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Notebook[{

Cell[CellGroupData[{
Cell["Differential Calculus", "Title",
  Evaluatable->False,
  TextAlignment->Center,
  AspectRatioFixed->True],

Cell["Simplified Version", "Subsubtitle",
  TextAlignment->Center,
  TextJustification->0],

Cell[TextData[{
  "Sean Mauch\nsean@caltech.edu\n",
  ButtonBox["http://www.its.caltech.edu/~sean",
    ButtonData:>{
      URL[ "http://www.its.caltech.edu/~sean"], None},
    ButtonStyle->"Hyperlink"],
  "\n",
  "This work is distributed under the GNU FDL.  See ",
  ButtonBox["license.nb ",
    ButtonData:>{"license.nb", None},
    ButtonStyle->"Hyperlink"],
  "for details."
}], "Text",
  Evaluatable->False,
  AspectRatioFixed->True],

Cell[CellGroupData[{

Cell["Limits", "Section",
  Evaluatable->False,
  AspectRatioFixed->True],

Cell["\<\
To evaluate an expression at a certain value of a formal variable, \
we use the substitution operator followed by a rule.\
\>", "Text",
  Evaluatable->False,
  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(Sin[x] /. x \[Rule] 0\)], "Input",
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Cell[TextData[{
  "The syntax of rules in ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " is a little misleading.  The following looks very close to the \
mathematical notation for the limit as ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " goes to zero.  However, the following is only a substitution."
}], "Text",
  Evaluatable->False,
  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(Sin[x]\/x /. x \[Rule] 0\)], "Input",
  AspectRatioFixed->True],

Cell[TextData[{
  "In order to evaluate an expression in the limit as a formal variable \
approached a limit, we must use the ",
  StyleBox["Limit[]", "Input"],
  " function."
}], "Text",
  Evaluatable->False,
  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(Limit[Sin[x]\/x, x \[Rule] 0]\)], "Input",
  AspectRatioFixed->True],

Cell["\<\
You can evaluate a limit as a formal variables goes to positive \
infinity.\
\>", "Text",
  Evaluatable->False,
  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(Limit[x\^2\ Exp[\(-x\)], x \[Rule] \[Infinity]]\)], "Input",
  AspectRatioFixed->True],

Cell["Mathematica will also tell you if a limit diverges.", "Text",
  Evaluatable->False,
  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(Limit[\(x + Sin[x]\)\/x\^2, x \[Rule] 0]\)], "Input",
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Cell[TextData[{
  "If the limit does not exists, ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " will tell you if the fluctuation of the expression is bounded in an \
interval."
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  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(Limit[Sin[x], x \[Rule] \[Infinity]]\)], "Input",
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Cell["\<\
It will also tell you if the expression fluctuates as the limit \
diverges.\
\>", "Text",
  Evaluatable->False,
  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(Limit[x\ Sin[x], x \[Rule] \[Infinity]]\)], "Input",
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Cell[TextData[{
  "By default ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " evaluates limits from above, (i.e. it approaches the specified value of \
the formal variable from the right), except for limits at \[Infinity] which \
it approaches from below.  You can change this default with the ",
  StyleBox["Direction", "Input"],
  " option."
}], "Text",
  Evaluatable->False,
  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[TextData[{
  "This is the default direction to evaluate the limit.  ",
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  " means to evaluate the limit from above."
}], "Text",
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  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(Limit[Exp[1\/x], x \[Rule] 0, Direction \[Rule] \(-1\)]\)], "Input",
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Cell["Here we evaluate the limit from below.", "Text",
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  TextAlignment->Left,
  TextJustification->1,
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Cell[BoxData[
    \(Limit[Exp[1\/x], x \[Rule] 0, Direction \[Rule] 1]\)], "Input",
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Cell[CellGroupData[{

Cell["Exercise 1", "Subsubsection"],

Cell["Evaluate the following limits.", "Text"],

Cell[BoxData[
    \(TraditionalForm\`lim\+\(x \[Rule] \[Infinity]\)\((1 + 1\/x)\)\^x, \ \ \ \
\ lim\+\(x \[Rule] 0\)\(a\ x\^2 + b\ x + c\)\/\(d\ x\^2 + e\ x + f\), \ \ \ \ \
lim\+\(x \[Rule] \[Infinity]\)\(a\ x\^2 + b\ x + c\)\/\(d\ x\^2 + e\ x + \
\(\(f\)\(.\)\)\)\)], "DisplayFormula",
  TextAlignment->Center]
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Cell[CellGroupData[{

