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

- anonymous

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- anonymous

the answer given is correct but not simplified....
use the formula \(\large [lny]'=\frac{y'}{y} \) to find the derivative...

- anonymous

then how do i use this formula for the problem

- anonymous

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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## More answers

- anonymous

no how do u find 1+1/x?
is the answer -1x-1/2

- anonymous

use the sum rule for derivatives...
\(\large [1+\frac{1}{x}]'=[1]'+[\frac{1}{x}]' \)
continue....

- anonymous

|dw:1357167466355:dw|
??

- anonymous

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...

- anonymous

wait but how do i simplify to get ...|dw:1357167740926:dw|

- anonymous

factor the top and bottom...

- PhoenixFire

\[[{x^n}]^{\prime }=nx^{n-1}\]
How did you get\[-x^{-{1\over 2}}\] shouldn't it be \[-x^{-2}=-{1\over x^2}\]

- anonymous

oops.... sorry.... @PhoenixFire is correct...

- anonymous

so what should i do should it look like this...|dw:1357167900420:dw|

- anonymous

yes.. ^^^,
now simplify....

- anonymous

how....|dw:1357167986477:dw|

- anonymous

try rewriting the denominator as...|dw:1357168045359:dw|

- anonymous

then can i just write the answer....|dw:1357168177116:dw| or no?!?!

- anonymous

how do i continue simplifying from what i have now?

- anonymous

i'm double checking....
looks like this is wrong....
|dw:1357168325298:dw|

- anonymous

well it says that is the derivative and answer and its never been wrong

- PhoenixFire

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.

- anonymous

the answer is wrong....
if you continue what we have, you'd get the answer produced by @PhoenixFire ...

- anonymous

and wolfram...
http://www.wolframalpha.com/input/?i=derivative+ln%281%2B1%2Fx%29

- anonymous

just continue what we have here...
|dw:1357168741789:dw|

- anonymous

yeah i believe both of u now. okay ill just tell my teacher that it is wrong.

- anonymous

so from the pic that u just put up

- anonymous

yes... just continue that and you'll get the correct answer..

- anonymous

idk exactly how to though

- anonymous

okay i was talking about the ln not log though to start the problem... is that original answer wrong still?

- anonymous

\(\huge \frac{\frac{-1}{x^2}}{\frac{x+1}{x}}=\frac{-1}{x^2}\cdot \frac{x}{x+1} \)

- anonymous

|dw:1357169117534:dw|

- anonymous

thanks for the help

- anonymous

\(\huge \frac{\frac{-1}{x^2}}{\frac{x+1}{x}}=\frac{-1}{\cancel {x^2}x}\cdot \frac{\cancel x1}{x+1} \)

- anonymous

@PhoenixFire , thanks for getting my back.... :)
i'm a little rusty in calc... :(

- anonymous

I like the complete lack of trying and still no understanding from this question.

- anonymous

this was funny to watch

- anonymous

When people don't try they don't succeed. It's that simple.

- anonymous

still funny seeing people trying to explain to someone thats lazy

- anonymous

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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- anonymous

There you go, I hope it's clear now

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