What is the (\sqrt[3]{x^2})^' ?

The (\sqrt[3]{x^2})^' is 2/(3x^{1/3)}

What is the first (\sqrt[3]{x^2})^' ?

The first (\sqrt[3]{x^2})^' is 2/(3x^{1/3)}

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(3 x^{2})^{^{'}}

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

Solution steps

(3 x^{2})^{'}

Apply the chain rule:

\frac{1}{3 ( x ^{2} ) ^{\frac{2}{3}}} (x^{2})^{'}

= \frac{1}{3 ( x ^{2} ) ^{\frac{2}{3}}} (x^{2})^{'}

(x^{2})^{'} = 2 x

= \frac{1}{3 ( x ^{2} ) ^{\frac{2}{3}}} \cdot 2 x

Simplify

\frac{1}{3 ( x ^{2} ) ^{\frac{2}{3}}} \cdot 2 x : \frac{2}{3 x ^{\frac{1}{3}}}

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

\int \frac{( 6 x - 1 )}{x ^{3} ( 2 x - 1 )} d x

4 a x - x

f (x) = \frac{5 x + 1}{5 x ^{2} + 1}

\int \frac{2}{x ( 1 + 2 x ) ^{4}} d x

f^{^{'}} (x) = 0

What is the (\sqrt[3]{x^2})^' ?
The (\sqrt[3]{x^2})^' is 2/(3x^{1/3)}
What is the first (\sqrt[3]{x^2})^' ?
The first (\sqrt[3]{x^2})^' is 2/(3x^{1/3)}