How do you differentiate $V\left( r \right)=\dfrac{4}{3}\pi {{r}^{3}}$?

Hint: The instantaneous rate of change of a curve is the rate of change of y with respect to rate of change of x. The instantaneous rate of change of a curve is something whose rate of change is possibly constantly changing and this is called the derivative of that curve. Thus, by applying the basic rules of differentiation, we shall differentiate the given equation.

Complete step-by-step solution:
Instantaneous rate is the central idea of differential calculus. If we have to find the rate of change of a curve at a particular point, then what we are actually finding is the tangent of the curve at that particular point. We call this the derivative of the curve at that particular point. To find this, we calculate the derivative and substitute the coordinates of that particular point in the derivative to find the actual rate of change at a particular point.
Differentiating $V\left( r \right)=\dfrac{4}{3}\pi {{r}^{3}}$ with respect to r, we get
$\dfrac{d}{dr}V\left( r \right)=\dfrac{d}{dr}\left( \dfrac{4}{3}\pi {{r}^{3}} \right)$
We take the constants outside differentiation and get,
$\Rightarrow {V}'\left( r \right)=\dfrac{4}{3}\pi \dfrac{d}{dr}{{r}^{3}}$
Applying the property of differentiation, $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ , we get
$\Rightarrow {V}'\left( r \right)=\dfrac{4}{3}\pi \left( 3{{r}^{2}} \right)$
$\Rightarrow {V}'\left( r \right)=4\pi {{r}^{2}}$
Therefore, the derivative of $V\left( r \right)=\dfrac{4}{3}\pi {{r}^{3}}$ is given as ${V}'\left( r \right)=4\pi {{r}^{2}}$.

Note: For a line, the slope of a line describes the rate of change of a vertical variable with respect to a horizontal variable. Any line is associated with a slope because it has a constant rate of change. If any two points are taken on a line, no matter how far apart or no matter how close together, and the slope is calculated by the change in y divided by the change in x, then we would get the same constant slope.