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

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

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

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

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

Solution steps

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (x y e^{- x^{2} - y^{2}})

\frac{d}{d x} (e^{(- 4 x)})

y^{^{'}} + y = \frac{1}{1 + e ^{x}}

\frac{d}{d x} (sin (x^{0}))

\int_{0}^{t} 5 e^{s} sin (t - s) d s

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