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Notebook[{
Cell[CellGroupData[{
Cell["\<\
Computing with Your Own
Rules for Differentiation\
\>", "Title",
Evaluatable->False,
AspectRatioFixed->True],
Cell["\<\
by
K. D. Stroyan
University of Iowa\
\>", "Subsubtitle",
Evaluatable->False,
AspectRatioFixed->True],
Cell["\<\
copyright 1997 by Academic Press, Inc. - All rights reserved.\
\>", "Text",
Evaluatable->False,
AspectRatioFixed->True,
FontFamily->"Times"],
Cell[CellGroupData[{
Cell[" Notebook Overview", "Section",
Evaluatable->False,
AspectRatioFixed->True],
Cell["\<\
Mathematica has a built in function called D[.] that will perform all the \
symbolic differentiations of calculus. We explain the use of D[.] in the \
Chapter 7 NoteBook DfDx. Before we use D[.], however, we will develop our \
own function \"diff\" to perform symbolic differentiation. In doing this you \
will be able to see how powerful each of these rules is. You will see what \
the various rules can do and what they can not do, by seeing the effect of \
adding new rules.\
\>", "Text",
Evaluatable->False],
Cell["\<\
We will be able to perform all the symbolic differentiations in basic \
calculus by writing only 16 general symbolic function rules in our program. \
Our \"diff\" program is built up in steps that correspond to the rules in \
Chapter 6 of the Main Text. As you cells, you are adding more rules \
to \"diff,\" just as you yourself are developing more powerful \
differentiation procedures in the corresponding sections of the text. For \
example, formulas that cannot be differentiated without the product rule also \
cannot be found with \"diff\" before you that section of the program. \
They can be computed with \"diff\" after you it. There are problems \
just like this in the text for you to compute or say why the rules are not \
powerful enough to do so.\
\>", "Text",
Evaluatable->False],
Cell[CellGroupData[{
Cell["References to the Text", "Subsection",
Evaluatable->False],
Cell["This NoteBook goes with Chapter 6 of the text.", "Text",
Evaluatable->False]
}, Closed]]
}, Closed]],
Cell[CellGroupData[{
Cell["Step-by-step", "Section",
Evaluatable->False,
AspectRatioFixed->True],
Cell[CellGroupData[{
Cell["Power Rule", "Subsection",
Evaluatable->False],
Cell["\<\
The first rules we define allow Mathematica to differentiate powers. The \
first rule below almost does this, but there are computer complications. \
Note the underscores after the expressions supplied to the diff function. \
The . after the n tells Mathematica that there doesn't need to be a number n \
given to the diff function. This provides for the case when we are \
differentiating x^1, which we usually just write as x. The second rule says \
the derivative of a constant c is zero, provided c doesn't depend on x. (See \
Mathematica's manual for more explanation of the peculiar \"?NumberQ\" \
command. Essentially, it asks if c is a number and not a function of x.)\
\>", "Text"],
Cell[BoxData[{
\(\(diff[x_\^n_. , x_] := n\ x\^\(n - 1\); \)\),
\(\(diff[c_?NumberQ, x_] := 0; \)\)}], "Input"],
Cell[TextData[{
"Once you have Entered the rules above you can differentiate powers. Enter \
the following function in the next cell: ",
Cell[BoxData[
\(x\^\(1/5\)\)]],
" . ( it now.)"
}], "Text"],
Cell[BoxData[
\(\(diff[?? ,x]\)\)], "Input"],
Cell[TextData[{
"Since the square root function can be expressed as a power, we can use the \
rules above to differentiate that as well. See this by typing ",
Cell[BoxData[
\(\@x\)]],
" in the next cell and entering."
