What is the derivative of (-6/(\sqrt[5]{x^3)}) ?

The derivative of (-6/(\sqrt[5]{x^3)}) is (18)/(5x^{8/5)}

What is the first derivative of (-6/(\sqrt[5]{x^3)}) ?

The first derivative of (-6/(\sqrt[5]{x^3)}) is (18)/(5x^{8/5)}

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\frac{18}{5 x ^{\frac{8}{5}}}

Solution steps

Simplify

Take the constant out:

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

Apply exponent rule:

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

= - 6 \frac{d}{d x} ((x^{3})^{- \frac{1}{5}})

Apply the chain rule:

- \frac{1}{5 ( x ^{3} ) ^{\frac{6}{5}}} \frac{d}{d x} (x^{3})

= - \frac{1}{5 ( x ^{3} ) ^{\frac{6}{5}}} \frac{d}{d x} (x^{3})

\frac{d}{d x} (x^{3}) = 3 x^{2}

= - 6 (- \frac{1}{5 ( x ^{3} ) ^{\frac{6}{5}}} \cdot 3 x^{2})

Simplify

- 6 (- \frac{1}{5 ( x ^{3} ) ^{\frac{6}{5}}} \cdot 3 x^{2}) : \frac{18}{5 x ^{\frac{8}{5}}}

= \frac{18}{5 x ^{\frac{8}{5}}}

cos^{3} (12 x)

y = 3 x^{2} + 2 x, (2, 4)

\int sin^{2} (x co) s^{2} x d x

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 6, y^{^{'}} (0) = 0

f (x) = \frac{sin ( x )}{2 + cos ( x )}

What is the derivative of (-6/(\sqrt[5]{x^3)}) ?
The derivative of (-6/(\sqrt[5]{x^3)}) is (18)/(5x^{8/5)}
What is the first derivative of (-6/(\sqrt[5]{x^3)}) ?
The first derivative of (-6/(\sqrt[5]{x^3)}) is (18)/(5x^{8/5)}