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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 56 "Secant Lines, Tangent Lin
es, Derivatives(\"D\" and \"diff\")" }}{PARA 256 "" 0 "" {TEXT -1 262 
"In this lab we investigate the relationship between secant lines and \+
average rates of change, tangent lines and instantaneous rates of chan
ge, how Maple can aid in the limiting process that connects these two \+
concepts, and finally the derivative of a function.  " }}}{SECT 1 
{PARA 3 "" 0 "" {TEXT -1 47 "Plotting Secant Lines / Average rates of \+
change" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 685 "As we discussed in clas
s, the tangent line to a graph of f(x) at x = a , has slope describing
 the instantaneous rate of change of y with respect to x, at x = a. We
 further illustrated that this tangent line can be considered the limi
ting case of  a converging sequence of secant lines. Furthermore,  the
 slope of a secant line gives the average rate of change of y with res
pect to x over an interval.   Therefore,  the slope of the tangent lin
e (the derivative) at x = a, is the limiting value of a converging seq
uence of average rates of change, leading to an instantaneous rate of \+
change.   In this section we illustrate this concept graphically. Firs
t we define the function f(x) = " }{XPPEDIT 18 0 "(x-2)^2;" "6#*$,&%\"
xG\"\"\"\"\"#!\"\"F'" }{TEXT -1 8 " + 1 by:" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 22 "f := x -> (x-2)^2 + 1;" }{TEXT -1 0 "" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%\"fGf*6#%\"xG6\"6$%)operatorG%&arrowGF(,&*$),
&9$\"\"\"\"\"#!\"\"F2F1F1F1F1F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 322 "The goal is to plot a sequence of secant lines that approaches
 the tangent line of f(x) at x = 2.  Using the point slope definition \+
of a line, the secant line through (2, f(2)) and (2+h, f(2+h))  is des
cribed by  y - y(2) = m (x - 2), where m (slope) is determined from th
e two points. Writing this as a function yields:  " }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 69 "                                                 \+
             f(x) = " }{XPPEDIT 18 0 "(f(2+h)-f(2))/h;" "6#*&,&-%\"fG6
#,&\"\"#\"\"\"%\"hGF*F*-F&6#F)!\"\"F*F+F." }{TEXT -1 16 " (x - 2) + f(
2)." }}{PARA 0 "" 0 "" {TEXT -1 182 "We now define a function of two v
ariables, x and h,  that describes the secant line.  I make it a funct
ion of two variables just so that I don't have to re-enter different v
alues of " }{TEXT 259 0 "" }{TEXT -1 4 " h. " }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 55 "secf2h := (x,h) -> (f(2 + h) - f(2))/h *(x - 2) \+
+ f(2);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'secf2hGf*
6$%\"xG%\"hG6\"6$%)operatorG%&arrowGF),&*(,&-%\"fG6#,&\"\"#\"\"\"9%F5F
5-F16#F4!\"\"F5F6F9,&9$F5F4F9F5F5F7F5F)F)F)" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 117 "This cryptic notation is for \"sec\"ant line of \"f\" \+
at x = \"2\" for step size \"h\". You may call it anything you want.  \+
 " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "plot([secf2h(x,2), sec
f2h(x,1), secf2h(x,.5), secf2h(x,.1), f(x)], x = 0 .. 4);" }{TEXT -1 
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[pY)\\f\\$F27$Fby$\"3?XzA\")Gx$y$F27$Fgy$\"3ax32%43X0%F27$F\\z$\"3#3*)
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G6$Q\"x6\"Q!F^an-%%VIEWG6$;F(Ffz%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 
1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve \+
4" "Curve 5" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 126 "Notice that this
 produces a sequence of lines that approach the tangent line.  Take a \+
guess at the slope of the tangent line. " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "# end of section" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT 
-1 77 "Plotting Tangent Lines/  Instantaneous Rates of change  / The \+
\"limit\" command" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 150 " Now we try \+
