What is the derivative of x^4-5+x^{-2}+41x^{-4} ?

The derivative of x^4-5+x^{-2}+41x^{-4} is 4x^3-2/(x^3)-(164)/(x^5)

What is the first derivative of x^4-5+x^{-2}+41x^{-4} ?

The first derivative of x^4-5+x^{-2}+41x^{-4} is 4x^3-2/(x^3)-(164)/(x^5)

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\frac{d}{d x} (x^{4} - 5 + x^{- 2} + 41 x^{- 4})

4 x^{3} - \frac{2}{x ^{3}} - \frac{164}{x ^{5}}

Solution steps

\frac{d}{d x} (x^{4} - 5 + x^{- 2} + 41 x^{- 4})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{4}) - \frac{d}{d x} (5) + \frac{d}{d x} (x^{- 2}) + \frac{d}{d x} (41 x^{- 4})

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

\frac{d}{d x} (5) = 0

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

\frac{d}{d x} (41 x^{- 4}) = - \frac{164}{x ^{5}}

= 4 x^{3} - 0 - \frac{2}{x ^{3}} - \frac{164}{x ^{5}}

Simplify

= 4 x^{3} - \frac{2}{x ^{3}} - \frac{164}{x ^{5}}

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What is the derivative of x^4-5+x^{-2}+41x^{-4} ?
The derivative of x^4-5+x^{-2}+41x^{-4} is 4x^3-2/(x^3)-(164)/(x^5)
What is the first derivative of x^4-5+x^{-2}+41x^{-4} ?
The first derivative of x^4-5+x^{-2}+41x^{-4} is 4x^3-2/(x^3)-(164)/(x^5)