Cell["Exercise 2", "Subsubsection"],

Cell[TextData[{
  "Check that ",
  Cell[BoxData[
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      \`\(\[DifferentialD]\/\[DifferentialD]x\) x\^n = n\ x\^\(n - 1\)\)]],
  " using the definition of the derivative,"
}], "Text"],

Cell[BoxData[
    \(TraditionalForm\`\(\[DifferentialD]\/\[DifferentialD]x\) \(f(
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f(x)\)\/\(\[CapitalDelta]\ \(\(x\)\(.\)\)\)\)], "DisplayFormula",
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Cell[CellGroupData[{

Cell[TextData["Partial Derivatives"], "Section",
  Evaluatable->False,
  AspectRatioFixed->True],

Cell[BoxData[
    \(Clear[x, y, f]\)], "Input"],

Cell[TextData[{
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " can find partial derivatives of symbolic expressions.  If you haven't \
heard the term partial derivative before, it's just a derivative where all \
the formal variables in an expression except the one that your \
differentiating with respect to are assumed to be constant.  The mathematical \
form that you'll see in books for partial derivatives is ",
  Cell[BoxData[
      \(TraditionalForm\`\[PartialD]\ f\/\[PartialD]\ x\)]],
  ".  Partial derivatives in ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " are written:"
}], "Text",
  Evaluatable->False,
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Cell[BoxData[
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Cell[TextData[{
  "To do the typeset version of the above,\n\t \[EscapeKey]pd\[EscapeKey] \
\[ControlKey]\[LeftModified]-\[RightModified] x \[ControlKey]\[LeftModified]\
\[SpaceKey]\[RightModified]\t\tto get   ",
  Cell[BoxData[
      \(TraditionalForm\`\[PartialD]\_x\)]],
  "\nand enclose the function to be differentiated in parentheses if it has \
more than one term."
}], "Text",
  TextAlignment->Left,
  TextJustification->1],

Cell[BoxData[
    \(\[PartialD]\_x\((x\^2 + y\^2 + a\ Sin[x])\)\)], "Input",
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Cell[TextData[{
  "Above, ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " assumed the variables ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`a\)]],
  " were constant.  You can take second and third, etc. derivatives as \
follows:"
}], "Text",
  Evaluatable->False,
  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(D[x\^2 + Sin[x], {x, 2}]\)], "Input",
  AspectRatioFixed->True],

Cell[BoxData[
    \(\[PartialD]\_\(x, x\)\((x\^2 + Sin[x])\)\)], "Input",
  AspectRatioFixed->True],

Cell[BoxData[
    \(\[PartialD]\_{x, 2}\((x\^2 + Sin[x])\)\)], "Input",
  AspectRatioFixed->True],

Cell[BoxData[
    \(D[x\^2 + Sin[x], {x, 3}]\)], "Input",
  AspectRatioFixed->True],

Cell[BoxData[
    \(\[PartialD]\_{x, 3}\((x\^2 + Sin[x])\)\)], "Input",
  AspectRatioFixed->True],

Cell[TextData[{
  " ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " also allows you to take partial derivatives with respect to several \
variables."
}], "Text",
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  TextAlignment->Left,
  TextJustification->1,
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Cell[BoxData[
    \(D[x\^2\ y, x, y]\)], "Input",
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Cell[BoxData[
    \(\[PartialD]\_\(x, y\)\((x\^2\ y)\)\)], "Input",
  AspectRatioFixed->True],

Cell[BoxData[
    \(D[Sin[x\ y], {x, 2}, y]\)], "Input",
  AspectRatioFixed->True],

Cell[BoxData[
    \(\[PartialD]\_\({x, 2}, y\)Sin[x\ y]\)], "Input",
  AspectRatioFixed->True],

Cell[TextData[{
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    FontSlant->"Italic"],
  " can take derivatives of formal functions."
}], "Text",
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  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(\[PartialD]\_x f[x]\)], "Input",
  AspectRatioFixed->True],

