Question

Qual é a derivada de $\sqrt[3]{x}*cos(\frac{1}{x})$

Answer 1

Explicação passo a passo:

$\frac{d}{dx}\sqrt[3]{x}.cos(\frac{1}{x})$

Aplique a regra do produto, (f · g)' = f' · g + f · g' , onde  $f=\sqrt[3]{x}$  e  $g=cos(\frac{1}{x})$

    $\frac{d}{dx}(\sqrt[3]{x}).cos(\frac{1}{x})+\frac{d}{dx}(cos(\frac{1}{x})).\sqrt[3]{x}$

Transforme  $\sqrt[3]{x}$  em uma potência e aplique a regra da cadeia para derivar $cos(\frac{1}{x})=-sen(\frac{1}{x}).\frac{d}{dx}(\frac{1}{x})$

    $\frac{d}{dx}(x^{\frac{1}{3}}).cos(\frac{1}{x})+(-sen(\frac{1}{x})).\frac{d}{dx}(\frac{1}{x}).\sqrt[3]{x}$

    $\frac{1}{3}.x^{\frac{1}{3}-1}+(-sen(\frac{1}{x})).\frac{d}{dx}(x^{-1}).\sqrt[3]{x}$

    $\frac{1}{3}x^{-\frac{2}{3}}.cos(\frac{1}{x})+(-sen(\frac{1}{x})).(-1.x^{-1-1}).\sqrt[3]{x}$

    $\frac{1}{3x^{\frac{2}{3}}}.cos(\frac{1}{x})+(-sen(\frac{1}{x})).(-1x^{-2}).\sqrt[3]{x}$

    $\frac{cos(\frac{1}{x})}{3x^{\frac{2}{3}}}+sen(\frac{1}{x}).\frac{1}{x^{2}}.x^{\frac{1}{3}}$

    $\frac{cos(\frac{1}{x})}{3x^{\frac{2}{3}}}+sen(\frac{1}{x}).\frac{x^{\frac{1}{3}}}{x^{2}}$

    $\frac{cos(\frac{1}{x})}{3x^{\frac{2}{3}}}+sen(\frac{1}{x}).x^{\frac{1}{3}-2}$

    $\frac{cos(\frac{1}{x})}{3x^{\frac{2}{3}}}+sen(\frac{1}{x}).x^{-\frac{5}{3}}$

    $\frac{cos(\frac{1}{x})}{3x^{\frac{2}{3}}}+sen(\frac{1}{x}).\frac{1}{x^{\frac{5}{3}}}$

    $\frac{cos(\frac{1}{x})}{3x^{\frac{2}{3}}}+\frac{sen(\frac{1}{x})}{x^{\frac{5}{3}}}$