What is the derivative of 3/(\sqrt[3]{x)} ?

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

What is the first derivative of 3/(\sqrt[3]{x)} ?

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

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derivative of

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

Solution steps

Take the constant out:

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 (- \frac{1}{3} x^{- \frac{1}{3} - 1})

Simplify

3 (- \frac{1}{3} x^{- \frac{1}{3} - 1}) : - \frac{1}{x ^{\frac{4}{3}}}

= - \frac{1}{x ^{\frac{4}{3}}}

n = 1 \sum \infty n! (0)^{n}

\frac{\partial}{\partial x} (y^{3} - 1)

\int 2 x \cdot sin (x^{2}) d x

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

x \to 7 lim (x^{2})

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