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Clarkmonsterman Group Title

can someone explain how to get the solution to IF-7 problem d?

  • one year ago
  • one year ago

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  1. OBMD Group Title
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    Here is how I did the problem. First change \[Q=(at)(1+bt ^{2})^{-3}\] Then use the product rule to get \[a(1+bt ^{2})^{-3}+at(-3)(1+bt ^{2})^{-4}(2bt)\] I then used the fact that \[a(1+bt ^{2})^{-3}=\frac{ a }{(1+bt ^{2})^{3}}=\frac{a(1+bt ^{2})}{(1+bt ^{2})^{-3}}=\frac{ a+abt ^{2} }{ (1+bt ^{2})^{3} }\] to get \[\frac{ a+abt ^{2} }{ (1+bt ^{2})^{4} }+\frac{ -6abt ^{2} }{ (1+bt ^{2})^{4} }\] which simplifies to \[\frac{ a-5abt ^{2} }{ (1+bt ^{2})^{4} }=\frac{ a(1-5bt ^{2} }{ (1+bt ^{2})^{4} }\]

    • one year ago
  2. OBMD Group Title
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    Two corrections on the third line of equations (darn cut and paste). It should look like this \[a(1+bt ^{2})^{-3}=\frac{ a }{ (1+bt ^{2})^{3} }=\frac{ a(1+bt ^{2}) }{ (1+bt ^{2})^{4} }=\frac{ a+abt ^{2} }{ (1+bt ^{2})^{4} }\]

    • one year ago
  3. Clarkmonsterman Group Title
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    ok thanks

    • one year ago
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