What is the derivative of f(x)=6\sqrt[3]{x^2} ?

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

What is the first derivative of f(x)=6\sqrt[3]{x^2} ?

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

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derivative of

\frac{4 x}{( x ^{2} ) ^{\frac{2}{3}}}

Solution steps

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

= 6 \cdot \frac{1}{3 ( x ^{2} ) ^{\frac{2}{3}}} \cdot 2 x

Simplify

6 \cdot \frac{1}{3 ( x ^{2} ) ^{\frac{2}{3}}} \cdot 2 x : \frac{4 x}{( x ^{2} ) ^{\frac{2}{3}}}

= \frac{4 x}{( x ^{2} ) ^{\frac{2}{3}}}

\frac{d N}{d t} + N = Nt e^{t + 2}

\int \frac{x ^{4}}{x ^{3} - 1} d x

y = \frac{4}{4 - x ^{2}}, (0, 1)

t y^{^{'}} + y = 8

\frac{2 t}{t ^{2} + 4}

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