Answer
$y'=\frac{-u^2+2u+8}{(1-u)^2}$
Work Step by Step
$y=\frac{(u+2)^2}{1-u}$
Use the quotient rule. Use the chain rule to derive the numerator.
$y'=\frac{(1-u)2(u+2)(1)-(u+2)^2(0-1)}{(1-u)^2}$
Expand.
$y'=\frac{(1-u)(2u+4)-(u^2+4u+4)(-1)}{(1-u)^2}$
Distribute.
$y'=\frac{2u+4-2u^2-4u+u^2+4u+4}{(1-u)^2}$
Combine like terms.
$y'=\frac{-u^2+2u+8}{(1-u)^2}$
