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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

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\int cos (ln (x^{2})) d x

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