What is the derivative of y=6ln(x)-x^3+2\sqrt[3]{x} ?

The derivative of y=6ln(x)-x^3+2\sqrt[3]{x} is (18x^{2/3}+2x-9x^{11/3})/(3x^{5/3)}

What is the first derivative of y=6ln(x)-x^3+2\sqrt[3]{x} ?

The first derivative of y=6ln(x)-x^3+2\sqrt[3]{x} is (18x^{2/3}+2x-9x^{11/3})/(3x^{5/3)}

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derivative of

\frac{18 x ^{\frac{2}{3}} + 2 x - 9 x ^{\frac{11}{3}}}{3 x ^{\frac{5}{3}}}

Solution steps

Apply the Sum/Difference Rule:

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

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

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

= \frac{6}{x} - 3 x^{2} + \frac{2}{3 x ^{\frac{2}{3}}}

Simplify

\frac{6}{x} - 3 x^{2} + \frac{2}{3 x ^{\frac{2}{3}}} : \frac{18 x ^{\frac{2}{3}} + 2 x - 9 x ^{\frac{11}{3}}}{3 x ^{\frac{5}{3}}}

= \frac{18 x ^{\frac{2}{3}} + 2 x - 9 x ^{\frac{11}{3}}}{3 x ^{\frac{5}{3}}}

x \to \infty lim (arctan (8 x))

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

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

\frac{\partial}{\partial y} (y sin (x))

x \to 2 + lim (2^{x} - 4)

What is the derivative of y=6ln(x)-x^3+2\sqrt[3]{x} ?
The derivative of y=6ln(x)-x^3+2\sqrt[3]{x} is (18x^{2/3}+2x-9x^{11/3})/(3x^{5/3)}
What is the first derivative of y=6ln(x)-x^3+2\sqrt[3]{x} ?
The first derivative of y=6ln(x)-x^3+2\sqrt[3]{x} is (18x^{2/3}+2x-9x^{11/3})/(3x^{5/3)}