Respuesta :

gmany

[tex]f(t)=t^\frac{2}{3}-t^\frac{1}{3}+4\\\\f'(t)=\left(t^\frac{2}{3}-t^\frac{1}{3}+4\right)'=\left(t^\frac{2}{3}\right)'-\left(t^\frac{1}{3}\right)'+(4)'\\\\=\dfrac{2}{3}t^{\frac{2}{3}-1}-\dfrac{1}{3}t^{\frac{1}{3}-1}+0=\dfrac{2}{3}t^{-\frac{1}{3}}-\dfrac{1}{3}t^{-\frac{2}{3}}\\\\\boxed{f'(x)=\dfrac{2}{3\left(t^\frac{1}{3}-t^\frac{2}{3}\right)}}\\\\Used:\\\\\ [f(x)+g(x)]'=f'(x)+g'(x)\\\\(c)'=0\\\\\left(x^n\right)'=nx^{n-1}\\\\a^{-n}=\dfrac{1}{a^n}[/tex]