Just want to check if I did it right: Thanks y=2∜(x^3 ) =x^3/2 =1/2(x^3) =3x^2/2
Follow Math Help Forum on Facebook and Google+
$\displaystyle y=2\sqrt[4]{x^3}=2x^{3/4}$ so $\displaystyle y'=\frac{3}{2}x^{-1/4}$
View Tag Cloud