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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 55441, 1569]*) (*NotebookOutlinePosition[ 56077, 1592]*) (* CellTagsIndexPosition[ 56033, 1588]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell[TextData[{ "Examples from Section 6.4\n", Cell[BoxData[ \(a\^x\)]], " and ", Cell[BoxData[ \(\(log\_a\) x\)]] }], "Section"], Cell["p. 480", "Subsection"], Cell[CellGroupData[{ Cell["Derivatives", "Subsection"], Cell[TextData[{ "For any positive number ", Cell[BoxData[ \(a\ = \ e\^\(ln \((a)\)\)\)]], " so for any function ", Cell[BoxData[ \(f \((x)\)\)]], ", ", Cell[BoxData[ \(a\^\(f \((x)\)\) = \ e\^\(f \((x)\) ln \((a)\)\)\)]], ". Then the chain rule gives ", Cell[BoxData[ \(\[PartialD]\_x\ a\^\(f \((x)\)\)\ = \ \(\[PartialD]\_x\ e\^\(f \((x)\)\ ln \((a)\)\) = \ \(e\^\(x\ ln \((a)\)\)\ f' \((x)\) ln \((a)\)\ = \ \(a\^\(f \((x)\)\)\) f' \((x)\)\ ln \((a)\)\)\)\)]], "." }], "Text"], Cell[TextData[{ "Similarly, ", Cell[BoxData[ \(\(log\_a\) \((f \((x)\))\)\ = \ ln \((f \((x)\))\)/ln \((a)\)\)]], " and the derivative is ", Cell[BoxData[ \(\[PartialD]\_x\ log\_a \((f \((x)\))\)\ = \ \(\[PartialD]\_x\ \(\(ln \((f \((x)\))\)\)\/\(ln \((a)\)\)\)\ = \ \(f' \((x)\)\)\/\(f \((x)\) ln \((a)\)\)\)\)]], "." }], "Text"], Cell[CellGroupData[{ Cell["11.", "Subsubsection"], Cell[BoxData[ \(\(y\ = \ 2\^x; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(D[y, x]\)], "Input"], Cell[BoxData[ \(2\^x\ Log[2]\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["13. 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Substitute ", Cell[BoxData[ \(u\ = \ cos \((t)\)\)]], ", ", Cell[BoxData[ \(du\ = \ \(-sin\) \((t)\) dt\)]], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\(7\^Cos[t]\) Sin[t] \[DifferentialD]t\)], "Input"], Cell[BoxData[ \(\(-\(7\^Cos[t]\/Log[7]\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%\(\[Pi]/2\)\(7\^Cos[t]\) Sin[t] \[DifferentialD]t\)], "Input"], Cell[BoxData[ \(6\/Log[7]\)], "Output"] }, Open ]], Cell[TextData[{ "67. ", Cell[BoxData[ \(\[Integral]\_0\%9\(\( 2 \( log\_10\) \((x + 1)\)\)\/\(x + 1\)\) \[DifferentialD]x\)]], ". Substitute ", Cell[BoxData[ \(u\ = \ \(\(log\_10\) \((x + 1)\)\ = \ \(ln \((x)\)\)\/\(ln \((10)\)\)\)\)]], ", ", Cell[BoxData[ \(du\ = \ \(1\/\(ln \((10)\) \((x + 1)\)\)\) dx\)]], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\(\(2 Log[10, x + 1]\)\/\(x + 1\)\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(Log[1 + x]\^2\/Log[10]\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%9\(\( 2 Log[10, x + 1]\)\/\(x + 1\)\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(Log[10]\)], "Output"] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Theory and Applications", "Subsection"], Cell[CellGroupData[{ Cell["79.", "Subsubsection"], Cell[TextData[{ "By what factor ", Cell[BoxData[ \(k\)]], " do you have to multiply the intensity ", Cell[BoxData[ \(I\)]], " of the sound from your audio amplifier to add 10 db to the sound level.\n\ \n", StyleBox["Solution.", FontSlant->"Italic"], " Let L be the origininal sound level. 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Of course, ", Cell[BoxData[ \(2\^4 = \ \(4\^2 = \ 16\)\)]], " and ", Cell[BoxData[ \(4\^4\ = \ \(4\^4 = \ 256\)\)]], ". We use ", StyleBox["NSolve[]", "Input"], " to find a numberical approximation to the third number." }], "Text"], Cell[BoxData[ \(\(Needs["\"]; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(solns\ = \ NSolve[x\^4 == \ 4\^x, x]\)], "Input"], Cell[BoxData[ \(InverseFunction::"ifun" \( : \ \) "Warning: Inverse functions are being used. Values may be lost for \ multivalued inverses."\)], "Message"], Cell[BoxData[ \(InverseFunction::"ifun" \( : \ \) "Warning: Inverse functions are being used. Values may be lost for \ multivalued inverses."\)], "Message"], Cell[BoxData[ \(InverseFunction::"ifun" \( : \ \) "Warning: Inverse functions are being used. Values may be lost for \ multivalued inverses."\)], "Message"], Cell[BoxData[ \(General::"stop" \( : \ \) "Further output of \!\(InverseFunction :: \"ifun\"\) will be suppressed \ during this calculation."\)], "Message"], Cell[BoxData[ \(Solve::"ifun" \( : \ \) "Inverse functions are being used by \!\(Solve\), so some solutions may \ not be found."\)], "Message"], Cell[BoxData[ \(Nonreal::"warning" \( : \ \) "Nonreal number encountered."\)], "Message"], Cell[BoxData[ \({{x \[Rule] \(-0.766664695962123055`\)}, {x \[Rule] 2.`}, { x \[Rule] 4.`}, { x \[Rule] \(-2.88539008177792677`\)\ ProductLog[Nonreal]}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(4\^x /. \ solns[\([1]\)]\)], "Input"], Cell[BoxData[ \(0.345479163831644831`\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(x^4\ /. \ solns[\([1]\)]\)], "Input"], Cell[BoxData[ \(0.345479163831644697`\)], "Output"] }, Open ]], Cell[TextData[{ "Another good numerical function for solving problems like this is ", StyleBox["FindRoot[]", "Input"], ". 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