## anonymous 5 years ago Differentiate the given function: f(x)= ln (x+1)/(x-1) I get -2/(x-1)^2

1. anonymous

-2/ (x^2-1)

2. anonymous

I used the quotient rule, but not sure what to do with the ln

3. anonymous

did u have? ln(x+1/x-1)?

4. anonymous

if yes, then d/dx (lnx)= 1/x ... easy

5. anonymous

the ln is before the fraction

6. anonymous

Quotient rule u= ln (x+1) v= ln (x-1)

7. anonymous

If ln is over the whole thing ln [(x+1)/(x-1)] rewrite as [ln (x+1)]/[ln (x-1)]

8. anonymous

I don't know if this will help ln x+1 x-1 x+1/x-1 one big parenthesis around this fraction

9. anonymous

Can be rewritten as $\ln (x+1)\div \ln (x-1)$

10. anonymous

Top is u, bottom v

11. anonymous

I have not done it that way

12. anonymous

the u and v

13. anonymous

what shorthand you use for quotient rule f(x) or what?

14. anonymous

ok for quotient rule, I can do that but not sure what to do with the ln. After the quotient rule I get (x-1)-(x+1) / (x-1)^2

15. anonymous

Excuse me for using u and v. Let u=ln (x+1), let v=ln (x-1) u'=?, v'=? u' means u prime or derivative of u The derivative of the whole thing is $(u'v-uv')\div(v ^{2})$

16. anonymous

I am sorry but I am having a real hard time understanding this math. I have a book that has this problem worked out. I just don't understand why certain things are done the way they are. The last section we went over was dervatives and I did really well with them. These I am not.

17. anonymous

OK, so write one or two steps from book and say what you don't understand

18. anonymous

1st step f'(x) = 1/(x+1/x-1) d/dx (x+1/x-1) why did they put 1 over the problem.

19. anonymous

next step x-1/x+1 [(x-1)(1) - (x+1)(1)//(x-1)^2 the 2nd half I get, that is the quotient rule

20. anonymous

they are treating the whole fraction thingy as a single number. Let (x+1)/(x-1) be u (sorry) f' of ln u=1/u

21. anonymous

Deal with one question at a time or everything would get confused

22. anonymous

ok with step one 1/(x+1/x-1) then I guess they took the bottom # and flipped and multiplied. Is that how the problem got switched for step 2

23. anonymous

OK, back to original question what might make it a little confusing is what is called chain rule. the f' (x) = 1/[x+1/(x-1)] du in addition to this you must find the derivative of the inner function (x+1)/(x-1). Find the derivative of that and that is your du. This is a complicated function. I can explain more after you look at it.

24. anonymous

why can't I use the quotient rule

25. anonymous

by putting the 1 / .... is that taking the ln out of the problem

26. anonymous

There is a lot going on in this problem. It is not that you can't use quotient rule. In fact they are using quotient rule. But there are some intricacies that make their answer difficult to understand. Start from the top. Ask one piece by one piece what you don't understand.

27. anonymous

ok 1st step 1/ (x+1/x-1) if I use the quotient rule derv. is x+1/x-1

28. anonymous

never mind that isn't correct.

29. anonymous

Just write the book line by line and I will do the play by play

30. anonymous

1st step f'(x) = 1/(x+1/x-1) d/dx (x+1/x-1) 2nd step x-1/x+1 [(x-1)(1) - (x+1)(1)/(x-1)^2 step 3 -2/(x+1)(x-1) That is all steps

31. anonymous

Are you familiar with the thing called the chain rule?

32. anonymous

yes

33. anonymous

There approach is good. Now that I see it. They considered the ln (that thing) ln (that thing)=[1/(that thing)][multiplied by derivative of that thing] They didn't find it necessary to do the quotient rule to find f'(x), but in finding the derivative of (that thing) they did the quotient rule. Derivative of ln doesn't require work, you just put 1/something

34. anonymous

ok I half way understand that but how did they get x-1/x+1 in step 2 before the quotient rule

35. anonymous

sometimes when they work the problems out in the book it doesn't explain what they are doing. If I understand the work it does not confuss me.

36. anonymous

1/(B/C)=C/B

37. anonymous

?

38. anonymous

You have fractional property at the bottom is flipped.

39. anonymous

ok you flip it to get rid of the 1

40. anonymous

yes

41. anonymous

ok on step 2 x-1/x+1 [(x-1)(1)-(x+1)(1)/(x-1)^2 ok I can see the quotient rule what are they doine with the x-1/x+1, are they multipling that by the quotient?

42. anonymous

never mind something just clicked, I see it now. They cross multiplied