## anonymous one year ago Need Help! Find derivative of f(x)=1-1/2sqrt(1-x)

1. anonymous

I'm really confused on how to find the derivative of fraction square root part..do I do the quotient rule?

2. anonymous

oh no !!

3. anonymous

as much as i lectured before about not using rational exponents, that was when you had $\frac{d}{dx}\sqrt{f(x)}$

4. Astrophysics

$f(x) = 1-\frac{ 1 }{ 2\sqrt{1-x} }?$

5. anonymous

Yep, that's it astrophysics!

6. anonymous

for this one you probably do want to use rational exponents

7. anonymous

LOL THANKS SATTELITE!

8. anonymous

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

9. Astrophysics

ye was gonna do that

10. anonymous

then go nuts

11. anonymous

i guess you are looking for the second derivative of the previous one right?

12. anonymous

wow...so like why couldnt I use the rational exponents for the other question we were doing? and yea

13. anonymous

of course you could, what i meant was it is not necessary once you know it because after you use the power rule, you are going to have to convert out of the rational exponents to do any sort of computation

14. anonymous

i wasn't saying you couldn't use it, i was saying you SHOULDN'T use it because you should just memorize the derivative of the square root function or its composition

15. anonymous

kk gotcha

16. anonymous

this one would be a bear to do without the power rule, but let me lecture once again

17. anonymous

|dw:1439263414965:dw|

18. Astrophysics

Power rule and chain rule

19. anonymous

DO NOT USE THE QUOTIENT RULE IF THE NUMERATOR IS A NUMBER !!

20. Astrophysics

^ this, I see so many people applying quotient rule to problems like that

21. anonymous

i will pontificate more on that in a moment, lets finish this one

22. anonymous

Okay enough about bashing on mwah, let's finish the problem

23. anonymous

$-\frac{1}{4}(1-x)^{-\frac{3}{2}}$ is your derivative right?

24. anonymous

Yea and then I solved it down to what I have above

25. anonymous

are you trying to set it equal to zero?

26. anonymous

yea...should I not

27. anonymous

well not you should not again the curse of rational exponents

28. anonymous

$-\frac{1}{4\sqrt{(1-x)^2}}$

29. anonymous

crap

30. anonymous

$-\frac{1}{4\sqrt{(1-x)^3}}$

31. anonymous

a fraction is only zero if the numerator is

32. anonymous

so forget about setting this equal to zero, it is never zero lesson: once you have the derivative (whatever method) get it out of exponential notation and write what it really is

33. anonymous

34. anonymous

yea

35. anonymous

ok the square root is always positive so minus the square root is always negative i.e. the second derivative is negative for all x in the domain

36. anonymous

btw$\left(\frac{1}{f}\right)'=-\frac{f'}{f^2}$ don't never ever use the quotient rule if the numerator is a constant$\left(\frac{1}{x^2+2x}\right)'=-\frac{2x+2}{x^2+2x}$

37. anonymous

oops $\left(\frac{1}{x^2+2x}\right)'=-\frac{2x+2}{(x^2+2x)^2}$

38. anonymous

Okay now im like super confused

39. anonymous

sorry what point?

40. anonymous

why cant I find its critical points, and find where f''(x) is undefined and thats it

41. anonymous

critical points are where the derivative is zero or undefined

42. anonymous

yea so, hollup lemme draw it for ya

43. anonymous

to find where the function is increasing resp decreasing you need to know where the derivative is positive resp negative

44. anonymous

|dw:1439264231886:dw|

45. anonymous

the problem here is the second derivative is NEVER 0

46. anonymous

|dw:1439264312465:dw|

47. anonymous

and then x=1 :)

48. anonymous

lost me on that one before we continue, let me point out that the function is always concave down that means the second derivative is always negative

49. anonymous

AND im right bubba because I just checked in the back of the book

50. anonymous

???

51. anonymous

what are you trying to find?

52. anonymous

where the function is CU or CD

53. anonymous

I guess I learned it a diff way than you

54. anonymous

ok good concave down for all x

55. anonymous

well for all x in the domain i.e. for all x in $$(-\infty, 1)$$

56. anonymous

yep thats it!

57. anonymous

don't forget that 1 is the endpoint of the domain of the original function that means the derivative is undefined there

58. anonymous

and since the derivative is undefined at $$x=1$$ it would be an amazing miracle if the second derivative was defined there !

59. anonymous

what you are interested in is not where the second derivative is zero, but rather where it is positive or negative in this case it is never zero, it is negative for all x

60. anonymous

yep, well thanks!

61. anonymous

yw