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what is the derivative of ln(1+1/x)? they say the answer is xsquared-1/ xcubed+x now can someone explain how you find the answer

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the answer given is correct but not simplified.... use the formula \(\large [lny]'=\frac{y'}{y} \) to find the derivative...
then how do i use this formula for the problem
let \(\large y=1+\frac{1}{x} \) so, \(\large [ln(1+\frac{1}{x})]'=\frac{[1+\frac{1}{x}]'}{1+\frac{1}{x}} \) can you find \(\large [1+\frac{1}{x}]'= \) ????

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no how do u find 1+1/x? is the answer -1x-1/2
use the sum rule for derivatives... \(\large [1+\frac{1}{x}]'=[1]'+[\frac{1}{x}]' \) continue....
|dw:1357167466355:dw| ??
ok... good... so, \(\large [ln(1+\frac{1}{x})]'=\frac{[1+\frac{1}{x}]'}{1+\frac{1}{x}}=\frac{-x^{(\frac{-1}{2})}}{1+\frac{1}{x}} \) now just simplify...
wait but how do i simplify to get ...|dw:1357167740926:dw|
factor the top and bottom...
\[[{x^n}]^{\prime }=nx^{n-1}\] How did you get\[-x^{-{1\over 2}}\] shouldn't it be \[-x^{-2}=-{1\over x^2}\]
oops.... sorry.... @PhoenixFire is correct...
so what should i do should it look like this...|dw:1357167900420:dw|
yes.. ^^^, now simplify....
try rewriting the denominator as...|dw:1357168045359:dw|
then can i just write the answer....|dw:1357168177116:dw| or no?!?!
how do i continue simplifying from what i have now?
i'm double checking.... looks like this is wrong.... |dw:1357168325298:dw|
well it says that is the derivative and answer and its never been wrong
Every time I do it I get \(\large -{1 \over x^2+x}\) I think the answer that it's given you is wrong. Unless you are accidentally looking at the wrong question in the answer section. I do that all the time.
the answer is wrong.... if you continue what we have, you'd get the answer produced by @PhoenixFire ...
and wolfram...
just continue what we have here... |dw:1357168741789:dw|
yeah i believe both of u now. okay ill just tell my teacher that it is wrong.
so from the pic that u just put up
yes... just continue that and you'll get the correct answer..
idk exactly how to though
okay i was talking about the ln not log though to start the problem... is that original answer wrong still?
\(\huge \frac{\frac{-1}{x^2}}{\frac{x+1}{x}}=\frac{-1}{x^2}\cdot \frac{x}{x+1} \)
thanks for the help
\(\huge \frac{\frac{-1}{x^2}}{\frac{x+1}{x}}=\frac{-1}{\cancel {x^2}x}\cdot \frac{\cancel x1}{x+1} \)
@PhoenixFire , thanks for getting my back.... :) i'm a little rusty in calc... :(
I like the complete lack of trying and still no understanding from this question.
this was funny to watch
When people don't try they don't succeed. It's that simple.
still funny seeing people trying to explain to someone thats lazy
I don't know if your teacher want's you to simplify that, that's not the kind of "simplifying exercise" it looks like it's just to learn the Ln derivative, I took a picture, sorry i'm very traditional, hope it helps you.
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There you go, I hope it's clear now
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