Question

Resolva a derivada por definição
$x^{1/3}$

Answer 1

Seja $y=x^{\frac{1}{3}}$, pela definição temos que:

$\lim_{x\rightarrow\,p}\frac{f(x)-f(p)}{x-p}=f'(x)\\\\f'(x)=\lim_{x\rightarrow\,p}\frac{f(x)-f(p)}{x-p}\\\\f'(x)=\lim_{x\rightarrow\,p}\frac{x^{\frac{1}{3}}-p^{\frac{1}{3}}}{x-p}\\\\f'(x)=\lim_{x\rightarrow\,p}\frac{(x^{\frac{1}{3}}-p^{\frac{1}{3}})}{x-p}\cdot\frac{(x^{\frac{2}{3}}+x^{\frac{1}{3}}p^{\frac{1}{3}}+p^{\frac{2}{3}})}{(x^{\frac{2}{3}}+x^{\frac{1}{3}}p^{\frac{1}{3}}+p^{\frac{2}{3}})}\\\\f'(x)=\lim_{x\to\,p}\frac{x^1+x^{\frac{2}{3}}p^{\frac{1}{3}}+x^{\frac{1}{3}}p^{\frac{2}{3}}-p^{\frac{1}{3}}x^{\frac{2}{3}}-x^{\frac{1}{3}}p^{\frac{2}{3}}-p^1}{(x-p)(x^{\frac{2}{3}}+x^{\frac{1}{3}}p^{\frac{1}{3}}+p^{\frac{2}{3}})}\\\\f'(x)=\lim_{x\to\,p}\frac{x-p}{(x-p)(x^{\frac{2}{3}}+x^{\frac{1}{3}}p^{\frac{1}{3}}+p^{\frac{2}{3}})}$

$f'(x)=\lim_{x\to\,p}\frac{1}{x^{\frac{2}{3}}+x^{\frac{1}{3}}p^{\frac{1}{3}}+p^{\frac{2}{3}}}\\\\f'(p)=\frac{1}{p^{\frac{2}{3}}+p^{\frac{1}{3}}p^{\frac{1}{3}}+p^{\frac{2}{3}}}\\\\f'(p)=\frac{1}{3\cdot\,p^{\frac{2}{3}}}\\\\\boxed{f'(p)=\frac{1}{3\sqrt[3]{p^2}}}$