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

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

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

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

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

x^{3} - \frac{1}{4 x ^{3}}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (\frac{1}{8 x ^{2}}) = - \frac{1}{4 x ^{3}}

= x^{3} - \frac{1}{4 x ^{3}}

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

f (x) = x^{3} + 1, (1, 2)

\int \frac{5 - e ^{- x}}{e ^{- x}} d x

\int \frac{x ^{2}}{x - 3} d x

\int \frac{6 x}{4 x ^{2} + 4 x + 10} d x

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