## anonymous 5 years ago Hey guys I'm having problems with derivatives if someone could explain the logx/lnx derivatives to me that would be great heres a few examples y = -2x^2(lnx)^4 or y = ln((x +1)/x)

1. anonymous

to start, do you know the derivative of $\ln(x)$ ?

2. anonymous

$d/dx [ \ln (x) ] = 1/x$

3. anonymous

yes i just dont understand how it applies when you have things attached to x

4. anonymous

such as the (lnx)^4 or the ln(x +1/x)

5. anonymous

ok

6. anonymous

let's try $\ln((x+1)/x)$

7. anonymous

alright

8. anonymous

have you used substitution rule before?

9. anonymous

That example is very simple, though the process might be tedious. Simply put a 1 over the expression then multiply by derivative of the expression

10. anonymous

No, i got that wrong, i was describing log.

11. anonymous

I'm contradicting myself, I think i told you right

12. anonymous

so, ln((x+1)/x)

13. anonymous

using the substitution rule: let's have $u = (x+1)/x , du = -1/x^2$

14. anonymous

du = =1/x^2 * dx

15. anonymous

16. anonymous

ok sorry the substitution thing is confusing me

17. anonymous

ok, well, the concept is thus: essentially you can simply things by taking a difficult expression to immeditaely differentitate, like here ln((x+1)/x), and simplify the first steps a bit

18. anonymous

outright, [ln((x+1)/x)]' looks difficult, but [ln(u)]'

19. anonymous

that is like a calc identity

20. anonymous

ok i get it

21. anonymous

the trick is keeping track of this substitution, you'll need to find the derivtative of u. as above, u=(x+1)/x, while du = (-1/x^2)*dx

22. anonymous

ahhh ok so if you set it as a variable it complicates things less

23. anonymous

so we have our first equation: ln((x+1)/x) dx

24. anonymous

yeah, you do less internal variable manipulation

25. anonymous

alright got it sorry

26. anonymous

cool, you got it?

27. anonymous

yep that helped a lottt

28. anonymous

so, you want to solve it? i can stay to help if you need it...

29. anonymous

but just to be sure once I have the x + 1 over x I have to then use quotient rule...?

30. anonymous

so, u = (x+1)/x ... but that equals x/x + 1/x

31. anonymous

so u = 1 + 1/x

32. anonymous

du = (1/x) dx

33. anonymous

er, (1/x)'

34. anonymous

which is ($x^-1$)'

35. anonymous

so, whats the derivative with respect to x of 1 + x^-1 ?

36. anonymous

-1x^-2...?

37. anonymous

yeah

38. anonymous

so, ln(u) du ---> but du = (-1/x^2)*dx

39. anonymous

ln(u)' --> 1/u, where u is (1 + 1/x) ....so you have $1/[(1 + 1/x)]$....then, you can't forget that du was -1/x^2 * dx, so when you now substitute back in for u so you have the original variable x, you have to multiply by (-1/x^2)

40. anonymous

$\frac{1}{(1+1/x)} * \frac{-1}{(x^2)}$

41. anonymous

which you can simplify

42. anonymous

to : -1/(x^2 + x)

43. anonymous

ok sorry my computers really crappy so i cant reply fast but i completely get it now thanks!

44. anonymous

no problem, hopefully now you can use the book examples much better too!

45. anonymous

also, here on OpenStudy, if someone gives you good help, you can award them with a medal here in the conversation

46. anonymous

so, ask more questions and maybe you can help some other math learners if you see a question you know to answer

47. anonymous

i'm off, but good luck with the calc ;)

48. anonymous

thank you very much