What is the derivative of 1/8 x^4+1/4 x^{-2} ?

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

What is the first derivative of 1/8 x^4+1/4 x^{-2} ?

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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