What is the derivative of 3/7 \sqrt[5]{x^3} ?

The derivative of 3/7 \sqrt[5]{x^3} is 9/(35x^{2/5)}

What is the first derivative of 3/7 \sqrt[5]{x^3} ?

The first derivative of 3/7 \sqrt[5]{x^3} is 9/(35x^{2/5)}

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\frac{d}{d x} (\frac{3}{7} 5 x^{3})

\frac{9}{35 x ^{\frac{2}{5}}}

Solution steps

\frac{d}{d x} (\frac{3}{7} 5 x^{3})

Take the constant out:

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

= \frac{3}{7} \frac{d}{d x} (5 x^{3})

Apply the chain rule:

\frac{1}{5 ( x ^{3} ) ^{\frac{4}{5}}} \frac{d}{d x} (x^{3})

= \frac{1}{5 ( x ^{3} ) ^{\frac{4}{5}}} \frac{d}{d x} (x^{3})

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

= \frac{3}{7} \cdot \frac{1}{5 ( x ^{3} ) ^{\frac{4}{5}}} \cdot 3 x^{2}

Simplify

\frac{3}{7} \cdot \frac{1}{5 ( x ^{3} ) ^{\frac{4}{5}}} \cdot 3 x^{2} : \frac{9}{35 x ^{\frac{2}{5}}}

= \frac{9}{35 x ^{\frac{2}{5}}}

x \to 0 + lim (7 x ln (x))

x \to 1 lim (e^{7 x^{5} - 7 x})

\frac{d}{d x} ((\frac{2 x}{x + 1})^{4})

\frac{d}{d x} (ln (ln (9 x)) + ln (ln (9)))

\frac{d}{d y} (e^{3 y})

What is the derivative of 3/7 \sqrt[5]{x^3} ?
The derivative of 3/7 \sqrt[5]{x^3} is 9/(35x^{2/5)}
What is the first derivative of 3/7 \sqrt[5]{x^3} ?
The first derivative of 3/7 \sqrt[5]{x^3} is 9/(35x^{2/5)}