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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "Lab 12 Maple Activities fo
r October 17:  (Lesson 9 - The Product and Quotient Rules)" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 0 "" }{TEXT 
257 8 "Examples" }{TEXT 258 0 "" }{TEXT -1 111 "  Use Maple to differe
ntiate the following functions from the lecture notes for the Product \+
and Quotient Rules." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 4 "1.  " }{TEXT 259 1 "f" }{TEXT -1 1 "(" }{TEXT 
260 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "(6*x+5)*(2*x-1);" "6#*&,&
*&\"\"'\"\"\"%\"xGF'F'\"\"&F'F',&*&\"\"#F'F(F'F'\"\"\"!\"\"F'" }{TEXT 
-1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 14 "    Using the " }{TEXT 261 4 "diff" }{TEXT -1 9 " command
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "diff((6*x + 5)*(2*x - \+
1),x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 262 5 "NOTE:" }{TEXT -1 381 "  Maple has already \+
simplified the answer for this problem.  This is probably because each
 factor is a linear factor.  That is, each factor is a polynomial of d
egree one.  We will see in our next examples that Maple does not simpl
ify the answer unless we asked it to simplify the answer.  It appears \+
that Maple will only simply the answer if both factors in the product \+
are linear." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 8 "2.  y = " }{XPPEDIT 18 0 "(x^2+x)*(2-3*x^3);" "6#*&,&*$%
\"xG\"\"#\"\"\"F&F(F(,&\"\"#F(*&\"\"$F(*$F&\"\"$F(!\"\"F(" }{TEXT -1 
1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 14 "    Using the " }{TEXT 263 1 "D" }{TEXT -1 9 " command:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "y := x -> (x^2+x)*(2-3*x^3);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "D(y)(x); simplify(%);" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }{TEXT 264 5 "NOTE:" }{TEXT -1 475 "  Maple applied the Product Rule
 to differentiate the function.  Which factor did it differentiate fir
st?  Notice that Maple did not simplify the answer for this function a
s it had in the first example above.  Both factors are not linear here
.  The first factor is quadratic; that is, a polynomial of degree two.
  And the second factor is cubic; that is, a polynomial of degree thre
e.  Unfortunately, Maple did not give the simplified answer in descend
ing or ascending order." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 4 "3.  " }{TEXT 265 1 "s" }{TEXT -1 1 "(" }
{TEXT 266 1 "t" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "(2*t^4-t^2+3*t+4)*(
5*t^2-8*t);" "6#*&,**&\"\"#\"\"\"*$%\"tG\"\"%F'F'*$F)\"\"#!\"\"*&\"\"$
F'F)F'F'\"\"%F'F',&*&\"\"&F'*$F)\"\"#F'F'*&\"\")F'F)F'F-F'" }{TEXT -1 
1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 14 "    Using the " }{TEXT 267 1 "D" }{TEXT -1 9 " command:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "s := t -> (2*t^4 - t^2 + 3*t
 + 4)*(5*t^2 - 8*t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "D(s
)(t); simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 8 "4.  y = " }{XPPEDIT 18 0 "(8-2*u-3*u^2)*(2*u-4*u^5
+3*u^9);" "6#*&,(\"\")\"\"\"%\"uG!\"#*$)F'\"\"#\"\"\"!\"$F&,(F'F+*$)F'
\"\"&F,!\"%*$)F'\"\"*F,\"\"$F&" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "    Using the " }
{TEXT 268 4 "diff" }{TEXT -1 9 " command:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 62 "diff((8 - 2*u - 3*u^2)*(2*u - 4*u^5 + 3*u^9), u); s
implify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 4 "5.  " }{TEXT 269 1 "f" }{TEXT -1 1 "(" }{TEXT 271 1 "x" 
}{TEXT -1 4 ") = " }{XPPEDIT 18 0 "4*x^2*sin(x);" "6#,$*&)%\"xG\"\"#\"
\"\"-%$sinG6#F&\"\"\"\"\"%" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "    Using the " }{TEXT 
272 4 "diff" }{TEXT -1 9 " command:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "diff(4*x^2*sin(x),x); simplify(%);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 273 0 "
" }{TEXT 274 7 "Example" }{TEXT 275 0 "" }{TEXT -1 5 "  If " }{TEXT 
276 1 "f" }{TEXT -1 1 "(" }{TEXT 277 1 "x" }{TEXT -1 4 ") = " }
{XPPEDIT 18 0 "(x^3-3*x^2+4*x-5)*(4*x^2+5*x-10);" "6#*&,**$)%\"xG\"\"$
\"\"\"\"\"\"*$)F'\"\"#F)!\"$F'\"\"%!\"&F*F*,(F+F/F'\"\"&!#5F*F*" }
{TEXT -1 12 ", then find " }{TEXT 278 1 "f" }{TEXT -1 7 " '(2). " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "  \+
  Using the " }{TEXT 280 1 "D" }{TEXT -1 9 " command:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "f := x -> (x^3 - 3*x^2 + 4*x - 5)*(
