[tex] Find \frac {dy} {dx}[/tex]

[tex]y = \frac {(2x+3)^3} {(4x^2-1)^8}[/tex]

I know that the answer is (from the textbook, but I don't know how it got there)

[tex] -\frac {2(2x+3)^2(52x^2+96x+3)} {(4x^2-1)^9}[/tex]

I'll attempt to show my work in the next post.