Derivative of the Tangent

Name:  Jo Anne Kenyon

Title:  Derivative of the Tangent

Description:  Using the definition of derivative and graphing to find the derivative of the tangent

Introduction

In the previous lesson, we saw that if f[x]=sin[x], then f'[x]=cos[x].  Now, let's find f'[x] when f[x]=tan[x].
Remember, f'[x] represents instantaneous growth.  We need to examine the limiting case of average growth rates as h closes in on 0.

Clear[x, f]f[x_] = Tan[x]

Remember, the average growth rate on the interval [x,x+h] in units on the y-axis per unit on the x-axis is:

((f[x + h] - f[x])/h)

(f[x+h] - f[x])/h as h->0

As with sin[x], the limiting case is hard to see from what we have above and it is going to be difficult to spot f'[x] by looking at this.  Let's look at some  plots of (f[x+h] - f[x])/h as h approaches 0.

Let's start with h=.1.  Here is the plot.

\(\(plot1 = 
      Plot[\(f[x +  .1] - f[x]\)\/ .1, {x, 0, 2\ π}, 
        PlotStyle -> {RGBColor[1, 0, 0]}]; \)\)

Now, let h get closer to 0.  Here is a graph when h=.01.

\(\(plot2 = 
      Plot[\(f[x +  .01] - f[x]\)\/ .01, {x, 0, 2\ π}, 
        PlotStyle -> {RGBColor[0, 1, 0]}]; \)\)

One last plot.  Let h=001.

\(\(plot3 = 
      Plot[\(f[x +  .001] - f[x]\)\/ .001, {x, 0, 2\ π}, 
        PlotStyle -> {RGBColor[0, 0, 1]}]; \)\)

A Possible Choice for f'[x]

Do these curves look like the curves of another trigonometric function?

That's correct, it does remind us of the graph of the secant except for the fact that these curves are all above the x-axis.  Here is the graph of the secant.

Plot[Sec[x], {x, 0, 2 π}];

What might we do to the secant in order to stay above the x-axis?

Let's look at the graph of secant squared.

secantsqplot = Plot[Sec[x]^2, {x, 0, 2 π}];

Verification of our Choice

Let's check this out by superimposing a plot of secant squared on each of the three average rate of growth graphs of the tangent.

Here is the graph of  (f[x+.1] - f[x])/.1 and secant squared on the same axis:

Show[plot1, secantsqplot];

These graphs are pretty close.

Here is a plot of (f[x+.01]-f[x])/.01 and secant squared:

Show[plot2, secantsqplot];

The two curves are almost concurrent.  Last, but not least, here is the graph of (f[x+.001] - f[x])/.001 and secant squared on the same axes:

Show[plot3, secantsqplot];

The evidence tells us that the limiting case of the average growth rates
        (f[x+h]-f[x])/h = (tan[x+h] - tan[x])/h
as h closes in on 0 is
         f'[x] = (((sec[x]))^2).
Let's check this numerically.

Clear[f, x]f[x_] = Tan[x]f'[x]

Hurray!!  We now know the derivative of tangent is secant squared.