}], "Text"],
Cell[BoxData[
\(\(diff[?? ,x]\)\)], "Input"],
Cell["\<\
The function f[x] = 1/x can also be expressed as a power. Use diff[.] to \
differentiate it.\
\>", "Text"],
Cell["", "Input"],
Cell[TextData[
"One odd thing about Mathematica is that it does not consider \[Pi] and \
(Euler's) \[ExponentialE] (= E \[TildeTilde] 2.71828...) to be `numbers.' (\
\[Pi] and \[ExponentialE] have many symbolic uses in Mathematica.) Thus the \
rule for numbers above will not apply to \[Pi] or \[ExponentialE]. We must \
explicitly define the derivates of \[Pi] and \[ExponentialE] to be 0 as \
follows. the next rule now, so our power rule is complete."], "Text"],
Cell[BoxData[{
\(\(diff[\[Pi], x_] := 0; \)\),
\(\(diff[E, x_] := 0; \)\)}], "Input"],
Cell[CellGroupData[{
Cell["Practice Exercise", "Subsubsection"],
Cell[TextData[{
"Use ",
StyleBox[" ",
FontSlant->"Italic"],
"diff[.] to differentiate the functions "
}], "Text"],
Cell[TextData[{
"\t",
Cell[BoxData[
\(y = \(f[x] = 1\/x\^2\)\)]]
}], "Text"],
Cell[TextData["\ty=f[x]=\[ExponentialE]"], "Text"],
Cell[TextData[{
"\t",
Cell[BoxData[
\(y = \(f[x] = x\^\[Pi]\)\)]]
}], "Text"],
Cell[BoxData[""], "Input"]
}, Closed]]
}, Closed]],
Cell[CellGroupData[{
Cell["Other Variables", "Subsection",
Evaluatable->False],
Cell["\<\
The variable name \"x\" in our program for \"diff\" can be changed to any \
convenient variable. For example,\
\>", "Text"],
Cell[BoxData[
\(V\ = \ r^3; \ndiff[V, r]\)], "Input"],
Cell[CellGroupData[{
Cell["Powers vs Sums and Multiples", "Subsubsection"],
Cell["\<\
Notice that Mathematica is still unable to differentiate some simple \
expressions. It still doesn't know that the derivative of a sum is the sum \
of the derivatives, because we haven't told it the superposition rule yet. \
Enter the next computation to see what happens when Mathematica doesn't know \
how to compute an answer.\
\>", "Text"],
Cell[BoxData[{
\(diff[3\ x + 2, x]\),
\(diff[\(\(4\ \[Pi]\)\/3\) r\^3, r]\),
\(diff[\[ExponentialE]\ x\^\[Pi], x]\)}], "Input"],
Cell["\<\
You know the superposition rule for differentiation, so what are the \
derivatives of the last two functions? How do you apply the power rule and \
superposition to get YOUR answer?\
\>", "Text"],
Cell["Ans.:", "Text"]
}, Closed]]
}, Closed]],
Cell[CellGroupData[{
Cell["Sine and Cosine Rules", "Subsection",
Evaluatable->False],
Cell["\<\
The rules below will enable us to differentiate Sines and Cosines. \
them now.\
\>", "Text"],
Cell[BoxData[{
\(\(diff[Sin[x_], x_] := Cos[x]; \)\),
\(\(diff[Cos[x_], x_] := \(-Sin[x]\); \)\)}], "Input"],
Cell[CellGroupData[{
Cell["Practice Exercise", "Subsubsection"],
Cell["\<\
After ing the sine and cosine rules, check them on a new variable by \
differentiating\
\>", "Text"],
Cell[TextData["\ty = f[ \[Theta] ] = Sin[ \[Theta] ]"], "Text"],
Cell["and", "Text"],
Cell[TextData["\ty = f[ \[Omega] ] = Cos[ \[Omega] ]"], "Text"],
Cell["\<\
\
\>", "Input"]
}, Closed]]
}, Closed]],
Cell[CellGroupData[{
Cell["Log and Exponential Rules", "Subsection",
Evaluatable->False],
Cell["\<\
The rules below enable us to differentiate Logs and Exponentials. \
them.\
\>", "Text"],
Cell[BoxData[{
\(\(diff[Exp[x_], x_] := Exp[x]; \)\),
\(\(diff[Log[x_], x_] := 1\/x; \)\)}], "Input"],
Cell[CellGroupData[{
Cell["Practice Exercise", "Subsubsection"],
Cell["\<\
After ing the sine and cosine rules, check them on a new variable by \
differentiating\
\>", "Text"],
Cell["\tx = f[ y ] = Log[ y ]", "Text"],
Cell["and", "Text"],
Cell[TextData[{
"\t",
Cell[BoxData[
