What is the derivative of 1/2 x^3+1/(6x) ?

The derivative of 1/2 x^3+1/(6x) is (3x^2)/2-1/(6x^2)

What is the first derivative of 1/2 x^3+1/(6x) ?

The first derivative of 1/2 x^3+1/(6x) is (3x^2)/2-1/(6x^2)

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\frac{d}{d x} (\frac{1}{2} x^{3} + \frac{1}{6 x})

\frac{3 x ^{2}}{2} - \frac{1}{6 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{2} x^{3} + \frac{1}{6 x})

Simplify

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

= \frac{d}{d x} (\frac{x ^{3}}{2} + \frac{1}{6 x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{3}}{2}) + \frac{d}{d x} (\frac{1}{6 x})

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

\frac{d}{d x} (\frac{1}{6 x}) = - \frac{1}{6 x ^{2}}

= \frac{3 x ^{2}}{2} - \frac{1}{6 x ^{2}}

\frac{\partial}{\partial x} (3^{4} - 6 5^{3})

y^{^{'}} = 3 y - 2 x

5 cos (7 x), 5 - 5 cos (7 x), [0, \frac{π}{7}]

y^{^{'}} + \frac{y}{6.96} = 0.0025862069

\frac{d}{d x} (ln (x^{\frac{1}{5}}))

What is the derivative of 1/2 x^3+1/(6x) ?
The derivative of 1/2 x^3+1/(6x) is (3x^2)/2-1/(6x^2)
What is the first derivative of 1/2 x^3+1/(6x) ?
The first derivative of 1/2 x^3+1/(6x) is (3x^2)/2-1/(6x^2)