derivative, you see any mistake?

derivative $$\large \frac{x-2}{1+\log|x-2|} f \prime (x) = \frac{1(1+\log|x-2|) - \frac{x-2}{x-2}}{(1+\log|x-2|)^2}$$

$$\large \frac{\log|x-2|}{(1+\log|x-2|)^2}$$

Question

derivative, you see any mistake?

derivative $$\large \frac{x-2}{1+\log|x-2|} f \prime (x) = \frac{1(1+\log|x-2|) - \frac{x-2}{x-2}}{(1+\log|x-2|)^2}$$

$$\large \frac{\log|x-2|}{(1+\log|x-2|)^2}$$

anonymous · Answer

No, that looks right to me.

anonymous · Answer

Assuming that log indicates the natural log, of course.

alfie · Answer

fact is, I type it in wolframalpha... http://www.wolframalpha.com/input/?i=derivative+%28x-2%29%2F%281%2Blog%7Cx-2%7C%29 I get this thingy, fine, it does some other step, but if I type in my thingy and its thingy and put them equal to each other it doesn't really return "true" even though the plots, now that I notice, are pretty much the same. o_O http://www.wolframalpha.com/input/?i=%282+log%28%28-2%2Bx%29%5E2%29%29%2F%282%2Blog%28%28-2%2Bx%29%5E2%29%29%5E2+%3D+%28log%7Cx-2%7C%29%2F%281%2Blog%7Cx-2%7C%29%5E2

anonymous · Answer

i had wolfram alpha give me a bad integral yesterday, sometimes it does weird stuff

alfie · Answer

Oh well, anyway I just noticed the two plots match, so I am just going to assume I did it correctly and I'm still able to take a derivative :D... thank you ;)

anonymous · Answer

It's entirely possible that it gets confused with the absolute value part when you use the chain rule on the log function because |x-2| is not differentiable at x=2.