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

The derivative of \sqrt[3]{3x^3+4x} is (9x^2+4)/(3(3x^3+4x)^{2/3)}

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

The first derivative of \sqrt[3]{3x^3+4x} is (9x^2+4)/(3(3x^3+4x)^{2/3)}

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

\frac{9 x ^{2} + 4}{3 ( 3 x ^{3} + 4 x ) ^{\frac{2}{3}}}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (3 x^{3} + 4 x) = 9 x^{2} + 4

= \frac{1}{3 ( 3 x ^{3} + 4 x ) ^{\frac{2}{3}}} (9 x^{2} + 4)

Simplify

\frac{1}{3 ( 3 x ^{3} + 4 x ) ^{\frac{2}{3}}} (9 x^{2} + 4) : \frac{9 x ^{2} + 4}{3 ( 3 x ^{3} + 4 x ) ^{\frac{2}{3}}}

= \frac{9 x ^{2} + 4}{3 ( 3 x ^{3} + 4 x ) ^{\frac{2}{3}}}

\int \frac{1}{( x - 2 ) ( x + 3 )} d x

\int_{- \infty}^{0} \frac{1}{4 - 9 x} d x

f (x) = x^{2} - 4

\int \frac{x ^{4}}{x ^{2} - 4} d x

\frac{d}{d x} (ln (18 - x^{2}))

What is the derivative of \sqrt[3]{3x^3+4x} ?
The derivative of \sqrt[3]{3x^3+4x} is (9x^2+4)/(3(3x^3+4x)^{2/3)}
What is the first derivative of \sqrt[3]{3x^3+4x} ?
The first derivative of \sqrt[3]{3x^3+4x} is (9x^2+4)/(3(3x^3+4x)^{2/3)}