Cell[BoxData[
    \(\[PartialD]\_{x, 3}\((f[x]\ g[x])\)\)], "Input",
  AspectRatioFixed->True],

Cell[TextData[{
  "Note that the first and second derivatives are denoted ",
  Cell[BoxData[
      \(TraditionalForm\`\(f'\)[x]\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(f'\)'\)[x]\)]],
  ".  Higher derivatives are denoted ",
  Cell[BoxData[
      \(TraditionalForm\`\(f\^\((n)\)\)[x]\)]],
  "."
}], "Text",
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Cell[TextData[{
  "You can specify that a formal variable depends on ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " by either explicitly writing it as a function of ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " or by using the ",
  StyleBox["NonConstants", "Input"],
  " option."
}], "Text",
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  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(\[PartialD]\_x\((x\^2\ y[x])\)\)], "Input",
  AspectRatioFixed->True],

Cell[BoxData[
    \(D[x\^2\ y, x, NonConstants \[Rule] y]\)], "Input",
  AspectRatioFixed->True],

Cell[BoxData[
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Cell[CellGroupData[{

Cell["Exercise 1", "Subsubsection"],

Cell[TextData[{
  "Verify that ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " knows the rules:"
}], "Text"],

Cell[BoxData[
    FormBox[GridBox[{
          {\(\(\[PartialD]\/\[PartialD]x\) \((u\ v)\) = u'\ v + u\ v', \), 
            \(Product\ rule\)},
          {
            \(\(\[PartialD]\/\[PartialD]x\) \((u\/v)\) = 
              \(u'\ v - u\ v'\)\/v\^2, \), \(Quotient\ rule\)},
          {\(\(\[PartialD]\/\[PartialD]x\) \((u\^a)\) = a\ u\^\(a - 1\)\ u', 
            \), \(Power\ rule\)},
          {\(\(\[PartialD]\/\[PartialD]x\) \((u(v(x)))\) = u' \((v)\)\ v', 
            \), \(Chain\ rule\)}
          }], TraditionalForm]], "DisplayFormula"]
}, Open  ]],

Cell[CellGroupData[{

Cell["Exercise 2", "Subsubsection"],

Cell["Find the local maxima and minima of the polynomial:", "Text"],

Cell[BoxData[
    \(TraditionalForm
    \`\((x - 1)\) \((x - 2)\) \((x - 3)\) \((x - 4)\) \((x - 5)\)\)], 
  "DisplayFormula"]
}, Open  ]],

Cell[CellGroupData[{

Cell["Exercise 3", "Subsubsection"],

Cell[TextData[{
  "Plot the function ",
  Cell[BoxData[
      \(TraditionalForm\`y(x)\)]],
  " defined by,"
}], "Text"],

Cell[BoxData[
    \(TraditionalForm\`x\^2 - x\ y + y\^2 = 3. \)], "DisplayFormula",
  TextAlignment->Center],

Cell[TextData[{
  "Use implicit differentiation to find ",
  Cell[BoxData[
      \(TraditionalForm\`y' \((x)\)\)]],
  " in terms of ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  ".  (That is, differentiate the equation with respect to ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " and solve for ",
  Cell[BoxData[
      \(TraditionalForm\`y' \((x)\)\)]],
  ".)  Differentiate ",
  Cell[BoxData[
      \(TraditionalForm\`y' \((x)\)\)]],
  " to determine ",
  Cell[BoxData[
      \(TraditionalForm\`\(y'\)' \((x)\)\)]],
  " in terms of ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  "."
}], "Text",
  TextAlignment->Left,
  TextJustification->1]
}, Open  ]],

Cell[CellGroupData[{

Cell["Exercise 4", "Subsubsection"],

Cell["\<\
Find the radius and height of a cylindrical cup of minimal surface \
whose volume is 64.\
\>", "Text",
  TextJustification->1]
}, Open  ]]
}, Closed]],

Cell[CellGroupData[{

Cell[TextData["Solutions"], "Section",
  Evaluatable->False,
  AspectRatioFixed->True],

Cell[CellGroupData[{

Cell["Solution 1", "Subsubsection"],

Cell[BoxData[
    \(Limit[\((1 + 1\/x)\)\^x, x \[Rule] \[Infinity]]\)], "Input"],

