anonymous
  • anonymous
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
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
the answer given is correct but not simplified.... use the formula \(\large [lny]'=\frac{y'}{y} \) to find the derivative...
anonymous
  • anonymous
then how do i use this formula for the problem
anonymous
  • 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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anonymous
  • anonymous
no how do u find 1+1/x? is the answer -1x-1/2
anonymous
  • anonymous
use the sum rule for derivatives... \(\large [1+\frac{1}{x}]'=[1]'+[\frac{1}{x}]' \) continue....
anonymous
  • anonymous
|dw:1357167466355:dw| ??
anonymous
  • 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
  • anonymous
wait but how do i simplify to get ...|dw:1357167740926:dw|
anonymous
  • anonymous
factor the top and bottom...
PhoenixFire
  • 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
  • anonymous
oops.... sorry.... @PhoenixFire is correct...
anonymous
  • anonymous
so what should i do should it look like this...|dw:1357167900420:dw|
anonymous
  • anonymous
yes.. ^^^, now simplify....
anonymous
  • anonymous
how....|dw:1357167986477:dw|
anonymous
  • anonymous
try rewriting the denominator as...|dw:1357168045359:dw|
anonymous
  • anonymous
then can i just write the answer....|dw:1357168177116:dw| or no?!?!
anonymous
  • anonymous
how do i continue simplifying from what i have now?
anonymous
  • anonymous
i'm double checking.... looks like this is wrong.... |dw:1357168325298:dw|
anonymous
  • anonymous
well it says that is the derivative and answer and its never been wrong
PhoenixFire
  • 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
  • anonymous
the answer is wrong.... if you continue what we have, you'd get the answer produced by @PhoenixFire ...
anonymous
  • anonymous
and wolfram... http://www.wolframalpha.com/input/?i=derivative+ln%281%2B1%2Fx%29
anonymous
  • anonymous
just continue what we have here... |dw:1357168741789:dw|
anonymous
  • anonymous
yeah i believe both of u now. okay ill just tell my teacher that it is wrong.
anonymous
  • anonymous
so from the pic that u just put up
anonymous
  • anonymous
yes... just continue that and you'll get the correct answer..
anonymous
  • anonymous
idk exactly how to though
anonymous
  • anonymous
okay i was talking about the ln not log though to start the problem... is that original answer wrong still?
anonymous
  • anonymous
\(\huge \frac{\frac{-1}{x^2}}{\frac{x+1}{x}}=\frac{-1}{x^2}\cdot \frac{x}{x+1} \)
anonymous
  • anonymous
|dw:1357169117534:dw|
anonymous
  • anonymous
thanks for the help
anonymous
  • anonymous
\(\huge \frac{\frac{-1}{x^2}}{\frac{x+1}{x}}=\frac{-1}{\cancel {x^2}x}\cdot \frac{\cancel x1}{x+1} \)
anonymous
  • anonymous
@PhoenixFire , thanks for getting my back.... :) i'm a little rusty in calc... :(
anonymous
  • anonymous
I like the complete lack of trying and still no understanding from this question.
anonymous
  • anonymous
this was funny to watch
anonymous
  • anonymous
When people don't try they don't succeed. It's that simple.
anonymous
  • anonymous
still funny seeing people trying to explain to someone thats lazy
anonymous
  • 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
  • anonymous
There you go, I hope it's clear now
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