What is the derivative of \sqrt[7]{x^3} ?

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

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

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

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

Solution steps

Apply the chain rule:

\frac{1}{7 ( x ^{3} ) ^{\frac{6}{7}}} \frac{d}{d x} (x^{3})

= \frac{1}{7 ( x ^{3} ) ^{\frac{6}{7}}} \frac{d}{d x} (x^{3})

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

= \frac{1}{7 ( x ^{3} ) ^{\frac{6}{7}}} \cdot 3 x^{2}

Simplify

\frac{1}{7 ( x ^{3} ) ^{\frac{6}{7}}} \cdot 3 x^{2} : \frac{3}{7 x ^{\frac{4}{7}}}

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

x \to - 5 lim (4 x^{2} - 8)

\int cos^{2} (x) + \frac{4}{4 - x ^{2}} d x

x \to 1 + lim (\frac{x ^{4} - 1}{x - 1})

n = 1 \sum \infty \frac{8 n}{- 2 n + 9}

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