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OBMD
 one year ago
Best ResponseYou've already chosen the best response.0Here 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 \[\frac{ a5abt ^{2} }{ (1+bt ^{2})^{4} }=\frac{ a(15bt ^{2} }{ (1+bt ^{2})^{4} }\]

OBMD
 one year ago
Best ResponseYou've already chosen the best response.0Two corrections on the third line of equations (darn cut and paste). It should look like this \[a(1+bt ^{2})^{3}=\frac{ a }{ (1+bt ^{2})^{3} }=\frac{ a(1+bt ^{2}) }{ (1+bt ^{2})^{4} }=\frac{ a+abt ^{2} }{ (1+bt ^{2})^{4} }\]
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