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mathcalculus Group Title

can someone help me find the derivative of a square root step by step please.

  • 11 months ago
  • 11 months ago

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  1. mathcalculus Group Title
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    Sqrt(2-5*x^2)

    • 11 months ago
  2. mathcalculus Group Title
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    i did this: 2-5x^2) ^1/2

    • 11 months ago
  3. mathcalculus Group Title
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    then 1/2(2-5x^2)

    • 11 months ago
  4. mathcalculus Group Title
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    @zepdrix

    • 11 months ago
  5. zepdrix Group Title
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    Woops you didn't apply the power rule correctly.\[\large (2-5x^2)^{1/2}\]Derivative gives us,\[\large \frac{1}{2}(2-5x^2)^{1/2-1} \qquad=\qquad \frac{1}{2}(2-5x^2)^{-1/2}\]But we also have to apply the chain rule after that, multiplying by the derivative of the inside.

    • 11 months ago
  6. mathcalculus Group Title
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    uh-oh got me there... may i ask how?

    • 11 months ago
  7. mathcalculus Group Title
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    i think i kind of do but im not sure how to proceed it.

    • 11 months ago
  8. zepdrix Group Title
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    \[\large \color{royalblue}{\left[(2-5x^2)^{1/2}\right]'} \qquad = \qquad \frac{1}{2}(2-5x^2)^{-1/2}\color{royalblue}{(2-5x^2)'}\]We took the derivative of the `outer` function, apply the power rule. Now we have to multiply by the derivative of the inside. The prime is to show that we still need to take the derivative of that part. Taking that derivative gives us,\[\large \frac{1}{2}(2-5x^2)^{-1/2}\color{orangered}{(-10x)}\]

    • 11 months ago
  9. zepdrix Group Title
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    Confused? :O Too much stuff going on?

    • 11 months ago
  10. mathcalculus Group Title
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    im a little confused with how you got to 10x

    • 11 months ago
  11. zepdrix Group Title
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    It's the derivative of that blue thing. Take the derivative of each term inside the brackets, separately.

    • 11 months ago
  12. mathcalculus Group Title
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    isn't it this: f(x)'*(g(x)+f(x)*g(x)' ?

    • 11 months ago
  13. mathcalculus Group Title
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    i see but what happens to the half?

    • 11 months ago
  14. zepdrix Group Title
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    That would be the product rule:\[\large \left[f(x) \cdot g(x)\right]' \qquad=\qquad f'(x)g(x)+f(x)g'(x)\] We're dealing with a composition of functions, a function within a function. Which tells us to apply the chain rule:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\]

    • 11 months ago
  15. zepdrix Group Title
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    The chain rule tells us... After you've taken care of the outer function... (In our case our outer function is "something" to the 1/2 power) make a copy of the inside, take it's derivative, and multiply by it.

    • 11 months ago
  16. zepdrix Group Title
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    I hate using function notation like this, but maybe it'll help you to understand the chain rule. I find it more confusing, but to each his own. So we have this,\[\large f(x)=x^{1/2}\]\[\large g(x)=2-5x^2\] \[\large f(g(x))=(2-5x^2)^{1/2}\] Taking the derivative using the chain rule:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\] \[\large =\left[(2-5x^2)^{1/2}\right]'\left[2-5x^2\right]'\]

    • 11 months ago
  17. zepdrix Group Title
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    |dw:1375768509806:dw|If we have a function of a potato... The chain rule tells us that after we take the derivative of the function, we make a copy of the potato and take it's derivative as well.

    • 11 months ago
  18. zepdrix Group Title
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    It's a tricky concept to get used to :(

    • 11 months ago
  19. Psymon Group Title
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    Haha

    • 11 months ago
  20. Psymon Group Title
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    I like that one :P

    • 11 months ago
  21. zepdrix Group Title
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    Uh oh, your brain esplode? :( Want to see a nice easy example? One which includes no potatoes? :o

    • 11 months ago
  22. mathcalculus Group Title
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    i understand the -10x but what happens to the -1/2 exponent.

    • 11 months ago
  23. mathcalculus Group Title
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    lol

    • 11 months ago
  24. Psymon Group Title
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    Next time we should do f ' (wal-mart)

    • 11 months ago
  25. zepdrix Group Title
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    We're making a copy of the `inside` and taking it's derivative. The -1/2 is on the `outside`.

    • 11 months ago
  26. zepdrix Group Title
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    We're not making a copy of the `outside`, if that makes sense :o

    • 11 months ago
  27. mathcalculus Group Title
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    i know the chain rule but i dont understand the other one

    • 11 months ago
  28. mathcalculus Group Title
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    after you find -10x then what

    • 11 months ago
  29. zepdrix Group Title
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    We have this delicious Chicken breast seasoned with rosemary and all this good stuff. Then we stuff it with cheese and maybe even potatoes, cause that's how we roll. When we take the derivative of this chicken, the chain rule tells us to make a copy of the cheese and potatoes. This copy will not include any of the rosemary or seasoning from the initial chicken. After the -10x? We're done!

