\frac{1}{5}$, the line will be tangent at the point $\left(5 , \ln \left(5\right)\right)$ and the line's slope of $\frac{1}{5}$ represents how fast the function $\ln \left(x\right)$ is changing at $x = 5$.

That line is $y - \ln \left(5\right) = \frac{1}{5} \left(x - 5\right)$, or:

graph{(y-ln(5)-1/5(x-5))(y-lnx)=0 [-0.365, 17.415, -3.44, 5.45]}