What is the derivative of ln(1/x+1/(x^2)) ?

The derivative of ln(1/x+1/(x^2)) is (-x-2)/(x(x+1))

What is the first derivative of ln(1/x+1/(x^2)) ?

The first derivative of ln(1/x+1/(x^2)) is (-x-2)/(x(x+1))

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{- x - 2}{x ( x + 1 )}

\int \frac{3 x - 5}{x ^{2} ( x + 5 )} d x

\int (2 x - 1) (3 x + 4) d x

\int 0.002 d x

x^{2} - 6 x + 5

y^{^{''}} - 4 y^{^{'}} = 16 t, y (0) = 4, y^{^{'}} (0) = 0

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