What is the derivative of 4/x-1/(6x^3+8/(x^5)) ?

The derivative of 4/x-1/(6x^3+8/(x^5)) is -4/(x^2)+1/(2x^4)-(40)/(x^6)

What is the first derivative of 4/x-1/(6x^3+8/(x^5)) ?

The first derivative of 4/x-1/(6x^3+8/(x^5)) is -4/(x^2)+1/(2x^4)-(40)/(x^6)

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\frac{d}{d x} (\frac{4}{x} - \frac{1}{6 x ^{3}} + \frac{8}{x ^{5}})

- \frac{4}{x ^{2}} + \frac{1}{2 x ^{4}} - \frac{40}{x ^{6}}

Solution steps

\frac{d}{d x} (\frac{4}{x} - \frac{1}{6 x ^{3}} + \frac{8}{x ^{5}})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{4}{x}) - \frac{d}{d x} (\frac{1}{6 x ^{3}}) + \frac{d}{d x} (\frac{8}{x ^{5}})

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

\frac{d}{d x} (\frac{1}{6 x ^{3}}) = - \frac{1}{2 x ^{4}}

\frac{d}{d x} (\frac{8}{x ^{5}}) = - \frac{40}{x ^{6}}

= - \frac{4}{x ^{2}} - (- \frac{1}{2 x ^{4}}) - \frac{40}{x ^{6}}

Simplify

= - \frac{4}{x ^{2}} + \frac{1}{2 x ^{4}} - \frac{40}{x ^{6}}

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6 x e^{x}

10 (x^{3} - x)^{4}

\frac{d}{d x} (sin (9 cos (x)))

\int \frac{2}{1 + x ^{2}} d x

What is the derivative of 4/x-1/(6x^3+8/(x^5)) ?
The derivative of 4/x-1/(6x^3+8/(x^5)) is -4/(x^2)+1/(2x^4)-(40)/(x^6)
What is the first derivative of 4/x-1/(6x^3+8/(x^5)) ?
The first derivative of 4/x-1/(6x^3+8/(x^5)) is -4/(x^2)+1/(2x^4)-(40)/(x^6)