Question

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

Answer 1

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

Answer 2

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

Answer 3

ok thanks