## anonymous 5 years ago I am stuck with an easy problem - I must be missing something. I am trying to use the limit process to find the derivative of f(x) = 8 / sqrt of x. Can you help me see where I am messing up?

1. anonymous

$f(x) = 8/\sqrt{x}$ find the derivative using the limit process

2. anonymous

f'(x) = lim $f^\prime(x) =\lim_{\Delta x \rightarrow 0} [(8/\sqrt{x+c}- 8/\sqrt{x})/c]$

3. anonymous

I am banging my head against this problem and getting a bad answer every time. We can tell by looking at it that the answer is$-(4/x^(3/2))$

4. anonymous

Well, I think you see what I mean...

5. anonymous

But I haven't figured out the algebra to leave me with a 4 in the numerator. I keep getting stuck with things that don't work out in the end.

6. anonymous

Weeku, what do you have for me so far?

7. anonymous

Whoops, I made a mistake. Darn editor. . . Hold on.

8. anonymous

thanks

9. anonymous

Seems no matter what I try, I end up with delta x /[( sqrt x + delta x) + sqrt x]

10. anonymous

I am going in circles.

11. anonymous

are you still there?

12. anonymous

Yeah. I'm stuck as well. Hrm. . . I'm still trying, though. Sorry xD Interesting problem, though.

13. anonymous

Ah. Got it. Took a whole loose-leaf paper to work out. . . Pretty nasty.

14. anonymous

Did you get up to $(8\sqrt{x} - 8\sqrt{x+h}) / (h \sqrt{x}\sqrt{x+h})$? Or, do you know how I got there? At that point, multiple the numerator and denominator by the conjugate of the numerator.

15. anonymous

multiply*

16. anonymous

Then it becomes a lot of distributing and factoring, until you can finally divide out the h. Once that happens, substitute 0 for h. Phew. I dunno if I can type out everything I did. . . I probably did it the long way, but I saw no other, shorter way. Do you want me to type it out?

17. anonymous

Yes, I got that far. I see you have h = delta x and you canceled out one h already.

18. anonymous

I would be grateful if you typed it out.

19. anonymous

Before that, all I did was find the common denominator of 8/sqrt[x+h] and 8/sqrt[x], add them together, and instead of dividing by h, multiply by 1/h to get $8\sqrt{x} - 8\sqrt{x+h} / h \sqrt{x} \sqrt{x+h}$ Then I factored out the 8 to get the above ^ Oh. Okay.

20. anonymous

Whoops. Never mind about factoring out the 8.

21. anonymous

I tried doing that, with no luck. :D

22. anonymous

$\frac{\frac{8}{\sqrt{x+h}} - \frac{8}{\sqrt{x}}}{h}$Firs Only the Numerator.... $8\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x+h}\times \sqrt{x}}$ Rationalization of Numerator $\frac{-8}{\sqrt{x+h}\times\sqrt{x}\times(\sqrt{x} + \sqrt{x+h})}$ $\frac{-8}{2\times x^{\frac{3}{2}}}$

23. anonymous

Ishaan94, I am looking over what you have done to see where I diverged. Thank you.

24. anonymous

I lost you after "Rationalization of Numerator," but nicely done. I did it the very, very long way, I see. . . Sorry about that, FreeTrader.

25. anonymous

$\Huge{\frac{\frac{-8\cancel{h}}{\sqrt{x+h}\times\sqrt{x}\times(\sqrt{x} + \sqrt{x+h})}}{\cancel{h}}}$

26. anonymous

$8\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x} \times \sqrt{x+h}} \times \frac{\sqrt{x} + \sqrt{x+h}}{\sqrt{x} + \sqrt{x+h}}$

27. anonymous

I see how Ishaan94 got to the "first only the numerator..." expression. But I am lost on how he got rid of the terms in the numerator.

28. anonymous

Right, multiply by conjugate.

