First principle of calculus

A secant line is a straight line that connects two points on a curve. A tangent line is a straight line that “just touches” the curve at a single point.

The slope of the secant line gives us the average rate of change of the function on the interval between the two points. The slope of the tangent line gives us the instantaneous rate of change.

Finding the slope of the secant line is easy:

$m_{\text{sec}}=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{y_2-y_1}{x_2-x_1}=\frac{3-4}{4-2}=-\frac{1}{2}\text{.}$

We can generalize this to give us the average rate on the interval $[x_1,x_2]$ for any function $f$ with

$m_{\text{sec}}=\frac{f(x_2)-f(x_1)}{x_2-x_1}\text{,}$

but what about the tangent line? We could draw it with a ruler and pick two points that is passes through, but this method is tedious and inaccurate. No, instead we can use our newly-acquired limit skills.

The key thing to realize here is that the secant line and the tangent line are related. A tangent line is just a secant line whose interval happens to be very, very small—infinitesimal, to be exact (this means indefinitely small). In other words, the slope of the tangent line at $x=a$ is equal to the slope of the secant line on the interval $[a,a+h]$ as $h$ approaches zero:

$m_{\text{tan}}=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}\text{.}$

This is the first principle of calculus. As with most other things, graphing will help us understand it:

As we move the second point closer to the first point, thereby making $h$ approach zero, the slope of the secant line will converge on the slope of the tangent line at $a\text{.}$

Suppose we want to find the slope of the tangent line at 3 on $f(x)=x^2\text{.}$ We simply plug it into our first principle of calculus, giving us

$m_{\text{tan}}=\underset{h\to 0}{\lim }\frac{f(3+h)-f(3)}{h}=\underset{h\to 0}{\lim }\frac{(3+h{)}^2-3^2}{h}\text{,}$

which simplifies to

$\underset{h\to 0}{\lim }\frac{9+6h+h^2-9}{h}=\underset{h\to 0}{\lim }\frac{6h+h^2}{h}=\underset{h\to 0}{\lim }6+h=6+0=6\text{,}$

therefore the instantaneous rate of change at $x=3$ is 6.