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

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

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

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

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

Solution steps

Take the constant out:

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

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

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

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

f (x) = - x + \frac{6}{x} + 1

\frac{s - 1}{s ^{2} - 2 s + 5}

\int \frac{4}{t} d t

2 y^{^{''}} + 32 y = 0

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