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Math Help - differentiation ..... :(

  1. #1
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    differentiation ..... :(

    can someone help me sove this ??????

    Differentiate (x^2 + 1)^8;
    Hence Find Sx(x^2 + 1)^7 dx

    thanks all ... ps can you post it step by step so i can understand it as well
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  2. #2
    MHF Contributor
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    (x^2 +1)^8

    You know d/dx (u)^n
    = n*[(u)^(n-1)]*du ?

    Let y = (x^2 +1)^8
    Differentiate both sides with respect to x,
    dy/dx = 8*(x^2 +1)^7 *(2x +0)
    dy/dx = 8(x^2 +1)^7 *(2x)
    dy/dx = 16x(x^2 +1)^7 ----****

    ------
    Sx(x^2 + 1)^7 dx

    INT.[x(x^2 +1)^7]dx

    You also know
    INT.[u^n]du = 1/(n+1) *[u^(n+1)] ?

    Be sure that the "du" outside the bracket is really the "du" of the "u" inside the bracket.

    INT.[x(x^2 +1)^7]dx
    The dx, or 1, outside the bracket is not the d/dx of x(x^2 +1)^7 inside the bracket.

    What if we bring out the x?

    = INT.[(x^2 +1)^7](x dx)
    The (x) is not the d/dx of the (x^2 +1).

    What is the d/dx of (x^2 +1)?
    It is 2x.
    So if we make the (x dx) outside the bracket into (2x dx), then this (2x) is already the d/dx of (x^2 +1).
    So we can integrate now?

    Not yet.
    In making (x dx) into (2x dx) we multiplied the (x dx) by 2. So we have to divide by 2, or multiply by (1/2), the new (2x dx) so that the original (x dx) is maintained.

    Hence,
    = INT.[(x^2 +1)^7](2x/2 dx)
    = INT.[(x^2 +1)^7](2x dx)(1/2)
    = (1/2)INT.[(x^2 +1)^7](2x dx)
    Now we can integrate,

    = (1/2)*(1/8)(x^2 +1)^8 +C
    = (1/16)(x^2 +1)^8 +C ----answer.

    -----
    You can ask if you want more explanations re this answer, as I do not know your "strength" in Calculus.
    Last edited by ticbol; May 13th 2005 at 05:54 AM.
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  3. #3
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    Quote Originally Posted by ticbol
    I do not know your "strength" in Calculus.
    well we sure know his strength in video games
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