Differentiation II: The Gory Details of Calculating Derivatives

In this lesson we will show how to find the slope of a function that is not a straight line. Let's demonstrate how we do this with the function <Sign in to see all the formulas>. First, the most important thing to know is what a secant is. A secant is a straight line between two points on a function. So the first thing we do is draw a straight line from the point $x_0$ where we want to know the derivative to the point $x_0+h$ where $h$ is arbitrary and can be either positive or negative. Next, lets find the slope of the secant line. This is simply its "rise" over its "run" and is given by

<Sign in to see all the formulas>

Now we need to do some algebra to simplify the mathematical expression. For the function $x^2$ it is easy. The steps are:

<Sign in to see all the formulas>

Next we "slide" the point $x_0+h$ towards the point $x_0$ by making $h$ smaller and smaller. In math speak we say that we take the function limit as $h$ goes to zero. We denote this by

<Sign in to see all the formulas>

But when we do this, the secant line that results is the secant line between $x_0$ and the closest point to it $x_0+h$. We call this particular secant line a tangent line. The slope of the tangent line is the slope of the function at the point $x_0$. We call this the function's derivative at point $x_0$. In practice we do not specify an actual value for $x_0$ and therefore just refer to the derivative as

<Sign in to see all the formulas>

knowing that we can plug in a specific value into $x$ at any time.

Now let's finish up our example with $x^2$ by writing everything out:

<Sign in to see all the formulas>

It is important to note that we needed to know the function limits <Sign in to see all the formulas> and <Sign in to see all the formulas> as well as our properties of function limits.

Now that we have the derivative, we can find the slopes at different points along the curve. For example, the picture below shows the tangent lines at $x=-5$, $x=0$, and $x=2$. The slopes are found by plugging the values of $x$ into <Sign in to see all the formulas>.

Here is a summary of the steps:
Draw the secant from the point where you want the slope to another point $x+h$.
Construct the function <Sign in to see all the formulas>. This is the slope of the secant line.
Do some algebra to simplify the function <Sign in to see all the formulas>.
Find the function limit <Sign in to see all the formulas> if it exists. This is the slope of the tangent line at $x$. We call this the derivative of the function $f(x)$ and denote it by either $f'(x)$ or $df/dx$.

We know that the straight line is defined by the equation <Sign in to see all the formulas> where $m$ is the slope of the function and $b$ is the $y$-intercept. We would expect that using the methodology above will yield a result of $m$. Let's see how this works by constructing the function <Sign in to see all the formulas>. First, <Sign in to see all the formulas>. Now we have
\begin{eqnarray*}
\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}&=&\lim_{h\to 0}\frac{m\,x+m\,h+b-m\,x-b}{h}\\
&=&\lim_{h\to 0}\frac{m\,h}{h}\\
&=&\lim_{h\to 0}m\\
&=&m
\end{eqnarray*}
It worked! <Sign in to see all the formulas> as we know it should.