What is the derivative of \sqrt[3]{x^2}+4sqrt(x^3) ?

The derivative of \sqrt[3]{x^2}+4sqrt(x^3) is 2/(3x^{1/3)}+6sqrt(x)

What is the first derivative of \sqrt[3]{x^2}+4sqrt(x^3) ?

The first derivative of \sqrt[3]{x^2}+4sqrt(x^3) is 2/(3x^{1/3)}+6sqrt(x)

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\frac{2}{3 x ^{\frac{1}{3}}} + 6 x

Solution steps

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

\frac{d}{d x} (4 x^{3}) = 6 x

= \frac{2}{3 x ^{\frac{1}{3}}} + 6 x

sin (\frac{1}{x})

x \to 0 + lim (\frac{ln ( x )}{x})

\int - 4 x^{3} - 8 x^{2} - 4 d x

\frac{d}{d x} (- x^{2} + 5 x)

\frac{\partial ^{2}}{\partial x ^{2}} (x)

What is the derivative of \sqrt[3]{x^2}+4sqrt(x^3) ?
The derivative of \sqrt[3]{x^2}+4sqrt(x^3) is 2/(3x^{1/3)}+6sqrt(x)
What is the first derivative of \sqrt[3]{x^2}+4sqrt(x^3) ?
The first derivative of \sqrt[3]{x^2}+4sqrt(x^3) is 2/(3x^{1/3)}+6sqrt(x)