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Thread: 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 $\displaystyle [(x+(\Delta)x]^3$. 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 $\displaystyle (x+(\Delta)x)$.

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

    f (x)= $\displaystyle x^3+x^2$

    My work thus far:

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

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

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

    The $\displaystyle (\Delta)x$ will cancel leaving:

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

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

    $\displaystyle x^2+2x$

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

    **To clarify, DIVIDED BY $\displaystyle (\Delta)x$ means everything on that line is divided by $\displaystyle (\Delta)x$. I can't figure out how to write $\displaystyle (\Delta)x$ without the parentheses being around Delta. Nor can I figure out how to get $\displaystyle (\Delta)x$ 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 $\displaystyle [(x+(\Delta)x]^3$. 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 $\displaystyle (x+(\Delta)x)$.

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

    f (x)= $\displaystyle x^3+x^2$

    My work thus far:

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

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

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

    The $\displaystyle (\Delta)x$ will cancel leaving:

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

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

    $\displaystyle x^2+2x$

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

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

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

    $\displaystyle = a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3$

    $\displaystyle = a^3 + 3a^2b + 3ab^2 + b^3$

    -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:
    $\displaystyle x^3 +x^2y + xy^2 + y^3$

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

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

    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 $\displaystyle [x+(\Delta)x]$. I know how to do $\displaystyle (a+b)^3$.

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

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

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

    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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