## anonymous 2 years ago can someone explain how to get the solution to IF-7 problem d?

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1. anonymous

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 $\frac{ a-5abt ^{2} }{ (1+bt ^{2})^{4} }=\frac{ a(1-5bt ^{2} }{ (1+bt ^{2})^{4} }$

2. anonymous

Two corrections on the third line of equations (darn cut and paste). It should look like this $a(1+bt ^{2})^{-3}=\frac{ a }{ (1+bt ^{2})^{3} }=\frac{ a(1+bt ^{2}) }{ (1+bt ^{2})^{4} }=\frac{ a+abt ^{2} }{ (1+bt ^{2})^{4} }$

3. anonymous

ok thanks