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

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

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

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

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

Solution steps

Take the constant out:

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

Apply exponent rule:

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

= 3 \frac{d}{d x} ((x^{4})^{- \frac{1}{3}})

Apply the chain rule:

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

= - \frac{1}{3 ( x ^{4} ) ^{\frac{4}{3}}} \frac{d}{d x} (x^{4})

\frac{d}{d x} (x^{4}) = 4 x^{3}

= 3 (- \frac{1}{3 ( x ^{4} ) ^{\frac{4}{3}}} \cdot 4 x^{3})

Simplify

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

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

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

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

\int \frac{1}{18} x d x

\frac{d}{d x} (- 4 x^{3} - sin (x))

x \to 1 lim (∣ x + 1 ∣)

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