Question

$\frac{dy}{dx} onde \: y = \sqrt[3]{x {}^{2} } + 3$
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Answer 1

Explicação passo-a-passo:

$\sf y=\sqrt[3]{x^2}+3$

$\sf y=x^{\frac{2}{3}}+3$

$\sf \dfrac{dy}{dx}=\dfrac{2}{3}\cdot x^{\frac{2}{3}-1}+0$

$\sf \dfrac{dy}{dx}=\dfrac{2}{3}\cdot x^{\frac{2-3}{3}}$

$\sf \dfrac{dy}{dx}=\dfrac{2}{3}\cdot x^{\frac{-1}{3}}$

$\sf \dfrac{dy}{dx}=\dfrac{2}{3}\cdot\dfrac{1}{\sqrt[3]{x}}$

$\sf \dfrac{dy}{dx}=\dfrac{2}{3\sqrt[3]{x}}$