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anonymous

  • one year ago

Need Help! Find derivative of f(x)=1-1/2sqrt(1-x)

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  1. anonymous
    • one year ago
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    I'm really confused on how to find the derivative of fraction square root part..do I do the quotient rule?

  2. anonymous
    • one year ago
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    oh no !!

  3. anonymous
    • one year ago
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    as much as i lectured before about not using rational exponents, that was when you had \[\frac{d}{dx}\sqrt{f(x)}\]

  4. Astrophysics
    • one year ago
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    \[f(x) = 1-\frac{ 1 }{ 2\sqrt{1-x} }?\]

  5. anonymous
    • one year ago
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    Yep, that's it astrophysics!

  6. anonymous
    • one year ago
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    for this one you probably do want to use rational exponents

  7. anonymous
    • one year ago
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    LOL THANKS SATTELITE!

  8. anonymous
    • one year ago
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    \[1-\frac{1}{2}(1-x)^{-\frac{1}{2}}\]

  9. Astrophysics
    • one year ago
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    ye was gonna do that

  10. anonymous
    • one year ago
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    then go nuts

  11. anonymous
    • one year ago
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    i guess you are looking for the second derivative of the previous one right?

  12. anonymous
    • one year ago
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    wow...so like why couldnt I use the rational exponents for the other question we were doing? and yea

  13. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    kk gotcha

  16. anonymous
    • one year ago
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    this one would be a bear to do without the power rule, but let me lecture once again

  17. anonymous
    • one year ago
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    |dw:1439263414965:dw|

  18. Astrophysics
    • one year ago
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    Power rule and chain rule

  19. anonymous
    • one year ago
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    DO NOT USE THE QUOTIENT RULE IF THE NUMERATOR IS A NUMBER !!

  20. Astrophysics
    • one year ago
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    ^ this, I see so many people applying quotient rule to problems like that

  21. anonymous
    • one year ago
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    i will pontificate more on that in a moment, lets finish this one

  22. anonymous
    • one year ago
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    Okay enough about bashing on mwah, let's finish the problem

  23. anonymous
    • one year ago
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    \[-\frac{1}{4}(1-x)^{-\frac{3}{2}}\] is your derivative right?

  24. anonymous
    • one year ago
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    Yea and then I solved it down to what I have above

  25. anonymous
    • one year ago
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    are you trying to set it equal to zero?

  26. anonymous
    • one year ago
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    yea...should I not

  27. anonymous
    • one year ago
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    well not you should not again the curse of rational exponents

  28. anonymous
    • one year ago
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    \[-\frac{1}{4\sqrt{(1-x)^2}}\]

  29. anonymous
    • one year ago
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    crap

  30. anonymous
    • one year ago
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    \[-\frac{1}{4\sqrt{(1-x)^3}}\]

  31. anonymous
    • one year ago
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    a fraction is only zero if the numerator is

  32. anonymous
    • one year ago
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    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
    • one year ago
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    my guess is you are asked about concavity right?

  34. anonymous
    • one year ago
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    yea

  35. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    oops \[\left(\frac{1}{x^2+2x}\right)'=-\frac{2x+2}{(x^2+2x)^2}\]

  38. anonymous
    • one year ago
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    Okay now im like super confused

  39. anonymous
    • one year ago
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    sorry what point?

  40. anonymous
    • one year ago
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    why cant I find its critical points, and find where f''(x) is undefined and thats it

  41. anonymous
    • one year ago
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    critical points are where the derivative is zero or undefined

  42. anonymous
    • one year ago
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    yea so, hollup lemme draw it for ya

  43. anonymous
    • one year ago
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    to find where the function is increasing resp decreasing you need to know where the derivative is positive resp negative

  44. anonymous
    • one year ago
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    |dw:1439264231886:dw|

  45. anonymous
    • one year ago
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    the problem here is the second derivative is NEVER 0

  46. anonymous
    • one year ago
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    |dw:1439264312465:dw|

  47. anonymous
    • one year ago
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    and then x=1 :)

  48. anonymous
    • one year ago
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    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
    • one year ago
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    AND im right bubba because I just checked in the back of the book

  50. anonymous
    • one year ago
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    ???

  51. anonymous
    • one year ago
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    what are you trying to find?

  52. anonymous
    • one year ago
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    where the function is CU or CD

  53. anonymous
    • one year ago
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    I guess I learned it a diff way than you

  54. anonymous
    • one year ago
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    ok good concave down for all x

  55. anonymous
    • one year ago
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    well for all x in the domain i.e. for all x in \((-\infty, 1)\)

  56. anonymous
    • one year ago
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    yep thats it!

  57. anonymous
    • one year ago
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    don't forget that 1 is the endpoint of the domain of the original function that means the derivative is undefined there

  58. anonymous
    • one year ago
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    and since the derivative is undefined at \(x=1\) it would be an amazing miracle if the second derivative was defined there !

  59. anonymous
    • one year ago
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    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
    • one year ago
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    yep, well thanks!

  61. anonymous
    • one year ago
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    yw

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