## kobe24 2 years ago 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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1. dpaInc

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

2. kobe24

then how do i use this formula for the problem

3. dpaInc

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

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

5. dpaInc

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

6. kobe24

|dw:1357167466355:dw| ??

7. dpaInc

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

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

9. dpaInc

factor the top and bottom...

10. PhoenixFire

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

11. dpaInc

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

12. kobe24

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

13. dpaInc

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

14. kobe24

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

15. dpaInc

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

16. kobe24

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

17. kobe24

how do i continue simplifying from what i have now?

18. dpaInc

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

19. kobe24

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

20. 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.

21. dpaInc

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

22. dpaInc
23. dpaInc

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

24. kobe24

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

25. kobe24

so from the pic that u just put up

26. dpaInc

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

27. kobe24

idk exactly how to though

28. kobe24

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

29. dpaInc

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

30. kobe24

|dw:1357169117534:dw|

31. kobe24

thanks for the help

32. dpaInc

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

33. dpaInc

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

34. malevolence19

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

35. Rowa

this was funny to watch

36. malevolence19

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

37. Rowa

still funny seeing people trying to explain to someone thats lazy

38. GCR92

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

There you go, I hope it's clear now