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

The derivative of (1/x-4/(x^3)(x^3-5x-1)) is 2x+1/(x^2)-(40)/(x^3)-(12)/(x^4)

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

The first derivative of (1/x-4/(x^3)(x^3-5x-1)) is 2x+1/(x^2)-(40)/(x^3)-(12)/(x^4)

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

2 x + \frac{1}{x ^{2}} - \frac{40}{x ^{3}} - \frac{12}{x ^{4}}

Solution steps

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

Simplify

(\frac{1}{x} - \frac{4}{x ^{3}}) (x^{3} - 5 x - 1) : x^{2} + \frac{20}{x ^{2}} + \frac{4}{x ^{3}} - \frac{1}{x} - 9

= \frac{d}{d x} (x^{2} + \frac{20}{x ^{2}} + \frac{4}{x ^{3}} - \frac{1}{x} - 9)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{2}) + \frac{d}{d x} (\frac{20}{x ^{2}}) + \frac{d}{d x} (\frac{4}{x ^{3}}) - \frac{d}{d x} (\frac{1}{x}) - \frac{d}{d x} (9)

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

\frac{d}{d x} (\frac{20}{x ^{2}}) = - \frac{40}{x ^{3}}

\frac{d}{d x} (\frac{4}{x ^{3}}) = - \frac{12}{x ^{4}}

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

\frac{d}{d x} (9) = 0

= 2 x - \frac{40}{x ^{3}} - \frac{12}{x ^{4}} - (- \frac{1}{x ^{2}}) - 0

Simplify

= 2 x + \frac{1}{x ^{2}} - \frac{40}{x ^{3}} - \frac{12}{x ^{4}}

x \to 1 lim (x^{5} - 2 x - 6)

\frac{d}{d x} (\frac{csc ( x )}{1 + csc ( x )})

y^{^{''}} + 4 y = e^{4 x}

\int e^{e^{x}} d x

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

What is the derivative of (1/x-4/(x^3)(x^3-5x-1)) ?
The derivative of (1/x-4/(x^3)(x^3-5x-1)) is 2x+1/(x^2)-(40)/(x^3)-(12)/(x^4)
What is the first derivative of (1/x-4/(x^3)(x^3-5x-1)) ?
The first derivative of (1/x-4/(x^3)(x^3-5x-1)) is 2x+1/(x^2)-(40)/(x^3)-(12)/(x^4)