What is the derivative of f(x)=-1/9 (x^{-9}-x^{18}) ?

The derivative of f(x)=-1/9 (x^{-9}-x^{18}) is (2x^{27}+1)/(x^{10)}

What is the first derivative of f(x)=-1/9 (x^{-9}-x^{18}) ?

The first derivative of f(x)=-1/9 (x^{-9}-x^{18}) is (2x^{27}+1)/(x^{10)}

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derivative of

f (x) = - \frac{1}{9} (x^{- 9} - x^{18})

\frac{2 x ^{27} + 1}{x ^{10}}

Solution steps

\frac{d}{d x} (- \frac{1}{9} (x^{- 9} - x^{18}))

Take the constant out:

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

= - \frac{1}{9} \frac{d}{d x} (x^{- 9} - x^{18})

Apply the Sum/Difference Rule:

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

= - \frac{1}{9} (\frac{d}{d x} (x^{- 9}) - \frac{d}{d x} (x^{18}))

\frac{d}{d x} (x^{- 9}) = - \frac{9}{x ^{10}}

\frac{d}{d x} (x^{18}) = 18 x^{17}

= - \frac{1}{9} (- \frac{9}{x ^{10}} - 18 x^{17})

Simplify

- \frac{1}{9} (- \frac{9}{x ^{10}} - 18 x^{17}) : \frac{2 x ^{27} + 1}{x ^{10}}

= \frac{2 x ^{27} + 1}{x ^{10}}

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(4 - x) y^{2} = x^{3}, (2, 2)

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What is the derivative of f(x)=-1/9 (x^{-9}-x^{18}) ?
The derivative of f(x)=-1/9 (x^{-9}-x^{18}) is (2x^{27}+1)/(x^{10)}
What is the first derivative of f(x)=-1/9 (x^{-9}-x^{18}) ?
The first derivative of f(x)=-1/9 (x^{-9}-x^{18}) is (2x^{27}+1)/(x^{10)}