What is the ((x^4)/8+1/(4x^2))^' ?

The ((x^4)/8+1/(4x^2))^' is (x^3)/2-1/(2x^3)

What is the first ((x^4)/8+1/(4x^2))^' ?

The first ((x^4)/8+1/(4x^2))^' is (x^3)/2-1/(2x^3)

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(\frac{x ^{4}}{8} + \frac{1}{4 x ^{2}})^{^{'}}

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

Solution steps

(\frac{x ^{4}}{8} + \frac{1}{4 x ^{2}})^{'}

Apply the Sum/Difference Rule:

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

= (\frac{x ^{4}}{8})^{^{'}} + (\frac{1}{4 x ^{2}})^{^{'}}

(\frac{x ^{4}}{8})^{'} = \frac{x ^{3}}{2}

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

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

\frac{d}{d x} (e^{\frac{x ^{2}}{2}} x)

\int e^{- 3 y} d y

x \to 6 lim (x + 6)

\int \frac{1}{3 9 x - 1} d x

\int 81 - x^{2} d x

What is the ((x^4)/8+1/(4x^2))^' ?
The ((x^4)/8+1/(4x^2))^' is (x^3)/2-1/(2x^3)
What is the first ((x^4)/8+1/(4x^2))^' ?
The first ((x^4)/8+1/(4x^2))^' is (x^3)/2-1/(2x^3)