What is the derivative of 7/(\sqrt[3]{x}) ?

The derivative of 7/(\sqrt[3]{x}) is -7/(3x^{4/3)}

What is the first derivative of 7/(\sqrt[3]{x}) ?

The first derivative of 7/(\sqrt[3]{x}) is -7/(3x^{4/3)}

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- \frac{7}{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}

= 7 \frac{d}{d x} (x^{- \frac{1}{3}})

Apply the Power Rule:

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

= 7 (- \frac{1}{3} x^{- \frac{1}{3} - 1})

Simplify

7 (- \frac{1}{3} x^{- \frac{1}{3} - 1}) : - \frac{7}{3 x ^{\frac{4}{3}}}

= - \frac{7}{3 x ^{\frac{4}{3}}}

\frac{d y}{d x} = (y - 1)^{2}

\frac{1}{( 1 - x )}

\int sin (x) cos^{4} (d) x d x

y^{^{'}} = \frac{( e ^{x - y} )}{( 1 + e ^{x} )}

What is the derivative of 7/(\sqrt[3]{x}) ?
The derivative of 7/(\sqrt[3]{x}) is -7/(3x^{4/3)}
What is the first derivative of 7/(\sqrt[3]{x}) ?
The first derivative of 7/(\sqrt[3]{x}) is -7/(3x^{4/3)}