to plot the tangent line to this curve at x = 2.  Recall that the slop
e of the tangent line , at x = 2, is defined by the following limit:" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "                               \+
   " }{XPPEDIT 18 0 "m[tan];" "6#&%\"mG6#%$tanG" }{TEXT -1 4 "  = " }
{XPPEDIT 18 0 "limit((f(2+h)-f(2))/h,h = 0);" "6#-%&limitG6$*&,&-%\"fG
6#,&\"\"#\"\"\"%\"hGF-F--F)6#F,!\"\"F-F.F1/F.\"\"!" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 107 "The following command in Maple will find this li
mit. Here we want the limit as h goes to zero so we type:  " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "mtanf2 := limit((f(2+h)-f(2)
)/h, h = 0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'mtanf2G\"\"!" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 306 "This notation is for the slope of
 the tangent line of  the function f(x), at x = 2.  Again, you can cal
l it anything you want.  Using a similar procedure as above, we plot t
he tangent line (from the point slope formula for a line) by:  f(x) = \+
mtanf2 (x - 2) + f(2).  And use this to plot the tangent line. " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "tanf2 := x -> mtanf2*(x - 2)
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LL$)G[kJF2$\"358jd&Q?gN#F27$$\"3#)****\\7yh]KF2$\"3U%GUH\"\\/kDF27$$\"
3xmmm')fdLLF2$\"3#R(R@7\\UyFF27$$\"3bmmm,FT=MF2$\"34K$\\Af%*=,$F27$$\"
3FLL$e#pa-NF2$\"3W+7LkskdKF27$$\"3!*******Rv&)zNF2$\"35^[pY)\\f\\$F27$
$\"3ILLLGUYoOF2$\"3?XzA\")Gx$y$F27$$\"3_mmm1^rZPF2$\"3ax32%43X0%F27$$
\"34++]sI@KQF2$\"3#3*)RIu/qN%F27$$\"34++]2%)38RF2$\"3w&)3\"\\D2*fYF27$
$\"\"%F)F*-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-F$6$7S7$F($\"\"\"F)7$F-Fb[l
7$F4Fb[l7$F:Fb[l7$F?Fb[l7$FDFb[l7$FIFb[l7$FNFb[l7$FSFb[l7$FXFb[l7$FgnF
b[l7$F\\oFb[l7$FaoFb[l7$FfoFb[l7$F[pFb[l7$F`pFb[l7$FepFb[l7$FjpFb[l7$F
_qFb[l7$FdqFb[l7$FiqFb[l7$F^rFb[l7$FcrFb[l7$FhrFb[l7$F]sFb[l7$FbsFb[l7
$FgsFb[l7$F\\tFb[l7$FatFb[l7$FftFb[l7$F[uFb[l7$F`uFb[l7$FeuFb[l7$FjuFb
[l7$F_vFb[l7$FdvFb[l7$FivFb[l7$F^wFb[l7$FcwFb[l7$FhwFb[l7$F]xFb[l7$Fbx
Fb[l7$FgxFb[l7$F\\yFb[l7$FayFb[l7$FfyFb[l7$F[zFb[l7$F`zFb[l7$FezFb[l-F
hz6&FjzF(F[[lF(-%+AXESLABELSG6$Q\"x6\"Q!Fj^l-%%VIEWG6$;F(Fez%(DEFAULTG
" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "C
urve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 165 "Notice above that the
 tanf2 = 0, and hence the tangent line is horizontal. This means that \+
the instantaneous rate of change of y with respect to x at x = 2 is ze
ro. " }}{PARA 0 "" 0 "" {TEXT -1 114 "Plot the tangent line to f (x) a
t x = 3 by defining a new slope (mtan3) and a new function tanf3. How'
s it look?  " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 148 "It should be poi
nted out here that the limit command need not be evaluated at a specif
ic value of x. For example, try evaluating the derivative of  " }}
{PARA 0 "" 0 "" {TEXT -1 7 "y = 13 " }{XPPEDIT 18 0 "x^4;" "6#*$%\"xG
\"\"%" }{TEXT -1 5 " - 5 " }{XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"#" }
{TEXT -1 36 " + 2 x - 1 by the limit definition: " }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 35 "q := x -> 13*x^4 - 5*x^2 + 2*x - 1;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"qGf*6#%\"xG6\"6$%)operatorG%&arrow
GF(,**&\"#8\"\"\")9$\"\"%F/F/*&\"\"&F/)F1\"\"#F/!\"\"*&F6F/F1F/F/F/F7F
(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "limit((q(x+h) - q(
x))/h, h = 0);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*&
\"#_\"\"\")%\"xG\"\"$F&F&\"\"#F&*&\"#5F&F(F&!\"\"" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 345 "Recognize this?  This is why we love Maple.  It p
erforms symbolic manipulations.  Ie. It doesn't just evaluate a sequen
ce of numbers, it can do all of the algebra that would be required to \+
determine this limit.  If Maple can perform the above limit calculatio
n then it should be able to symbolically determine derivatives. Right?
