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Math Help - Cubing (x+∆x)

  1. #1
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    Cubing (x+∆x)

    We just started differential equations in my AP Calculus class, and the only thing I am having trouble with is [(x+(\Delta)x]^3. I tried using the binomial theorem (for cubing) but I get the wrong answer at the end. The only place I think I could be messing up is cubing (x+(\Delta)x).

    The problem:
    Use the definition of the derivative to find f'(x).

    f (x)= x^3+x^2

    My work thus far:

    f' (x)= \lim_{(\Delta)x\rightarrow0} [(x+(\Delta)x]^3 + [(x+(\Delta)x]^2 - (x^3+x^2) DIVIDED BY (\Delta)x

    = x^3+x^2(\Delta)x+x(\Delta)x^2+(\Delta)x^3+x^2+2x(\  Delta)x+(\Delta)x^2-x^3-x^2 DIVIDED BY (\Delta)x

    =  (\Delta)x[x^2+x(\Delta)x+(\Delta)x^2+2x+(\Delta)x] DIVIDED BY (\Delta)x

    The (\Delta)x will cancel leaving:

    [x^2+x(\Delta)x+(\Delta)x^2+2x+(\Delta)x]

    I plugged 0 in for (\Delta)x and was left with

    x^2+2x

    But I know this is the wrong answer (the right answer should be 3x^2+2x)

    **To clarify, DIVIDED BY (\Delta)x means everything on that line is divided by (\Delta)x. I can't figure out how to write (\Delta)x without the parentheses being around Delta. Nor can I figure out how to get (\Delta)x in the denominator of the whole expression. I tried though.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by yeloc View Post
    We just started differential equations in my AP Calculus class, and the only thing I am having trouble with is [(x+(\Delta)x]^3. I tried using the binomial theorem (for cubing) but I get the wrong answer at the end. The only place I think I could be messing up is cubing (x+(\Delta)x).

    The problem:
    Use the definition of the derivative to find f'(x).

    f (x)= x^3+x^2

    My work thus far:

    f' (x)= \lim_{(\Delta)x\rightarrow0} [(x+(\Delta)x]^3 + [(x+(\Delta)x]^2 - (x^3+x^2) DIVIDED BY (\Delta)x

    = x^3+x^2(\Delta)x+x(\Delta)x^2+(\Delta)x^3+x^2+2x(\  Delta)x+(\Delta)x^2-x^3-x^2 DIVIDED BY (\Delta)x

    =  (\Delta)x[x^2+x(\Delta)x+(\Delta)x^2+2x+(\Delta)x] DIVIDED BY (\Delta)x

    The (\Delta)x will cancel leaving:

    [x^2+x(\Delta)x+(\Delta)x^2+2x+(\Delta)x]

    I plugged 0 in for (\Delta)x and was left with

    x^2+2x

    But I know this is the wrong answer (the right answer should be 3x^2+2x)

    **To clarify, DIVIDED BY (\Delta)x means everything on that line is divided by (\Delta)x. I can't figure out how to write (\Delta)x without the parentheses being around Delta. Nor can I figure out how to get (\Delta)x in the denominator of the whole expression. I tried though.
    You got the wrong answer because
    (a + b)^3 = (a + b) \cdot (a + b)^2

    = (a + b) \cdot (a^2 + 2ab + b^2)

    = a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3

    = a^3 + 3a^2b + 3ab^2 + b^3

    -Dan
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  3. #3
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    When I expand, I usually use Pascal's triangle. I know that the powers of the first term descend and second term ascend. For cubes, the coefficients are:
    1 3 3 1

    (x+y)^3 expands into:
    x^3 +x^2y + xy^2 + y^3

    Multiply by the coefficients:
    x^3 + 3x^2y + 3xy^2 + y^3

    Your mistake was that you didn't multiply by the coefficients. The binomial theorem is:
    (x+y)^n = \sum_{k=0}^{n}\ nCk\ x^{n-k}y^k

    You can use combinations to find the coefficients as well.
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  4. #4
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    The thing that was throwing me off was cubing [x+(\Delta)x]. I know how to do  (a+b)^3.

    Would it be a big deal if I used another variable for (\Delta)x and just plugged 0 in for that variable at the end?

    Thanks for the responses and I figured out by trial and error that [x+(\Delta)x]^3 equals:

     x^3+3x^2(\Delta)x+3x(\Delta)x^2+(\Delta)x^3

    I guess I am seeing the two x's in the problem and getting confused.
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  5. #5
    Super Member 11rdc11's Avatar
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    If it helps you not get confused do it. Just sub the delta x back when you finish
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