Question

How to find instantaneous rate of change for $f\left( x \right)=\ln x$ at $x=5?$

Answer 1

Hint: Instantaneous rate of change of an object is the rate of the object at a specific instance of time. It is the slope of the tangent line at a specific point at which the object lies at a specific instance of time. We will find the first derivative of the given function and then find the value of the derivative at the given point.

Complete step by step solution:
Consider the given function, $f\left( x \right)=\ln x.$
We need to find the instantaneous rate of change of the function at the point $x=5.$
We know that the instantaneous rate of change of an object is the rate of change of the object at a specific instance of time.
The rate of change of an object is found by differentiating the object with respect to time.
So, we are going to differentiate the given function with respect to $x$ to get $\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\ln x.$
Now, we know that the first derivative of the given function is $\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\ln x=\dfrac{1}{x}.$
So, the rate of change of the given function is $\dfrac{1}{x}.$
Now we need to find the value of this derivative at $x=5.$
So, we will apply the value of $x$ in the derivative to find the value of derivative when $x=5.$
Therefore, the derivative at $x=5$ is ${{\left. \dfrac{1}{x} \right|}_{x=5}}=\dfrac{1}{5}.$
Hence the instantaneous rate of change of the given function at the given point is $\dfrac{1}{5}.$

Note: The graph of the given function has a tangent line at the point $\left( 5,\ln 5 \right)$ with a slope of $\dfrac{1}{5}.$ The instantaneous rate of change is different from the average rate of change. The average rate of change is over a range and the instantaneous rate of change is over a particular instance.