## Calculus 10th Edition

$f'(v)=\dfrac{-9(1-2v)^2}{(1+v)^4}.$
Using the Quotient Rule and the Chain Rule: $f'(v)=(\dfrac{y(v)}{z(v)})'=\dfrac{y'(v)z(v)-z'(v)y(v)}{(z(v))^2}$ $y(v)=(1-2v)^3; y'(v)=(3)(-2)(1-2v)^{3-1}=-6(1-2v)^2.$ $z(v)=(1+v)^3; z'(v)=(3)(1)(1+v)^{3-1}=3(1+v)^2$ $f'(v)=\dfrac{-6(1-2v)^2(1+v)^3-3(1+v)^2(1-2v)^3}{((1+v)^3)^2}$ $=\dfrac{-9(1-2v)^2}{(1+v)^4}$