What is the derivative of f(x)=-1/2 (x^{-2}-x^4) ?

The derivative of f(x)=-1/2 (x^{-2}-x^4) is (2x^6+1)/(x^3)

What is the first derivative of f(x)=-1/2 (x^{-2}-x^4) ?

The first derivative of f(x)=-1/2 (x^{-2}-x^4) is (2x^6+1)/(x^3)

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derivative of

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

\frac{2 x ^{6} + 1}{x ^{3}}

Solution steps

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

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

- \frac{1}{2} (- \frac{2}{x ^{3}} - 4 x^{3}) : \frac{2 x ^{6} + 1}{x ^{3}}

= \frac{2 x ^{6} + 1}{x ^{3}}

\int sin (2 x) \cdot cos (2 x) d x

n = 1 \sum \infty \frac{1}{n} x^{n}

n = 1 \sum \infty (\frac{6}{4})^{n}

\int (1 - x) d x

\frac{\partial}{\partial x} (sin (4 x - 7 y))

What is the derivative of f(x)=-1/2 (x^{-2}-x^4) ?
The derivative of f(x)=-1/2 (x^{-2}-x^4) is (2x^6+1)/(x^3)
What is the first derivative of f(x)=-1/2 (x^{-2}-x^4) ?
The first derivative of f(x)=-1/2 (x^{-2}-x^4) is (2x^6+1)/(x^3)