Question

derivada de
$y=6 \sqrt[3]{x^2}- \frac{4}{ \sqrt{x} }$

Answer 1

Vamos tirar da raiz:

$\\ y = 6x^2^/^3 - \frac{4}{x^1^/^2} \\ \\ y = 6x^2^/^3 -4x^-^1^/^2$

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$\\ y' = 6* \frac{2}{3} *x^2^/^3^-^1-4* \frac{-1}{2} *x^-^1^/^2^-^1 \\ \\y' = 4x^2^/^3^-^3^/^3 +2x^-^1^/^2^-^2^/^2 \\ \\ y' = 4x^-^1^/^3 +2x^-^3^/^2 \\ \\ y' = \frac{4}{x^1^/^3 } + \frac{2}{x^3^/^2} \\ \\ y' = \frac{4*x^3^/^2+2*x^1^/^3}{x^1^/^3 *x^3^/^2} \\ \\ y' = \frac{4 \sqrt[2]{x^3} +2 \sqrt[3]{x} }{x^1^/^3^+^3^/^2} \\ \\ y' = \frac{4 \sqrt[2]{x^3} +2 \sqrt[3]{x}}{x^1^1^/^9} \\ \\ y' = \frac{4 \sqrt[2]{x^3} +2 \sqrt[3]{x}}{ \sqrt[9]{x^1^1} }$