\(y = \(f[x] = \[ExponentialE]\^x\)\)]]
}], "Text"],
Cell[BoxData[""], "Input"]
}, Closed]]
}, Closed]],
Cell[CellGroupData[{
Cell["Superposition Rule", "Subsection",
Evaluatable->False],
Cell["\<\
Now we define rules that will differentiate linear combinations of functions. \
We break this into two parts. Given a function times a constant a, \"diff\" \
is defined simply as the constant times \"diff\" of the function. Next, we \
define \"diff\" of a sum as the sum of the \"diff\"s. Notice that g itself \
can be any function, even another sum of functions. If g itself is a sum, \
Mathematica will simply apply the rule again.\
\>", "Text"],
Cell[" these rules now and continue. ", "Text"],
Cell[BoxData[{
\(\(diff[a_?NumberQ\ f_, x_] := a\ diff[f, x]; \)\),
\(\(diff[f_ + g_, x_] := diff[f, x] + diff[g, x]; \)\)}], "Input"],
Cell["\<\
Now Mathematica is able to differentiate the simple sum we gave it above: 3 x \
+ 2 . it..\
\>", "Text"],
Cell[BoxData[
\(\(diff[\ ?? ??????,\ x\ ]\)\)], "Input"],
Cell["\<\
To see how powerful these rules are, enter the following function and \
the cell. \
\>", "Text"],
Cell[TextData[{
"\t",
Cell[BoxData[
\(y = Cos[x] + 7\ Sin[x] - 0.5\ \@x + \(3\ Log[x]\)\/4\)]]
}], "Text"],
Cell[BoxData[
\(y = \(?? ?;\ndiff[y, x]\)\)], "Input"],
Cell["\<\
Notice, however, there are still many expressions Mathematica still can't \
handle. Try a product\
\>", "Text"],
Cell["diff[ x Sin[x], x ]", "Input"],
Cell[CellGroupData[{
Cell["Practice Exercise", "Subsubsection"],
Cell["\<\
1) Insert cells and show that diff[.] can compute rather complicated \
expressions that you could do using only the rules above here in this \
NoteBook, but would just as soon let the computer do for you.
2) Give an example of your own construction of a function y = f[x] that \
YOU CAN differentiate, but that diff[.] can not differentiate with only the \
rules above here in this NoteBook.\
\>", "Text"]
}, Closed]]
}, Closed]],
Cell[CellGroupData[{
Cell["The Product Rule", "Subsection",
Evaluatable->False],
Cell["Next we will define the product rule for differentiation.", "Text"],
Cell[BoxData[
\(\(diff[f_\ g_, x_] := diff[f, x]\ g + f\ diff[g, x]; \)\)], "Input"],
Cell["\<\
Now we can differentiate the product given above: x Sin[x] . Do it:\
\>", "Text"],
Cell[BoxData[
\(\(diff[\ ?? ???,\ x\ ]\)\)], "Input"],
Cell[TextData[{
"Again to emphasize how powerful this rule is, note that g above itself can \
be a product of two functions. If it is, ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" will simply use the same rule again. Enter the following function in the \
appropriate place to see: 3 Sin[x] (Cos[x] Exp[x])"
}], "Text"],
Cell[BoxData[
\(\(diff[\ ?? ????????,\ x\ ]\)\)], "Input"],
Cell["\<\
On the other hand, diff[.] does not yet have a rule to differentiate the \
composition of two functions. Not even very simple compositions. diff[.] \
can differentiate 3 x, it:\
\>", "Text"],
Cell[BoxData[
\(diff[3\ x, x]\)], "Input"],
Cell["diff[.] can differentiate Sin[u], it:", "Text"],
Cell[BoxData[
\(diff[Sin[u], u]\)], "Input"],
Cell["\<\
But diff[.] can not differentiate the simple composition (knowing only the \
rules above here)\
\>", "Text"],
Cell["\ty = Sin[ 3 x ]", "Text"],
Cell["See for yourself", "Text"],
Cell["diff[ Sin[ 3 x],x]", "Input"],
Cell[CellGroupData[{
Cell["Practice Exercise", "Subsubsection"],
Cell["\<\
1) Insert cells and show that diff[.] can compute rather complicated \
expressions that you could do with only the rules so far, but would just as \
soon let the computer do for you.