29. anonymous

I really think you are going about this the wrong way. You should change the square root into a power (1/2) and then bring the denominator to the top by changing the power to negative. Then you can simply use the limit process of: f(x)= 8x^(-1/2) f(x+h)=8(x+h)^(1/2) And plug this into the derivative formula. You should get rid of the terms in the denominator.

30. anonymous

Then factor out a h and cancel.

31. anonymous

$8\frac{x - (x+h)}{\sqrt{x}\times \sqrt{x+h} \times (\sqrt{x} + \sqrt{x+h})}$

32. anonymous

I think it's dawning on me that I missed distributing the -1 on the h, too.

33. anonymous

Stick with me, I am going to check that on my paper.

34. anonymous

Ishaan94, when I perform the operation 8 *[ x -(x+h) ] I am left with 8x -8(x+h) which gives 8x -8x -8h

35. anonymous

Right

36. anonymous

Which gives -8h. divide out the h's.

37. anonymous

I am onto this now... working the algebra. searching for the lightswitch for the bulb over my head...

38. anonymous

$f(x)=\frac{8}{\sqrt{x}}$ $f(x)=\frac{8}{x ^{1/2}}$ $f(x)=8x ^{\frac{-1}{2}}$ $f(x+h)=8(x+h)^{-1/2}$ $f \prime(x)\lim_{h \rightarrow 0}=I can't remember the formula$ Now just use you derivative formula, cancel where necisary, factor out a h, then cancel with the denominator.

39. anonymous

Lol@chaise xD

40. anonymous

Lol

41. anonymous

My problem is now in the denominator of the numerator. I don't see a h to cancel the one against the -8.

42. anonymous

You're forgetting that the entire equation is over h at this point. $\Huge{\frac{\frac{-8\cancel{h}}{\sqrt{x+h}\times\sqrt{x}\times(\sqrt{x} + \sqrt{x+h})}}{\cancel{h}}}$

43. anonymous

I have that now. Thanks. I am down to working with the remaining denom terms.

44. anonymous

All terms on the right hand side of the minus sign in your derivative formula that have a x term should cancel with the terms. I think you're going about this the wrong way. Instead of having a quotient you should really bring the denominator to the top by making the power negative. Remember this rule? am^-x = a/m^x

45. anonymous

Yes. I am already so deep in the prob that I am going to finish it with frax.

46. anonymous

Here's what I have in the denominator, which I am tempted to misspell as demon-ator. $\sqrt{x} * \sqrt{x} * \sqrt{x+h} + \sqrt{x} * \sqrt{x+h} * \sqrt{x+h}$

47. anonymous

Since the h values go to zero in the limit, the following is the denominator in the limit: $2\sqrt{x}\sqrt{x}\sqrt{x}$

48. anonymous

$f(x)=\frac{8}{\sqrt{x}}$ $f(x)=\frac{8}{x ^{1/2}}$ $f(x)=8x ^{\frac{-1}{2}}$ $f(x+h)=8(x+h)^{-1/2}$ $\lim_{h \rightarrow 0}f'(x)=\frac{f(x+h)-f(x)}{h}$ lim{h-->0) of f'(x) = (8(x+h)^(-1/2) - 8x^(-1/2))/h You can do the rest, I'm sure.

49. anonymous

50. anonymous

Okay, I am losing my sanity. Can you help me remember if $\sqrt{x^3} = \sqrt{x}\sqrt{x}\sqrt{x}$ ???

51. anonymous

Yes - and this should be in the denominator of the equation once it has been solved. Notice how the ^(-1/2) becomes 1.5 when you increase the power by one. And yes it does FreeTrader

52. anonymous

Yup. That is a true statement.

53. anonymous

Stop writing it as a square root until you get to the final stage of the question. sqrt(x)^3 = x^(3/2)

54. anonymous

I have temp insanity.

55. anonymous

'Tis okay.:)

56. anonymous

So, my answer took me 7 hours to reach.

57. anonymous

It was totally worth it!

58. anonymous

THANK YOU FRIENDS!

59. anonymous

:)