Cell[BoxData[
    \(Limit[\(a\ x\^2 + b\ x + c\)\/\(d\ x\^2 + e\ x + f\), 
      x \[Rule] \[Infinity]]\)], "Input"],

Cell[BoxData[
    \(Limit[\(a\ x\^2 + b\ x + c\)\/\(d\ x\^2 + e\ x + f\), x \[Rule] 0]\)], 
  "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["Solution 2", "Subsubsection"],

Cell[BoxData[
    \(Limit[\(\((x + \[CapitalDelta])\)\^n - x\^n\)\/\[CapitalDelta], 
      \[CapitalDelta] \[Rule] 0]\)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["Solution 3", "Subsubsection"],

Cell[BoxData[
    \(\[PartialD]\_x\((u[x] v[x])\)\)], "Input"],

Cell[BoxData[
    \(\[PartialD]\_x\((u[x]\/v[x])\)\)], "Input"],

Cell[BoxData[
    \(% // Together\)], "Input"],

Cell[BoxData[
    \(\[PartialD]\_x u[x]\^a\)], "Input"],

Cell[BoxData[
    \(\[PartialD]\_x u[v[x]]\)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["Solution 4", "Subsubsection"],

Cell[TextData[{
  "From the plot and what we know about polynomials, we know that there will \
be two local maxima and two local minima for the polynomial:\n\t",
  Cell[BoxData[
      \(TraditionalForm
      \`\((x - 1)\) \((x - 2)\) \((x - 3)\) \((x - 4)\) \((x - 5)\)\)]],
  "."
}], "Text",
  Evaluatable->False,
  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[TextData[{
  "First we define a function, ",
  Cell[BoxData[
      \(TraditionalForm\`f\)]],
  ", to be the above polynomial."
}], "Text",
  Evaluatable->False,
  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(f[x_] := 
      \((x - 1)\)\ \((x - 2)\)\ \((x - 3)\)\ \((x - 4)\)\ \((x - 5)\)\)], 
  "Input",
  AspectRatioFixed->True],

Cell[BoxData[
    \(\(Plot[f[x], {x, 0, 6}]; \)\)], "Input",
  AspectRatioFixed->True],

Cell[TextData[{
  "Here are the values of ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " where the local extrema occur."
}], "Text",
  Evaluatable->False,
  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(solns = Simplify[Solve[\[PartialD]\_x f[x] == 0, x]]\)], "Input",
  AspectRatioFixed->True],

Cell["Here is the numerical approximation.", "Text"],

Cell[BoxData[
    \(N[solns]\)], "Input"],

Cell["We sort the solutions in increasing numerical order.", "Text"],

Cell[BoxData[
    \(solns = Sort[solns, \((N[x /. #1] < N[x /. #2])\) &]\)], "Input"],

Cell[BoxData[
    \(N[solns]\)], "Input"],

Cell["Here are the values of the function at the local extrema.", "Text",
  Evaluatable->False,
  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(Simplify[f[x] /. solns]\)], "Input",
  AspectRatioFixed->True],

Cell[BoxData[
    \(N[%]\)], "Input"],

Cell["\<\
Here are the values of the second derivative of the function at the \
local extrema.\
\>", "Text",
  Evaluatable->False,
  TextAlignment->Left,
  TextJustification->1,
  AspectRatioFixed->True],

Cell[BoxData[
    \(Simplify[\(\(f'\)'\)[x] /. solns]\)], "Input",
  AspectRatioFixed->True],

Cell[BoxData[
    \(N[%]\)], "Input"],

Cell["\<\
Recalling that if the second derivative is (positive/negative) at a \
local extrema, it is a (minima/maxima), we see that the first and third \
extrema are maxima and the second and fourth extrema are minima.\
\>", "Text",\

  TextAlignment->Left,
  TextJustification->1],

Cell[BoxData[
    \(Clear[f, solns]\)], "Input",
  AspectRatioFixed->True]
}, Closed]],

Cell[CellGroupData[{

Cell["Solution 5", "Subsubsection"],

Cell[TextData[{
  "In order to graph ",
  Cell[BoxData[
      \(TraditionalForm\`y(x)\)]],
  " we solve the equation for ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  "."
}], "Text"],

Cell[BoxData[
    \(Clear[x, y]; \nSolve[x\^2 - x\ y + y\^2 == 3, y]\)], "Input"],

