y=3x3+13=(3x3+1)13y = \sqrt[3]{3x^{3}+1} = (3x^{3}+1)^{\frac{1}{3}}y=33x3+1=(3x3+1)31
Let u = 3x3+13x^{3}+13x3+1; y = u13u^{\frac{1}{3}}u31
dydx=(dydu)(dudx)\frac{dy}{dx} = \left(\frac{dy}{du}\right)\left(\frac{du}{dx}\right)dxdy=(dudy)(dxdu)
dydu=13u−23\frac{dy}{du} = \frac{1}{3}u^{-\frac{2}{3}}dudy=31u−32
dudx=9x2\frac{du}{dx} = 9x^{2}dxdu=9x2
dydx=(13(3x3+1)−23)(9x2)\frac{dy}{dx} = \left(\frac{1}{3}(3x^{3}+1)^{-\frac{2}{3}}\right)(9x^{2})dxdy=(31(3x3+1)−32)(9x2)
= 3x2(3x3+1)23\frac{3x^{2}}{\sqrt[3]{(3x^{3}+1)^{2}}}3(3x3+1)23x2
