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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

(\frac{x ^{3}}{6})^{'} = \frac{x ^{2}}{2}

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

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

\int_{6}^{\infty} \frac{1}{( x - 5 ) ^{\frac{3}{2}}} d x

\int_{0}^{5} x^{2} + 25 d x

y = sin (x y)

\frac{\partial}{\partial t} (x ln (t))

\int cos^{3} (\frac{x}{10}) d x

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