  ... Yes Sir!!!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "# end o
f section" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 42 "Derivatives: The \+
\"D\" and \"diff\" commands. " }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "M
aple has two commands to evaluate derivatives of functions.  D(f) and \+
diff(f(x), x). " }}{PARA 0 "" 0 "" {TEXT -1 42 "D(f)  returns the deri
vative as  function." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "g :
= x -> x^3 - 4*x;" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
\"gGf*6#%\"xG6\"6$%)operatorG%&arrowGF(,&*$)9$\"\"$\"\"\"F1*&\"\"%F1F/
F1!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "D(g);" }
{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#f*6#%\"xG6\"6$%)operat
orG%&arrowGF&,&*&\"\"$\"\"\")9$\"\"#F-F-\"\"%!\"\"F&F&F&" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 57 "Notice this describes a function and we g
ive it a name by" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "derivg \+
:= D(g);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'derivgGf
*6#%\"xG6\"6$%)operatorG%&arrowGF(,&*&\"\"$\"\"\")9$\"\"#F/F/\"\"%!\"
\"F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "or " }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 23 "derivg := x -> D(g)(x);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%'derivgGf*6#%\"xG6\"6$%)operatorG%&arrowGF(--%\"DG
6#%\"gG6#9$F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "In either ca
se, you can now evaluate the derivative at a number such as 3 by" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "derivg(3);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#\"#B" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "You can
 also plot the derivative with" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 58 "plot([g(x), derivg(x)], x = -4 .. 4, color = [red, blue]);" }
{TEXT -1 0 "" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&
-%'CURVESG6$7S7$$!\"%\"\"!$!#[F*7$$!3ommmmFiDQ!#<$!3w42dS))poS!#;7$$!3
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y\")Q<qlEF07$F`r$!3s!3'H%\\a@C$F07$Fer$!3EzuGW7'4n$F07$Fjr$!3[%*f(e<H%
4RF07$F_s$!3%*e+\"HA`***RF07$Fes$!3X_^7UH!3\"RF07$Fjs$!3)>lbneJ7o$F07$
F_t$!33\"ezl26WF$F07$Fdt$!3)f'\\npU&3o#F07$Fit$!3=#z:&4;xH>F07$F^u$!3u
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A>u(Gr\"GF07$Fbv$\"3#>Dv_6ejI%F07$Fgv$\"3x*))eZG(o-hF07$F\\w$\"3A9IS,?
L`zF07$Faw$\"36T_vArS05F37$Ffw$\"33w:piWAF7F37$F[x$\"3oT2`&*Q&oZ\"F37$
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2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 682 "In this graph we see the original functi
on in red and the derivative in blue.  From this graph you should be a
ble to estimate approximately where the tangent line is horizontal.  B
etter yet, if you click on the graph, you can then use the mouse to ha
ve Maple show you the coordinates of any point on the graph. Click on \+
the graph and notice that a window opens up on the menu bar showing th
e location of the cursor.  If you click on the cursor anywhere on the \+
graph these coordinates will change.  Try this and notice that the der
ivative is equal to zero at about x =   -1.15 and 1.15.  These are the
 square roots of 4/3 which you can find by setting the derivative equa
l to zero.  " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 127 "The command diff
(f(x), x) returns an expression describing the derivative of f(x) with
 respect to the independent variable x.  " }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 25 "h := x -> x^2 + 5*cos(x);" }{TEXT -1 0 "" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"hGf*6#%\"xG6\"6$%)operatorG%&arrowGF(,&*
$)9$\"\"#\"\"\"F1*&\"\"&F1-%$cosG6#F/F1F1F(F(F(" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 14 "diff(h(x), x);" }{TEXT -1 0 "" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#,&*&\"\"#\"\"\"%\"xGF&F&*&\"\"&F&-%$sinG6#F'F&!\"
\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 104 "Notice this returns an exp
ression for the derivative. In order to enter this as a function we mu
st type:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "derivh := x -> \+
diff(h(x), x);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'de
rivhGf*6#%\"xG6\"6$%)operatorG%&arrowGF(-%%diffG6$-%\"hG6#9$F2F(F(F(" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 91 "The big difference between diff
 and D is that you cannot evaluate the \"diff\" at a number.  " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "derivh(3);" }}{PARA 8 "" 1 "
" {TEXT -1 73 "Error, (in derivh) wrong number (or type) of parameters
 in function diff\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "but you ca