Cell[TextData[{
  "We note that ",
  Cell[BoxData[
      \(TraditionalForm\`y\)]],
  " is real-valued for ",
  Cell[BoxData[
      \(TraditionalForm\`x \[Element] \([\(-2\), 2]\)\)]],
  ".  Here is a plot."
}], "Text"],

Cell[BoxData[
    \(\(Plot[Evaluate[y /. %], {x, \(-2\), 2}, AspectRatio -> Automatic]; 
    \)\)], "Input"],

Cell[TextData[{
  "Differentiating the left-hand-side of ",
  Cell[BoxData[
      \(TraditionalForm\`x\^2 - x\ y + y\^2 = 3\)]],
  ","
}], "Text"],

Cell[BoxData[
    \(\[PartialD]\_x\((x\^2 - x\ y[x] + y[x]\^2)\)\)], "Input"],

Cell[TextData[{
  "Now we set this equal to the derivative of the right side and solve for ",
  Cell[BoxData[
      \(TraditionalForm\`y' \((x)\)\)]],
  "."
}], "Text"],

Cell[BoxData[
    \(Solve[% == 0, \(y'\)[x]]\)], "Input"],

Cell[TextData[{
  "The next statement defines ",
  Cell[BoxData[
      \(TraditionalForm\`\(y'\)[x]\)]],
  " as the above solution."
}], "Text"],

Cell[BoxData[
    \(\(y'\)[x_] = \(y'\)[x] /. %[\([1]\)]\)], "Input"],

Cell[TextData[{
  "Now we differentiate ",
  Cell[BoxData[
      \(TraditionalForm\`y' \((x)\)\)]],
  " to get ",
  Cell[BoxData[
      \(TraditionalForm\`\(y'\)' \((x)\)\)]],
  "."
}], "Text"],

Cell[BoxData[
    \(\[PartialD]\_x\ \(y'\)[x]\)], "Input"],

Cell[TextData[{
  "Note that ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " has automatically substituted in the definition of ",
  Cell[BoxData[
      \(TraditionalForm\`y' \((x)\)\)]],
  " into the above expression.  Simplifying the expression yields:"
}], "Text"],

Cell[BoxData[
    \(Simplify[%]\)], "Input"],

Cell["\<\
Finally we can simplify the numerator by using the original \
implicit equation.\
\>", "Text",
  TextJustification->1],

Cell[BoxData[
    \(% /. \((x\^2 - x\ y[x] + y[x]\^2)\) -> 3\)], "Input"],

Cell[TextData[{
  "This removes the definition of ",
  Cell[BoxData[
      \(TraditionalForm\`y' \((x)\)\)]],
  " that we made before.  (Using the ",
  StyleBox["Clear[]",
    FontWeight->"Bold"],
  " function wouldn't do the job.)"
}], "Text",
  TextJustification->1],

Cell[BoxData[
    \(Remove[y]\)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell["Solution 6", "Subsubsection"],

Cell[TextData[{
  "The volume of the cup is ",
  Cell[BoxData[
      \(TraditionalForm\`\[Pi]\ r\^2\ h = 64\)]],
  ".  Using this formula we relate the height to the radius.  The surface \
area of the cup is the sum of the areas of the circular bottom and \
cylindrical sides."
}], "Text",
  TextJustification->1],

Cell[BoxData[
    \(Clear[area, r, h]; \nh = 64\/\(\[Pi]\ r\^2\); \n
    area[r_] = \[Pi]\ r\^2 + 2  \[Pi]\ r\ h\)], "Input"],

Cell["Solving for the stationary points,", "Text"],

Cell[BoxData[
    \(Solve[\(area'\)[r] == 0, r]\)], "Input"],

Cell["\<\
we see that there are three solutions, however only the second one \
is real-valued.\
\>", "Text"],

Cell[BoxData[
    \(N[%]\)], "Input"],

Cell["Thus the radius of the cup is", "Text"],

Cell[BoxData[
    \(\(r = 4\/\[Pi]\^\(1/3\); \)\)], "Input"],

Cell["and the height is", "Text"],

Cell[BoxData[
    \(h\)], "Input"],

Cell[BoxData[
    \(Remove[f, r, h]\)], "Input"]
}, Closed]]
}, Closed]]
}, Open  ]]
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