What is the derivative of y=\sqrt[3]{x^2} ?

The derivative of y=\sqrt[3]{x^2} is (2x)/(3(x^2)^{2/3)}

What is the first derivative of y=\sqrt[3]{x^2} ?

The first derivative of y=\sqrt[3]{x^2} is (2x)/(3(x^2)^{2/3)}

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derivative of

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

Solution steps

Apply the chain rule:

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

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

\frac{d}{d x} (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 x}{3 ( x ^{2} ) ^{\frac{2}{3}}}

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

x \frac{d y}{d x} - y = 3 x^{4}

\frac{d}{d x} ((- 3 x + 1)^{3} (2 x - 1)^{4})

\frac{d}{d x} (\frac{1}{( x ^{2} + 3 x + 2 ) ^{3}})

\int cosh (d) x d x

What is the derivative of y=\sqrt[3]{x^2} ?
The derivative of y=\sqrt[3]{x^2} is (2x)/(3(x^2)^{2/3)}
What is the first derivative of y=\sqrt[3]{x^2} ?
The first derivative of y=\sqrt[3]{x^2} is (2x)/(3(x^2)^{2/3)}