f(Y) = (Y[sup]3[/sup] + 1 )[sup]1/3[/sup]
Bruk kjerneregelen:
f ' (Y) = (3y[sup]2[/sup])*(1/3)*(Y[sup]3[/sup] + 1)[sup](1/3)-1[/sup]
f ' (Y) = (Y[sup]2[/sup])*(Y[sup]3[/sup] + 1)[sup](-2/3)[/sup]
Derivasjons hjelp!
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
[tex]f(y) = (y^3 + 1)^{\frac{1}{3}}[/tex]
La [tex]u = y^3 + 1[/tex]
[tex] f \prime (y) = \frac{d f(y)}{d u}\frac{d u}{d y} = \frac{1}{3}(y^3 + 1)^{-\frac{2}{3}} \cdot 3y^2 = \frac{y^2}{(y^3 + 1)^{\frac{2}{3}}}[/tex]
La [tex]u = y^3 + 1[/tex]
[tex] f \prime (y) = \frac{d f(y)}{d u}\frac{d u}{d y} = \frac{1}{3}(y^3 + 1)^{-\frac{2}{3}} \cdot 3y^2 = \frac{y^2}{(y^3 + 1)^{\frac{2}{3}}}[/tex]