    • 11 months ago
  30. zepdrix Group Title
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    That last example was so ... unnecessary lol... I should delete that.

    • 11 months ago
  31. Psymon Group Title
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    Haha, we need the humor with these things sometimes, though : )

    • 11 months ago
  32. mathcalculus Group Title
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    noooo dont lol

    • 11 months ago
  33. mathcalculus Group Title
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    don't delete :) thank you

    • 11 months ago
  34. mathcalculus Group Title
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    now, the critical points are= x=10?

    • 11 months ago
  35. mathcalculus Group Title
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    i mean for x

    • 11 months ago
  36. mathcalculus Group Title
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    i know how to solve for y

    • 11 months ago
  37. zepdrix Group Title
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    Oh since addition is commutative, we can rearrange things a bit just to make it look prettier.\[\large \frac{1}{2}(2-5x^2)^{-1/2}\color{orangered}{(-10x)} \qquad=\qquad \frac{-5x}{\sqrt{2-5x^2}}\]

    • 11 months ago
  38. zepdrix Group Title
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    Since `multiplication` is commutative* Now we need to find critical points?

    • 11 months ago
  39. zepdrix Group Title
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    I kinda threw a bunch of steps in there when I simplified. Did any of that confuse you? :o

    • 11 months ago
  40. mathcalculus Group Title
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    the chain rule, yes

    • 11 months ago
  41. mathcalculus Group Title
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    oh simplied? uhhh no not really. i know how you did that.

    • 11 months ago
  42. mathcalculus Group Title
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    @zepdrix

    • 11 months ago
  43. mathcalculus Group Title
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    oh wait i understand th chain rule now. simply find the derivative of the "inner" which to me... is 2-5x^2

    • 11 months ago
  44. zepdrix Group Title
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    yah that's the inner thing :o take it's derivative, and multiply by it.

    • 11 months ago
  45. mathcalculus Group Title
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    rewrite and first and multiply by the inner's derivative

    • 11 months ago
  46. mathcalculus Group Title
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    okay i caught up with ya.

    • 11 months ago
  47. mathcalculus Group Title
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    now to find that x.....

    • 11 months ago
  48. mathcalculus Group Title
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    because they;re asking for the critical points.

    • 11 months ago
  49. mathcalculus Group Title
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    @zepdrix

    • 11 months ago
  50. mathcalculus Group Title
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    @Psymon

    • 11 months ago
  51. Psymon Group Title
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    I didn't want to interfere xD

    • 11 months ago
  52. mathcalculus Group Title
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    it's okay, team work =]

    • 11 months ago
  53. mathcalculus Group Title
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    @Psymon

    • 11 months ago
  54. zepdrix Group Title
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    To find critical points we set the derivative function equal to zero. Critical points exist anywhere the derivative is zero, or does not exist.\[\large 0=\frac{-5x}{\sqrt{2-5x^2}}\] So we can determine that a critical point exists when:\[\large 0=-5x \qquad\qquad\qquad 0=\sqrt{2-5x^2}\]

    • 11 months ago
  55. mathcalculus Group Title
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    x=5?

    • 11 months ago
  56. mathcalculus Group Title
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    those are not critical points..... :/ it should be written like this.. ex: (0,0)

    • 11 months ago
  57. zepdrix Group Title
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    From that first zero? Hmm it looks like if we divide both sides by -5, we get x=0.

    • 11 months ago
  58. zepdrix Group Title
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    I'm not sure where you're getting 5 from :o

    • 11 months ago
  59. zepdrix Group Title
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    No, normally you write critical points as x= Max and min are written as coordinate pairs. Maybe your teacher does it differently though :)

    • 11 months ago
  60. mathcalculus Group Title
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    oops! right x=0

    • 11 months ago
  61. zepdrix Group Title
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    How bout this: \[\large 0=\sqrt{2-5x^2}\] What critical points is it giving us?

    • 11 months ago
  62. Psymon Group Title
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    Yeah, I write crit points as x = as well, only a coordinate pair for min and max. Just to kinda back up zep, lol.

    • 11 months ago
  63. mathcalculus Group Title
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    does not exist?

    • 11 months ago
  64. zepdrix Group Title
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    Squaring both sides gives us,\[\large 0=2-5x^2\]Solving for x gives us,\[\large x=\pm \sqrt{2/5}\]

    • 11 months ago
  65. mathcalculus Group Title
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    wait a min i had that... :/ but it was x= sqrt(2/-5)

    • 11 months ago
  66. mathcalculus Group Title
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    because the - is there^..

    • 11 months ago
  67. zepdrix Group Title
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    Hmm you missed a negative somewhere :o Subtracting 2 from each side gives us -2 on the left. The dividing by -5 gives us positive 2/5 on the left.

    • 11 months ago
  68. mathcalculus Group Title
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    oops, right again.

    • 11 months ago
  69. mathcalculus Group Title
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    @ zepdrix thank you so much for your time and effort. i appreciate lot's! goodnight.

    • 11 months ago
  70. zepdrix Group Title
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    np c: cya

    • 11 months ago
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