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mathcalculus

  • one year ago

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

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  1. mathcalculus
    • one year ago
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    Sqrt(2-5*x^2)

  2. mathcalculus
    • one year ago
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    i did this: 2-5x^2) ^1/2

  3. mathcalculus
    • one year ago
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    then 1/2(2-5x^2)

  4. mathcalculus
    • one year ago
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    @zepdrix

  5. zepdrix
    • one year ago
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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.

  6. mathcalculus
    • one year ago
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    uh-oh got me there... may i ask how?

  7. mathcalculus
    • one year ago
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    i think i kind of do but im not sure how to proceed it.

  8. zepdrix
    • one year ago
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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,\[\large \frac{1}{2}(2-5x^2)^{-1/2}\color{orangered}{(-10x)}\]

  9. zepdrix
    • one year ago
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    Confused? :O Too much stuff going on?

  10. mathcalculus
    • one year ago
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    im a little confused with how you got to 10x

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

  12. mathcalculus
    • one year ago
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    isn't it this: f(x)'*(g(x)+f(x)*g(x)' ?

  13. mathcalculus
    • one year ago
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    i see but what happens to the half?

  14. zepdrix
    • one year ago
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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:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\]

  15. zepdrix
    • one year ago
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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.

  16. zepdrix
    • one year ago
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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:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\] \[\large =\left[(2-5x^2)^{1/2}\right]'\left[2-5x^2\right]'\]

  17. zepdrix
    • one year ago
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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.

  18. zepdrix
    • one year ago
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    It's a tricky concept to get used to :(

  19. Psymon
    • one year ago
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    Haha

  20. Psymon
    • one year ago
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    I like that one :P

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

  22. mathcalculus
    • one year ago
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    i understand the -10x but what happens to the -1/2 exponent.

  23. mathcalculus
    • one year ago
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    lol

  24. Psymon
    • one year ago
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    Next time we should do f ' (wal-mart)

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

  26. zepdrix
    • one year ago
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    We're not making a copy of the `outside`, if that makes sense :o

  27. mathcalculus
    • one year ago
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    i know the chain rule but i dont understand the other one

  28. mathcalculus
    • one year ago
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    after you find -10x then what

  29. zepdrix
    • one year ago
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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!

  30. zepdrix
    • one year ago
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    That last example was so ... unnecessary lol... I should delete that.

  31. Psymon
    • one year ago
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    Haha, we need the humor with these things sometimes, though : )

  32. mathcalculus
    • one year ago
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    noooo dont lol

  33. mathcalculus
    • one year ago
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    don't delete :) thank you

  34. mathcalculus
    • one year ago
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    now, the critical points are= x=10?

  35. mathcalculus
    • one year ago
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    i mean for x

  36. mathcalculus
    • one year ago
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    i know how to solve for y

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

  38. zepdrix
    • one year ago
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    Since `multiplication` is commutative* Now we need to find critical points?

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

  40. mathcalculus
    • one year ago
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    the chain rule, yes

  41. mathcalculus
    • one year ago
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    oh simplied? uhhh no not really. i know how you did that.

  42. mathcalculus
    • one year ago
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    @zepdrix

  43. mathcalculus
    • one year ago
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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

  44. zepdrix
    • one year ago
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    yah that's the inner thing :o take it's derivative, and multiply by it.

  45. mathcalculus
    • one year ago
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    rewrite and first and multiply by the inner's derivative

  46. mathcalculus
    • one year ago
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    okay i caught up with ya.

  47. mathcalculus
    • one year ago
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    now to find that x.....

  48. mathcalculus
    • one year ago
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    because they;re asking for the critical points.

  49. mathcalculus
    • one year ago
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    @zepdrix

  50. mathcalculus
    • one year ago
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    @Psymon

  51. Psymon
    • one year ago
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    I didn't want to interfere xD

  52. mathcalculus
    • one year ago
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    it's okay, team work =]

  53. mathcalculus
    • one year ago
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    @Psymon

  54. zepdrix
    • one year ago
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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:\[\large 0=-5x \qquad\qquad\qquad 0=\sqrt{2-5x^2}\]

  55. mathcalculus
    • one year ago
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    x=5?

  56. mathcalculus
    • one year ago
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    those are not critical points..... :/ it should be written like this.. ex: (0,0)

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

  58. zepdrix
    • one year ago
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    I'm not sure where you're getting 5 from :o

  59. zepdrix
    • one year ago
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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 :)

  60. mathcalculus
    • one year ago
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    oops! right x=0

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

  62. Psymon
    • one year ago
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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.

  63. mathcalculus
    • one year ago
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    does not exist?

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

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

  66. mathcalculus
    • one year ago
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    because the - is there^..

  67. zepdrix
    • one year ago
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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.

  68. mathcalculus
    • one year ago
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    oops, right again.

  69. mathcalculus
    • one year ago
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    @ zepdrix thank you so much for your time and effort. i appreciate lot's! goodnight.

  70. zepdrix
    • one year ago
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    np c: cya

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