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

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

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

The first derivative of f(x)= 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}}}

\frac{\partial}{\partial x} (\frac{( x + y )}{( x - y ) ^{2}})

h \to 0 lim (3 h + 18)

f (x) = \frac{1}{2} x ln (x^{2}), (- 1, 0)

\int cos (θ) + sin (θ) d θ

\frac{d}{d x} (\frac{2 x}{x ^{2} + 9})

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