Calculus

Recall the formula for finding the slope of a line joining two points $(x_1,y_1)$ and $(x_2,y_2)$ 

$$m=\frac{y_2-y_1}{x_2-x_1}$$

Suppose these two points are points on a curve $y=f(x)$:

Notice that $y_1=f(x_1)$ and $y_2=f(x_2)$. Let's adjust our notation slightly. Instead of $x_1$ and $x_2$, let's use $x$ and $x+h$ instead. Therefore, instead of $y_1$ and $y_2$, we now have $f(x)$ and $f(x+h)$ as shown below

Notice that with the new notation, the formula for the slope of the line has now become

$$m=\frac{y_2-y_1}{x_2-x_1}=\frac{f(x+h)-f(x)}{x+h-x}=\frac{f(x+h)-f(x)}{h}$$

Now suppose we make these two points on the curve closer together (we do this by letting $h$ become very small).

The closer we push the second point towards the $(x,f(x))$, the closer the above formula comes to giving us the slope of the tangent line to the curve at the point $(x,f(x))$.

The expression 

$$\frac{f(x+h)-f(x)}{h}$$

is not defined for for $h=0$, but this is where limits come in handy. We can approximate the slope of the tangent line by taking the limit of the above expression as $h\to 0$ (if the limit exists). We call this limit the derivative of the function $f(x)$, and denote it $f^{\prime }(x)$ or ${\displaystyle \frac{d}{dx}}(f(x))$.

The point $P$ is $(x,f(x))$, and the point $Q$ is $(x+h,f(x+h))$. As point $Q$ moves closer to point $P$, the distance between the two points, $h$, also decreases.

Observe that as this happens, the approximated tangent, represented by the blue dashed line, approaches the actual tangent at point $P$, represented by the red line.

Try changing the function, $f(x)$, in the left side panel, and play around by moving points $P$ and $Q$.

Tip: You can drag the point $N$ horizontally.