What is the derivative of 6\sqrt[3]{x^2}+1 ?

The derivative of 6\sqrt[3]{x^2}+1 is 4/(x^{1/3)}

What is the first derivative of 6\sqrt[3]{x^2}+1 ?

The first derivative of 6\sqrt[3]{x^2}+1 is 4/(x^{1/3)}

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

\frac{4}{x ^{\frac{1}{3}}}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (6 3 x^{2}) = \frac{4}{x ^{\frac{1}{3}}}

\frac{d}{d x} (1) = 0

= \frac{4}{x ^{\frac{1}{3}}} + 0

Simplify

= \frac{4}{x ^{\frac{1}{3}}}

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What is the derivative of 6\sqrt[3]{x^2}+1 ?
The derivative of 6\sqrt[3]{x^2}+1 is 4/(x^{1/3)}
What is the first derivative of 6\sqrt[3]{x^2}+1 ?
The first derivative of 6\sqrt[3]{x^2}+1 is 4/(x^{1/3)}