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

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

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

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

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derivative of

y = \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}}

\int - \frac{x}{1} d x

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x} (1 + x)

y = 4^{x} (2 x^{4})

3 (e^{x} + e^{x} x)

\frac{d}{d θ} (6 cos (θ) + 8 sin (θ))

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