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

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

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

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

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derivative of

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

- \frac{1}{x ^{2}} - \frac{2}{x ^{3}} - \frac{3}{x ^{4}}

Solution steps

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

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

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

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

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

\frac{t d y}{d t} + (t + 1) y = t

y = b e^{e^{3} + x^{3}}

\frac{d}{d x} (2 lo g_{10} (x) - 3 x)

\int \frac{2 x ^{2}}{4 x - x ^{2}} d x

\frac{3 x + 4}{3 x - 4}

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