Evaluate: y' =(\sqrt[3]{x^4}-4x^2)/x

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Expression: $y\prime =\frac{\sqrt[3]{x^{4}}-4x^{2}}{x}$

If{\quad $}f^{\prime}(x)=g(x)${\quad $then{\quad}}f(x)=\int{g(x)}dx$

$y=\int \frac{\sqrt[3]{x^{4}}-4x^{2}}{x}dx$

$\int \frac{\sqrt[3]{x^{4}}-4x^{2}}{x}dx=\frac{3}{4}x^{\frac{4}{3}}-2x^{2}+c_{1}$

$=\frac{3}{4}x^{\frac{4}{3}}-2x^{2}+c_{1}$

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