What is the derivative of f(x)=2x^{1/3}+x/3-1/4 x^3 ?

The derivative of f(x)=2x^{1/3}+x/3-1/4 x^3 is 1/3+(8-9x^{8/3})/(12x^{2/3)}

What is the first derivative of f(x)=2x^{1/3}+x/3-1/4 x^3 ?

The first derivative of f(x)=2x^{1/3}+x/3-1/4 x^3 is 1/3+(8-9x^{8/3})/(12x^{2/3)}

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derivative of

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

\frac{1}{3} + \frac{8 - 9 x ^{\frac{8}{3}}}{12 x ^{\frac{2}{3}}}

Solution steps

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

Simplify

2 x^{\frac{1}{3}} + \frac{x}{3} - \frac{1}{4} x^{3} : 2 x^{\frac{1}{3}} + \frac{x}{3} - \frac{x ^{3}}{4}

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (\frac{x}{3}) = \frac{1}{3}

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

= \frac{2}{3 x ^{\frac{2}{3}}} + \frac{1}{3} - \frac{3 x ^{2}}{4}

Combine the fractions

\frac{2}{3 x ^{\frac{2}{3}}} - \frac{3 x ^{2}}{4} : \frac{8 - 9 x ^{\frac{8}{3}}}{12 x ^{\frac{2}{3}}}

= \frac{1}{3} + \frac{8 - 9 x ^{\frac{8}{3}}}{12 x ^{\frac{2}{3}}}

\int - 7 (- 7 x)^{3} d x

\frac{d}{d x} (x^{4} e^{9 x})

\frac{d}{d x} (\frac{6}{7 x ^{4}})

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

\frac{\partial}{\partial y} (\frac{2 x - 2 z}{3 y + 2 z})

What is the derivative of f(x)=2x^{1/3}+x/3-1/4 x^3 ?
The derivative of f(x)=2x^{1/3}+x/3-1/4 x^3 is 1/3+(8-9x^{8/3})/(12x^{2/3)}
What is the first derivative of f(x)=2x^{1/3}+x/3-1/4 x^3 ?
The first derivative of f(x)=2x^{1/3}+x/3-1/4 x^3 is 1/3+(8-9x^{8/3})/(12x^{2/3)}