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

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

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

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

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derivative of

\frac{x ^{4}}{8} + \frac{1}{4 x ^{2}}

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

\int_{0}^{1} ln (1 + x^{2}) d x

1 + x, 1

\int \frac{1}{4} cos^{2} (2 x) d x

y^{^{'}} + 0.2 \cdot y = 5

f (x) = x^{\frac{1}{6}}

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