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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derivative of

- \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{1}{1 + e ^{z}} d z

x + y^{2} = 6, x + y = 0

y = (x^{6} - 7 x + 4) (x^{4} - 5 x)

f (x) = x^{2} + 4

(\frac{y}{x} + 3 x) d x + (ln (x) - 2) d 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)}