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Name: Jo Anne Kenyon
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Title: Derivative of the Tangent
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Description: Using the definition of derivative and
graphing to find the derivative of the tangent
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Derivative of Tangent
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2:1,14,10,Courier,1,12,0,0,0;1,28,21,New York,1,24,0,65535,65535;
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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.
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Clear[x,f]
f[x_]=Tan[x]
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Remember, the average growth rate on the interval [x,x+h] in units on the y-axis per unit on the x-axis is:
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(f[x+h]-f[x])/h
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(f[x+h] - f[x])/h as h->0
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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.
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Let's start with h=.1. Here is the plot.
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plot1=Plot[(f[x+.1]-f[x])/.1,{x,0,2Pi},PlotStyle->{RGBColor[1,0,0]}];
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Now, let h get closer to 0. Here is a graph when h=.01.
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plot2=Plot[(f[x+.01]-f[x])/.01,{x,0,2Pi},PlotStyle->{RGBColor[0,1,0]}]
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One last plot. Let h=001.
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plot3=Plot[(f[x+.001]-f[x])/.001,{x,0,2Pi},PlotStyle->{RGBColor[0,0,1]}]
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A Possible Choice for f'[x]
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Do these curves look like the curves of another trigonometric function?
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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.
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Plot[Sec[x],{x,0,2Pi}]
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What might we do to the secant in order to stay above the x-axis?
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Let's look at the graph of secant squared.
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secantsqplot=Plot[(Sec[x])^2,{x,0,2Pi}]
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Verification of our Choice
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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.
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Here is the graph of (f[x+.1] - f[x])/.1 and secant squared on the same axis:
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Show[plot1,secantsqplot]
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-Graphics-
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These graphs are pretty close.
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Here is a plot of (f[x+.01]-f[x])/.01 and secant squared:
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Show[plot2,secantsqplot]
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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:
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Show[plot3,secantsqplot]
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when Unformatted text is generated.
;[o]
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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])Û.
Let's check this numerically.
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Clear [f,x]
f[x_]=Tan[x]
f'[x]
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Hurray!! We now know the derivative of tangent is secant squared.
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^*)