# Differentiat-log(x+1/x)?

## 3 Answers

- No01
- g(x) = (x+1/x)

g'(x) = 1 - 1/x²

f(x) = -ln(g(x)) = - g'(x)/g(x) = - (1-1/x²)/(x+1/x)

= -(x²-1)/(x³+x)

= (1-x²)/(x + x³)

Why ask the same question so many times?00 - I guess (?) that's a minus sign just after the 2nd t ?

d[ -log(x + 1/x) ]/dx

= -[1/(x + 1/x)]*(1 - 1/x^2)

= -[x/(x^2 + 1)]*[(x^2-1)/x^2]

= (1 - x^2)/(x^3 + x).00

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