Intuitive explanation of a derivative with an example

Question

I am having a hard time grasping a concept of derivative intuitively, perhaps due to a lack of a good example of how it can be used in practice. I am looking for an explanation in laymen terms with a practical example that can be deconstructed and would give an idea of how a derivative can be used in practice. I am not looking for mathematical proof or strict mathematical definition.

Here is my current understanding, please point out where it is correct or incorrect intuitively:

Let's say that we have $y$ (dependent variable or output) and $x$ (independent variable or input). If we have a function of $y=x^{3}$. Does derivative tell us by how much the output of a function (dependent variable $y$) when we change input (independent variable $x$) by a certain amount ($dx$)? In other words, derivative tells us how sensitive the function is to the changes in its input.

P.S. I could not find a satisfactory explanation of this question anywhere on stack exchange.

Answer 1

Yes of course the derivative of a function represents the rate of change $\Delta y$ of the y coordinates for a change $\Delta x$ of x coordinates as the increment becomes "small", that is

$$f'(x)=lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$

Slope could be a very intuitive concept to understand derivatives. We can indeed think to the function as a mountain track, then the derivative at a point is the slope (i.e. the slope of the tangent line) exactly at that point.

Answer 2

Suppose $x$ is a measurement of time and $y=y(x)$ is the distance travelled by a car at time $x$ along a straight line from a starting point. The car may stop or reverse; it need not be always moving forward, or even moving at all. The $average$ speed in the forward direction, over the time interval from time $x$ to time $x+d$ (with $d\ne 0$) is $\frac {y(x+d)-y(d)}{(x+d)-x}=\frac {y(x+d)-y(x)}{d}.$ Keeping $x$ fixed and letting $d\to 0,$ this tends to $y'(x),$ which we call the speed "at time $x.$ " It is the rate of change of the car's position at time $x.$ If $y'(x)=0$ it means the car was stopped at that moment. If $y'(x)<0$ the car was moving in reverse at time $x.$