2) Give an example of your own construction of a function y = f[x] that \
YOU CAN differentiate, but that diff[.] can not differentiate with only the \
rules above here in this NoteBook.\
\>", "Text"]
}, Closed]]
}, Closed]],
Cell[CellGroupData[{
Cell["The Chain Rule", "Subsection",
Evaluatable->False],
Cell["\<\
The last rule we need is the Chain Rule. Doing this is a little more \
complicated, but if we remember what the chain rule says, the rules below \
will be clear.\
\>", "Text"],
Cell["Suppose we wanted to differentiate the following function:", "Text"],
Cell[TextData[{
"\t",
Cell[BoxData[
\(y = \((x\^2 + 2\ x - 1)\)\^3\)]]
}], "Text"],
Cell["\<\
To use the chain rule we would first differentiate using the power rule and \
then multiply by the derivative of the function x^2 + 2 x - 1 to get:\
\>", "Text"],
Cell[TextData[{
"\t",
Cell[BoxData[
\(dy\/dx = 3\ \((x\^2 + 2\ x - 1)\)\^2\ \((2\ x + 2)\)\)]]
}], "Text"],
Cell["\<\
This suggests the following rule for differentiation.\t it.\
\>", "Text"],
Cell[BoxData[
\(\(diff[f_\^n_, x_] := n\ f\^\(n - 1\)\ diff[f, x]; \)\)], "Input"],
Cell["\<\
This is the Chain Rule applied to a function to a power. Test it with the \
cell below.\
\>", "Text"],
Cell[BoxData[
\(diff[\((x\^2 + 2\ x - 1)\)\^3, x]\)], "Input"],
Cell["\<\
Once the form of the above rule is clear, it should be clear how to finish \
defining the chain rule. We simply modify the rules above for special \
functions of x to apply to functions of functions of x. Below are the Sine, \
Cosine, Log, and Exponential Rules redefined for arbitrary functions f.\
\>", "Text"],
Cell[BoxData[{
\(\(diff[Sin[f_], x_] := Cos[f]\ diff[f, x]; \)\),
\(\(diff[Cos[f_], x_] := \(-Sin[f]\)\ diff[f, x]; \)\),
\(\(diff[Exp[f_], x_] := Exp[f]\ diff[f, x]; \)\),
\(\(diff[Log[f_], x_] := diff[f, x]\/f; \)\)}], "Input"],
Cell["\tTest the rules above on this expression: ", "Text"],
Cell[TextData[{
Cell[BoxData[
\(\(Exp[2\ x]\ Sin[3\ x]\)\/Log[x]\)]],
"= ",
Cell[BoxData[
\(Exp[2\ x]\ Sin[3\ x] \((Log[x])\)\^\(-1\)\)]],
" "
}], "Text"],
Cell[BoxData[
\(\(diff[?? ,x]\)\)], "Input"],
Cell["\<\
The Superposition Rule and Product Rule remain unaltered and Mathematica can \
use them all. The function \"diff\" can now do the quotient rule as follows.\
\
\>", "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell["The Quotient Rule", "Subsection",
Evaluatable->False],
Cell["\<\
There is not much need to use the Quotient Rule once we have the above rules \
working properly. As another illustration of how powerful the above rules \
are, however, we will use the above rules to obtain the Quotient Rule in \
explicit form.\
\>", "Text"],
Cell[BoxData[
\(diff[f\/g, x]\)], "Input"],
Cell["\<\
This looks a little odd, but if we use the Together command we will see the \
Quotient Rule in the text.\
\>", "Text"],
Cell[BoxData[
\(Together[diff[f\/g, x]]\)], "Input"]
}, Closed]],