4*x^2 + 5*x - 10);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "D(f)(2
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 14 "    Using the " }{TEXT 281 4 "diff" }{TEXT -1 9 " command:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "diff((x^3 - 3*x^2 + 4*x - 5)
*(4*x^2 + 5*x - 10), x); subs(x=2, %);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 0 "" }{TEXT 
257 8 "Examples" }{TEXT 258 0 "" }{TEXT -1 111 "  Use Maple to differe
ntiate the following functions from the lecture notes for the Product \+
and Quotient Rules." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 4 "1.  " }{TEXT 285 1 "f" }{TEXT -1 1 "(" }{TEXT 
286 1 "x" }{TEXT -1 6 ")  =  " }{XPPEDIT 18 0 "2*x/(x-4);" "6#,$*&%\"x
G\"\"\",&F%\"\"\"!\"%F(!\"\"\"\"#" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "    Using the " }
{TEXT 287 1 "D" }{TEXT -1 9 " command:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "f := x -> (2*x)/(x-4);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "D(f)(x); simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 288 5 "NOTE:" }
{TEXT -1 366 "  Maple gives some strange answers when fractions are in
volved.  You will have to do a little of work with the Maple answer to
 realize that it is the same answer that you get when you apply the Qu
otient Rule.  Or, you will have to do a little work with your answer o
btained from using the Quotient Rule to realize that it is the same an
swer that Maple is giving you." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 4 "2.  " }{TEXT 289 1 "y" }{TEXT -1 8 "  \+
=  3/(" }{XPPEDIT 18 0 "x^2+5*x;" "6#,&*$)%\"xG\"\"#\"\"\"F(*&\"\"&F(F
&F(F(" }{TEXT -1 1 ")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 13 "   Using the " }{TEXT 290 4 "diff" }
{TEXT -1 9 " command:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "di
ff(3/(x^2 + 5*x), x); simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "3.  " }{TEXT 291 1 "g" }{TEXT 
-1 1 "(" }{TEXT 292 1 "w" }{TEXT -1 6 ")  =  " }{XPPEDIT 18 0 "(2*w^3-
3*w)/(1-w^4);" "6#*&,&*$)%\"wG\"\"$\"\"\"\"\"#F'!\"$F),&\"\"\"F-*$)F'
\"\"%F)!\"\"!\"\"" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "    Using the " }{TEXT 297 1 "D" }
{TEXT -1 9 " command:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "g \+
:= w -> (2*w^3 - 3*w)/(1 - w^4);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "D(g)(w); simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 4 "4.  " }{TEXT 294 1 "s" }{TEXT -1 1 "(" }{TEXT 295 1 "t" }
{TEXT -1 5 ") =  " }{XPPEDIT 18 0 "(4-t-t^2)/(8*t^3-2*t+5);" "6#*&,(\"
\"%\"\"\"%\"tG!\"\"*$)F'\"\"#\"\"\"F(F,,(*$)F'\"\"$F,\"\")F'!\"#\"\"&F
&!\"\"" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 14 "    Using the " }{TEXT 296 1 "D" }{TEXT 
-1 9 " command:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "s := t -
> (4 - t - t^2)/(8*t^3 - 2*t +5);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "D(s)(t); simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 298 0 "" }{TEXT 
299 7 "Example" }{TEXT 300 0 "" }{TEXT -1 11 "  Find the " }{TEXT 301 
1 "x" }{TEXT -1 54 "-coordinate of the point(s) on the graph of functi
on  " }{TEXT 302 1 "y" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "(2*x-5)/(x^2-
4);" "6#*&,&%\"xG\"\"#!\"&\"\"\"\"\"\",&*$)F%F&F)F(!\"%F(!\"\"" }
{TEXT -1 39 "  where the tangent line is horizontal." }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "From lecture, w
e know that we want to solve the equation " }{TEXT 303 1 "y" }{TEXT 
-1 13 " ' = 0 since " }{TEXT 304 1 "y" }{TEXT -1 65 " ' gives the slop
e of tangent lines on the graph of the function " }{TEXT 312 1 "y" }
{TEXT -1 0 "" }{TEXT -1 41 " and the slope of a horizontal line is 0.
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
14 "    Using the " }{TEXT 305 1 "D" }{TEXT -1 9 " command:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "y := x -> (2*x - 5)/(x^2 - 4
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "solve( D(y)(x)=0 );" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 
"    Using the " }{TEXT 306 4 "diff" }{TEXT -1 9 " command:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "solve( diff((2*x - 5)/(x^2 -
 4), x) = 0 );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 4 "    " }{TEXT 307 7 "Answer " }{TEXT 308 2 "  " }
{TEXT 309 1 "x" }{TEXT 310 9 " = 1, 4  " }{TEXT 311 0 "" }{TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "1 0 0" 8 }
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