What is the derivative of \sqrt[3]{6x} ?

The derivative of \sqrt[3]{6x} is (2^{1/3})/(3^{2/3)x^{2/3}}

What is the first derivative of \sqrt[3]{6x} ?

The first derivative of \sqrt[3]{6x} is (2^{1/3})/(3^{2/3)x^{2/3}}

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derivative of

\frac{2 ^{\frac{1}{3}}}{3 ^{\frac{2}{3}} x ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( 6 x ) ^{\frac{2}{3}}} \frac{d}{d x} (6 x)

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

\frac{d}{d x} (6 x) = 6

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

Simplify

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

= \frac{2 ^{\frac{1}{3}}}{3 ^{\frac{2}{3}} x ^{\frac{2}{3}}}

f (θ) = 2 tan (θ)

\int 4 sin (\frac{x}{2}) + 8 d x

f (x) = x^{2} + 6, (1, 7)

\int csc^{6} (x) d x

\frac{( 1 - v )}{( 1 + v )}

What is the derivative of \sqrt[3]{6x} ?
The derivative of \sqrt[3]{6x} is (2^{1/3})/(3^{2/3)x^{2/3}}
What is the first derivative of \sqrt[3]{6x} ?
The first derivative of \sqrt[3]{6x} is (2^{1/3})/(3^{2/3)x^{2/3}}