What is the derivative of 4/(\sqrt[5]{x^2}) ?

The derivative of 4/(\sqrt[5]{x^2}) is -8/(5x^{7/5)}

What is the first derivative of 4/(\sqrt[5]{x^2}) ?

The first derivative of 4/(\sqrt[5]{x^2}) is -8/(5x^{7/5)}

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- \frac{8}{5 x ^{\frac{7}{5}}}

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^{2})^{- \frac{1}{5}})

Apply the chain rule:

- \frac{1}{5 ( x ^{2} ) ^{\frac{6}{5}}} \frac{d}{d x} (x^{2})

= - \frac{1}{5 ( x ^{2} ) ^{\frac{6}{5}}} \frac{d}{d x} (x^{2})

\frac{d}{d x} (x^{2}) = 2 x

= 4 (- \frac{1}{5 ( x ^{2} ) ^{\frac{6}{5}}} \cdot 2 x)

Simplify

4 (- \frac{1}{5 ( x ^{2} ) ^{\frac{6}{5}}} \cdot 2 x) : - \frac{8}{5 x ^{\frac{7}{5}}}

= - \frac{8}{5 x ^{\frac{7}{5}}}

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{9} d x

\int_{0}^{5} x + 1 d x

\frac{d y}{d t} = - \frac{t}{y}

\frac{\partial}{\partial x} (\frac{y}{2 x} - 5)

\int sin^{3} (17 x) d x

What is the derivative of 4/(\sqrt[5]{x^2}) ?
The derivative of 4/(\sqrt[5]{x^2}) is -8/(5x^{7/5)}
What is the first derivative of 4/(\sqrt[5]{x^2}) ?
The first derivative of 4/(\sqrt[5]{x^2}) is -8/(5x^{7/5)}