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Differentiation problem.

  1. Oct 24, 2007 #1
    This isnt a homework problem but something that stumped me in selfstudy, however i see this to be the most approperiate place to ask for help nevertheless.

    This is straight out of Adan's calculus 6e, Chain rule section.

    y= (x^5)sqrt(3+x^6) / (4+x^2)^3

    To me it looks like a combination of the chain rule, product rule and quotient rule. But i donät see how to go about solving it, whats the proper order to apply the rules?
    Also I don¨t see what the outside function is supposed to be for the whole expression.

    Any help would be much appreciated, and general tips for problems requiring multiple differentiation rules would be even more so.

    Thanks.
     
  2. jcsd
  3. Oct 24, 2007 #2

    Integral

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    Apply the quotient rule, then the chain rule as required to complete the derivatives called for by the quotient rule.
     
  4. Oct 24, 2007 #3

    HallsofIvy

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    Since, here, the denominator is fairly complex, I would be inclined to write it as
    [itex]y= (x^5)\sqrt{3+x^6} (4+x^2)^{-3}[/itex] and use just the product rule (and, of course, the chain rule).
    Perhaps even better would be [itex]y= (x^5)(3+x^6)^{1/2} (4+x^2)^{-3}[/itex]
     
  5. Oct 25, 2007 #4
    Im not sure how to apply the product rule when the number of terms is more than 2.

    1. Find the derivatives of the components of den and num, namely
    X^5 by power rule = 5x^4
    (3+x^6)^1/2 by chain rule = 1/2(3+x^6)^-1/2 (6x^5)
    (4+x^2)^3 by chain rule = 3(4+x^2)^2 (2x)

    2. Apply product rule to numerator:
    f '(num) =(x^5)((1/2(3+x^6)^-1/2(6x^5)) + (5x^4)((3+x^6)^1/2)

    3. Plug parts into quotient formula:

    It looks like a big mess to me.
     
  6. Oct 25, 2007 #5

    HallsofIvy

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    It's a straight forward extension. If u, v, and w are all functions of x, to find (uvw)' think of it as ((uv)w)' and apply the product rule to the two functions uv and w:
    ((uv)w)'= (uv)'w+ uvw'. Now apply the product rule to (uv)': (uvw)'= (u'v+ uv')w+ uvw'= u'vw+ uv'w+ uvw'. The extension to any number of factors should be obvious: with n factors you will have n terms each with a different factor differentiated.
     
  7. Oct 25, 2007 #6
    Ah I see, thanks!
    Btw can I delete threads that I created, or is that a mod privilege?
     
  8. Oct 25, 2007 #7

    HallsofIvy

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    No, you have to leave them their so everyone can laugh at how foolish you were- just like they do the rest of us!
     
  9. Oct 26, 2007 #8
    Ah, the shame :C
     
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