differentiate with respect to x: ln(x-1/x)

Question

amistre64 · Answer

looks simple enough, whats the issue?

anonymous · Answer

Apply chain rule

anonymous · Answer

answer is 1/(x-1)

anonymous · Answer

apparently the answer is x^2+1/x(x^2-1)

mendicant_bias · Answer

Are you stuck on a process or something? If you know the answer, what exactly is your question?

mendicant_bias · Answer

?

mendicant_bias · Answer

(Also, as far as I'm aware, that's not the right answer; I'll check with wolframalpha after this, but here's how I would do it): The derivative of ln(x) is always 1/(x), or more generally, the derivative of a natural log is always one divided by the argument of the log, applying the chain rule if necessary. So,
$$\frac{ d }{ dx } (\ln[(x-1)/x]) = \frac{ 1 }{ [(x-1)/x] }\times \frac{ (x)(x-1)\frac{ d }{ dx }-(x-1)(x)\frac{ d }{ dx } }{ x ^{2} }$$
From there, just simplify. But unless I'm wrong, that's how it's done.

anonymous · Answer

the answers are provided.
but thank you! :)

mendicant_bias · Answer

I just checked with wolframalpha, WA got 1/(x-1)x, which I don't really get, but that may just be the simplified version of what I was doing, but I doubt it. Either way, the answer you were provided sounds totally wrong.