What is the derivative of x^{-3}+4x^{-5}-3x^{-8} ?

The derivative of x^{-3}+4x^{-5}-3x^{-8} is -3/(x^4)-(20)/(x^6)+(24)/(x^9)

What is the first derivative of x^{-3}+4x^{-5}-3x^{-8} ?

The first derivative of x^{-3}+4x^{-5}-3x^{-8} is -3/(x^4)-(20)/(x^6)+(24)/(x^9)

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\frac{d}{d x} (x^{- 3} + 4 x^{- 5} - 3 x^{- 8})

- \frac{3}{x ^{4}} - \frac{20}{x ^{6}} + \frac{24}{x ^{9}}

Solution steps

\frac{d}{d x} (x^{- 3} + 4 x^{- 5} - 3 x^{- 8})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{- 3}) + \frac{d}{d x} (4 x^{- 5}) - \frac{d}{d x} (3 x^{- 8})

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

\frac{d}{d x} (4 x^{- 5}) = - \frac{20}{x ^{6}}

\frac{d}{d x} (3 x^{- 8}) = - \frac{24}{x ^{9}}

= - \frac{3}{x ^{4}} - \frac{20}{x ^{6}} - (- \frac{24}{x ^{9}})

Simplify

= - \frac{3}{x ^{4}} - \frac{20}{x ^{6}} + \frac{24}{x ^{9}}

\frac{d}{d x} (ln (6 - 5 x))

t e^{, at}

\frac{d}{d x} (20 x^{2})

\frac{d ^{2}}{d x ^{2}} (\frac{sin ( x )}{x})

What is the derivative of x^{-3}+4x^{-5}-3x^{-8} ?
The derivative of x^{-3}+4x^{-5}-3x^{-8} is -3/(x^4)-(20)/(x^6)+(24)/(x^9)
What is the first derivative of x^{-3}+4x^{-5}-3x^{-8} ?
The first derivative of x^{-3}+4x^{-5}-3x^{-8} is -3/(x^4)-(20)/(x^6)+(24)/(x^9)