Cell[CellGroupData[{
Cell["The Tangent Rule", "Subsection",
Evaluatable->False],
Cell[TextData[{
"You know that Tan[x] = Sin[x]/Cos[x] and ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" knows that, too. Unfortunately, it knows it too well. Try to compute \
the derivative of tangent using the diff[] rules so far, ing:"
}], "Text"],
Cell[BoxData[
\(Sin[x]\/Cos[x]\)], "Input"],
Cell[BoxData[
\(diff[Sin[x]\/Cos[x], x]\)], "Input"],
Cell[CellGroupData[{
Cell["Exercise", "Subsubsection",
Evaluatable->False],
Cell[TextData[{
"Use the quotient rule to find the derivative of tangent and then expand \
diff[.] to make a new rule like the rules above for Sin[x] or Log[x] and \
their associated chain rules as above. (In Chapter 5 we also showed by direct \
geometric estimates that the derivative of Tangent is Secant squared. Note: \
It is also possible to tell ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" to turn the trig simplification rules off. If you do this, the \
expression Sin[x] * (Cos[x]^(-1) will be reduced by diff[.] to a derivable \
expression.)"
}], "Text",
Evaluatable->False],
Cell["", "Input"]
}, Closed]]
}, Closed]]
}, Closed]],
Cell[CellGroupData[{
Cell["Combined Computation", "Section",
Evaluatable->False,
AspectRatioFixed->True],
Cell["\<\
With all the cells above ed, you have written a general symbolic \
differentiation function that can do all the exercises that you are supposed \
to be able to do by the end of Chapter 6 on Symbolic Differentiation. \
(Mathematica's built-in D[.] also knows derivatives for non-elementary \
functions like Bessel functions.)\
\>", "Text",
Evaluatable->False]
}, Closed]],
Cell[CellGroupData[{
Cell["Exercises", "Section",
Evaluatable->False,
AspectRatioFixed->True],
Cell[CellGroupData[{
Cell["Exercise 6.2.4", "Subsection",
Evaluatable->False],
Cell[TextData[{
"Run this NoteBook and work through it line by line so that you can see how \
the addition of each of the Superposition Rule, Product Rule and Chain Rule \
makes symbolic differentiation more powerful. For example, we can't \
differentiate ",
Cell[BoxData[
\(\@x\ Cos[x]\)]],
" without the Product Rule and neither can the computer."
}], "Text",
Evaluatable->False]
}, Closed]],
Cell[CellGroupData[{
Cell["Silly Fun", "Subsection",
Evaluatable->False],
Cell["\<\
You could also use the ideas above defining diff[.] to give bogus \"rules\" \
for differentiation. You could define diff[ Tan[x],x] to be Logarithm, or \
diff[ f + g, x ] to be f * g. It's a little silly, but it might be fun to \
see what the computer does with some completely arbitrary meaningless \
rules.\
\>", "Text",
Evaluatable->False]
}, Closed]]
}, Closed]]
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*)
(***********************************************************************
End of Mathematica Notebook file.
***********************************************************************)