n still plot it. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot([
h(x), derivh(x)], x = -2*Pi .. 2*Pi, color = [red, blue]);" }{TEXT -1 
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9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "Again we have the original functi
on in red and the derivative in blue.  Is the slope of h(x) ever negat
ive? " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 143 "For now, the Major diff
erence between D(f) and diff(f(x), x) is  that the D(f) results in a f
unction and diff(f(x),x) results in an expression." }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 0 "" }{TEXT 258 6 "Aside:" }{TEXT -1 112 "  The benef
it of the \"diff\" command is that we may describe functions with more
 than one variable such as y = a " }{XPPEDIT 18 0 "t^2;" "6#*$%\"tG\"
\"#" }{TEXT -1 5 " + b " }{TEXT 260 1 "t" }{TEXT -1 65 " + c and take \+
the derivative with respect to a specific variable:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 37 "quad := (a,b,c,t) -> a*t^2 + b*t + c;" }
{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%quadGf*6&%\"aG%\"bG
%\"cG%\"tG6\"6$%)operatorG%&arrowGF+,(*&9$\"\"\")9'\"\"#F2F2*&9%F2F4F2
F29&F2F+F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "Notice this is ju
st a quadratic in " }{TEXT 261 1 "t" }{TEXT -1 98 ", but also contains
 the variables a, b, c.  We would expect to differentiate this with re
spect to " }{TEXT 262 1 "t" }{TEXT -1 3 " by" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 23 "diff(quad(a,b,c,t), t);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,&*(\"\"#\"\"\"%\"aGF&%\"tGF&F&%\"bGF&" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 100 "This gives expected derivative.  Suppose
 we wanted to differentiate this function with respect to a:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "diff(quad(a,b,c,t), a);" }
{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*$)%\"tG\"\"#\"\"\"" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 133 "This may seem strange but if \"
a\" is considered the independent variable then the derivative with re
spect to \"a\" gives the \"constant\" " }{XPPEDIT 18 0 "t^2;" "6#*$%\"
tG\"\"#" }{TEXT -1 92 ".  We will deal with differentiating with respe
ct to more than one variable in the future.  " }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 0 "" }{TEXT 263 9 "Summary: " }{TEXT -1 202 "The \"D\" com
mand is better when using the derivative as a function because you can
 evaluate the derivative at a number.  The \"diff\" command works well
 if you have more than one variable in the function. " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "# end of section" }}}}{SECT 1 
{PARA 3 "" 0 "" {TEXT -1 10 "Assignment" }}{PARA 0 "" 0 "" {TEXT -1 
203 "Remember: Course, Your Name, and Lab number in upper left.  You m
ay put all of this in one line if you want to save vertical space.  Th
ese two graphs and corresponding written portions must be put on to " 
}{TEXT 268 3 "one" }{TEXT -1 15 " printed page. " }}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 0 "" }{TEXT 256 11 "Problem # 1" }{TEXT -1 2 ": " }}
{PARA 0 "" 0 "" {TEXT -1 29 "Consider the function f(x) = " }{XPPEDIT 
18 0 "x/10;" "6#*&%\"xG\"\"\"\"#5!\"\"" }{TEXT -1 3 " ( " }{XPPEDIT 
18 0 "x^2/3;" "6#*&%\"xG\"\"#\"\"$!\"\"" }{TEXT -1 10 " - 4.1  x)" }}
{PARA 0 "" 0 "" {TEXT -1 4 "a)  " }{TEXT 264 6 "Graph:" }{TEXT -1 401 
" Plot the function and the tangent to the curve at x=6. Plot the seca
nt lines through (2,f(2)) and (2+h,f(2+h)) for h=3 and h = 2.  Put all
 this (function, tangent, 2 secants) on one graph plotted over the dom
ain [-2,12]. The secant lines should appear to be converging to the ta
ngent line. Label the curve and label the lines as secant or tangent. \+
This labelling is probably best done by hand.       " }}{PARA 0 "" 0 "
" {TEXT -1 4 "b)  " }{TEXT 265 10 "Questions:" }{TEXT -1 107 " For wha
t value(s) of x is the tangent to the curve y = f(x) a horizontal line
. Give these values exactly. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "
" }{TEXT 257 11 "Problem # 2" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 3 "a) " }{TEXT 266 6 "Graph:" }{TEXT -1 33 " Plot the functions cos
(x)  and  " }{XPPEDIT 18 0 "(sin(x+h)-sin(x))/h;" "6#*&,&-%$sinG6#,&%
\"xG\"\"\"%\"hGF*F*-F&6#F)!\"\"F*F+F." }{TEXT -1 139 "  for values of \+
 h = 1, 0.5, and 0.2. Put these on one graph plotted over one period. \+
Label the curves by hand indicating the value of h.  " }}{PARA 0 "" 0 
"" {TEXT -1 3 "b) " }{TEXT 267 10 "Question: " }{TEXT -1 60 "What do t
hese graphs suggest about the derivative of sin(x)?" }}}}}{MARK "4" 0 
}{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }