What is the derivative of f(x)= 4/x-2/(x^3)+1/(x^4) ?

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

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

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

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derivative of

f (x) = \frac{4}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}}

- \frac{4}{x ^{2}} + \frac{6}{x ^{4}} - \frac{4}{x ^{5}}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

= - \frac{4}{x ^{2}} - (- \frac{6}{x ^{4}}) - \frac{4}{x ^{5}}

Simplify

= - \frac{4}{x ^{2}} + \frac{6}{x ^{4}} - \frac{4}{x ^{5}}

y = - \frac{7}{2} x^{2} + 5 x + 5, (- 1, - \frac{7}{2})

\int cos^{3} (x) sin^{- 4} (x) d x

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

\int (4 e^{x} - x^{3} + 6 x - 8) d x

\int \frac{x ^{5}}{x ^{6} + 4} d x

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