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## kobe24 Group Title 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 one year ago one year ago

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1. dpaInc Group Title

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

2. kobe24 Group Title

then how do i use this formula for the problem

3. dpaInc Group Title

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 $$\large [1+\frac{1}{x}]'=$$ ????

4. kobe24 Group Title

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

5. dpaInc Group Title

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

6. kobe24 Group Title

|dw:1357167466355:dw| ??

7. dpaInc Group Title

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...

8. kobe24 Group Title

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

9. dpaInc Group Title

factor the top and bottom...

10. PhoenixFire Group Title

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

11. dpaInc Group Title

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

12. kobe24 Group Title

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

13. dpaInc Group Title

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

14. kobe24 Group Title

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

15. dpaInc Group Title

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

16. kobe24 Group Title

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

17. kobe24 Group Title

how do i continue simplifying from what i have now?

18. dpaInc Group Title

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

19. kobe24 Group Title

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

20. PhoenixFire Group Title

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.

21. dpaInc Group Title

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

22. dpaInc Group Title
23. dpaInc Group Title

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

24. kobe24 Group Title

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

25. kobe24 Group Title

so from the pic that u just put up

26. dpaInc Group Title

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

27. kobe24 Group Title

idk exactly how to though

28. kobe24 Group Title

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

29. dpaInc Group Title

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

30. kobe24 Group Title

|dw:1357169117534:dw|

31. kobe24 Group Title

thanks for the help

32. dpaInc Group Title

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

33. dpaInc Group Title

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

34. malevolence19 Group Title

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

35. Rowa Group Title

this was funny to watch

36. malevolence19 Group Title

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

37. Rowa Group Title

still funny seeing people trying to explain to someone thats lazy

38. GCR92 Group Title

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.

39. GCR92 Group Title

There you go, I hope it's clear now