What is the derivative of y=sqrt(3x)+\sqrt[3]{x}+1/x ?

The derivative of y=sqrt(3x)+\sqrt[3]{x}+1/x is (3sqrt(3)x^{13/6}+2x^2-6x^{2/3})/(6x^{8/3)}

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

The first derivative of y=sqrt(3x)+\sqrt[3]{x}+1/x is (3sqrt(3)x^{13/6}+2x^2-6x^{2/3})/(6x^{8/3)}

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derivative of

\frac{3 3 x ^{\frac{13}{6}} + 2 x ^{2} - 6 x ^{\frac{2}{3}}}{6 x ^{\frac{8}{3}}}

Solution steps

Apply the Sum/Difference Rule:

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

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

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

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

Simplify

\frac{3}{2 x} + \frac{1}{3 x ^{\frac{2}{3}}} - \frac{1}{x ^{2}} : \frac{3 3 x ^{\frac{13}{6}} + 2 x ^{2} - 6 x ^{\frac{2}{3}}}{6 x ^{\frac{8}{3}}}

= \frac{3 3 x ^{\frac{13}{6}} + 2 x ^{2} - 6 x ^{\frac{2}{3}}}{6 x ^{\frac{8}{3}}}

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

\frac{d y}{d x} (x^{2} + 1) + 7 x (y - 1) = 0

\int_{0}^{1} \frac{41}{x ^{5}} d x

\frac{d}{d x} (\frac{x}{x - 37})

f (x) = sin (3 x^{2})

What is the derivative of y=sqrt(3x)+\sqrt[3]{x}+1/x ?
The derivative of y=sqrt(3x)+\sqrt[3]{x}+1/x is (3sqrt(3)x^{13/6}+2x^2-6x^{2/3})/(6x^{8/3)}
What is the first derivative of y=sqrt(3x)+\sqrt[3]{x}+1/x ?
The first derivative of y=sqrt(3x)+\sqrt[3]{x}+1/x is (3sqrt(3)x^{13/6}+2x^2-6x^{2/3})/(6x^{8/3)}