# Thread: Finding Derivative problems

1. ## Finding Derivative problems

I have

(y^2/y+3)^4

The answer I keep getting is

(4y^8+6y)/(y+3)^5

But i keep getting told that's wrong so I'm not sure where I messed up.

2. $

\left(\frac{y^2}{y+3}\right)^4 = 4\left(\frac{y^2}{y+3}\right)^3\left(\frac{y^2}{y+ 3}\right)' =
$

$

4 \left( \frac{y^6}{(y+3)^3}\right) \left( \frac{2y(y+3)-y^2}{(y+3)^2}\right)
$

$
4 \left( \frac{y^6}{(y+3)^3}\right) \left( \frac{y^2+6y}{(y+3)^2}\right)
$

$
\frac{4y^7(y+6)}{(y+3)^5}
$