What is the derivative of-1/4 (x^{-4}-x^8) ?

The derivative of-1/4 (x^{-4}-x^8) is (2x^{12}+1)/(x^5)

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

The first derivative of-1/4 (x^{-4}-x^8) is (2x^{12}+1)/(x^5)

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derivative of

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

\frac{2 x ^{12} + 1}{x ^{5}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (x^{- 4}) = - \frac{4}{x ^{5}}

\frac{d}{d x} (x^{8}) = 8 x^{7}

= - \frac{1}{4} (- \frac{4}{x ^{5}} - 8 x^{7})

Simplify

- \frac{1}{4} (- \frac{4}{x ^{5}} - 8 x^{7}) : \frac{2 x ^{12} + 1}{x ^{5}}

= \frac{2 x ^{12} + 1}{x ^{5}}

e^{- \frac{π}{2} t}

\int e^{\frac{1}{2} x^{2}} \cdot x^{3} d x

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

\int (x^{8} - \frac{2 x ^{5}}{7} - 5) d x

y^{^{'}} = \frac{x ^{2} + x y + y ^{2}}{x ^{2}}

What is the derivative of-1/4 (x^{-4}-x^8) ?
The derivative of-1/4 (x^{-4}-x^8) is (2x^{12}+1)/(x^5)
What is the first derivative of-1/4 (x^{-4}-x^8) ?
The first derivative of-1/4 (x^{-4}-x^8) is (2x^{12}+1)/(x^5)