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Differentiation Help

  1. Jun 21, 2006 #1
    Greetings. I was just wondering if someone could take a look and tell me I did the following correctly:

    1)

    y=8x^4-5x^2-2/4x^3

    (32x^3-10x)(4x^3)-(12x^2)(8x^4-5x^2-2)/(4x^3)^2

    128x^6-40x^4+96x^6-60x^4-24x^2/(4x^3)^2

    224x^6-100x^4-24x/16x^6

    4x(56x^5-25x^3-24)/16x^6

    y’=56x^5-25x^3-24/4x^5

    2) y=sqroot 5x - sqroot x/5
    y=(5x)^1/2 - (x/5)^1/2
    y'=1/2(5x)^-1/2 (5) -1/2(x/5)^-1/2 (1/5)
    y'=2/5(5x)^-1/2 -1/10(x/5)^-1/2
    y'=2/5sqroot(5x) - 1/10(sqroot x/5)
     
    Last edited: Jun 21, 2006
  2. jcsd
  3. Jun 21, 2006 #2

    arildno

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    Dearly Missed

    1. You should use parentheses in order to indecate numerators and denominator in an unamiguous manner.
    2. In your first line, your numerator is 4x^2, not 4x^3

    3. In your third line, you've forgotten to change signs when removing a parenthesis
     
  4. Jun 21, 2006 #3
    I typoed on the first one. Let me edit it. Thanks.
     
  5. Jun 21, 2006 #4

    arildno

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    Well, it was item 3 that makes your answer wrong.
     
  6. Jun 21, 2006 #5
    Okay I got the 1st one now for sure. Is the 2nd one correct?
     
  7. Jun 21, 2006 #6
    Okay one more question:

    2x^3+2y^3-9xy=0

    dy/dx = 9y+6x^2/9x-6y^2?
     
  8. Jun 21, 2006 #7

    HallsofIvy

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    You were just "bawled out" for not using parentheses!

    No, the dy/dx is NOT (9y+ 6x^2)/(9x- 6y^2). You' missed a negative signs.
     
  9. Jun 22, 2006 #8

    VietDao29

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    You are wrong when going from the former line to the latter one.
    It should read:
    [tex]\frac{5}{2 \sqrt{5x}} - \frac{1}{10 \sqrt{\frac{x}{5}}}[/tex]
    It's 5/2, not 2/5. :)
    There's another way to do the first problem. Divide all by the denominator to get:
    [tex]y = \frac{8x ^ 4 - 5 x ^ 2 - 2}{4x ^ 3} = 2x - \frac{5}{4x} - \frac{1}{2x ^ 3}[/tex]
    Now let's differentiate it with respect to x, we have:
    [tex]y' = 2 + \frac{5}{4x ^ 2} + \frac{3}{2x ^ 4}[/tex]
    For the last problem, as HOI pointed out, you've missed quite a few negative signs. :)
     
  10. Jun 22, 2006 #9
    Thank you very much for your help! I found my errors now and found out what I did wrong. Thanks.

    Instead of making another thread, I have one last question:

    Given f(x) = x^3-0.5x^2+1, find the value of x where dy/dx = 4

    So I found the derivative: f'(x)=3x^2-x
    Tried setting it equal to 4

    4=3x^2-x

    But where do I go from there? Do I rearrange it and use quadratic formula to find the value of x?
     
  11. Jun 22, 2006 #10
    that should work :smile:
     
  12. Jun 22, 2006 #11

    Hootenanny

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    You could do that but an easier option would be to factorise;

    [tex]3x^2 - x - 4 = (3x-4)(x+1) = 0[/tex]

    :wink:
     
  13. Jun 22, 2006 #12

    HallsofIvy

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    "HOI"??

    I have a friend who uses the ename "Hog on Ice" which he abbreviates as HOI!
     
  14. Jun 23, 2006 #13

    Hootenanny

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    I've heard of Bamby on Ice, but never "Hog on Ice" :rofl: :rofl: . A quick google didn't reveal its meaning, perhaps you could enlighten us HOI? :wink:
     
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