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

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

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simplify derivative of 1/(x^2+1)+ln(1/x)

simplify

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

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

Solution steps

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

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{1}{x}) = ln (1) - ln (x)

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

\frac{7}{30} (ln (7) - ln (2))

lo g_{2} (x - 3) + lo g_{2} (x - 1)

ln (2) e^{- 6} + (- 2.268) \cdot ln (0.15)

lo g_{10} (525) 5

\frac{1}{2} ln (\frac{130}{100})

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