[m]y`=(x^{\frac{1}{3}})`-(lnx)`+2\cdot (e^{x})`-(sinx)`+(4)`=\frac{1}{3}\cdot(x^{\frac{1}{3}-1})-\frac{1}{x}+2\cdot (e^{x})-cosx+0=\frac{1}{3\sqrt[3]{x^2}}-\frac{1}{x}+2\cdot (e^{x})-cosx[/m]
