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Math Help - Differentiate by definition

  1. #1
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    Differentiate by definition

    Can someone please confirm I am doing this correctly? Thanks!
    Y=x2-2x+2 at P(x,y), let Q (x+Δx, y+Δy) be another point on the curve. Differentiate by definition
    y+
    Δy=(x+Δx)2 -2(x+Δx)+2
    =X2 +2x2 (
    Δx)+2x(Δx)2+(Δx)2-2x-2Δx+2
    Δy=2x2Δx+2x(Δx)2+(Δx)2-2Δx
    Δy/Δx=2x2+2xΔx+(Δx)2-2
    dy/dx=lim/
    Δx->0 [2x2+2xΔx+(Δx)2-2]
    dy/dx=2x2-2

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  2. #2
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    Re: Differentiate by definition

    Cal...

    (x+Δx)^2=Χ^2+2(Χ)(ΔΧ)+(ΔΧ)^2 ΑΝD -2(Χ+ΔΧ) = -2Χ-2ΔΧ
    CORRECT YOUR SECOND LINE (EXPANSION)
    Anyway the final result is ok.

    Minoas
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  3. #3
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    Re: Differentiate by definition

    Can someone please confirm I am doing this correctly? Thanks!
    Y=x^2-2x+2 at P(x,y), let Q (x+Δx, y+Δy) be another point on the curve. Differentiate by definition
    y+Δy=(x+Δx)^2 -2(x+Δx)+2
    =X^2 +2x^2 (Δx)+2x(Δx)^2+(Δx)^2-2x-2Δx+2
    Δy=2x^2Δx+2x(Δx)^2+(Δx)^2-2Δx
    Δy/Δx=2x^2+2xΔx+(Δx)^2-2
    dy/dx=lim/Δx->0 [2x2+2xΔx+(Δx)2-2]
    dy/dx=2x2-2
    y = x^2 - 2x + 2

    y + \Delta y = (x + \Delta x)^2 - 2(x + \Delta x) +2 = x^2 + 2x\Delta x + (\Delta x)^2 - 2x - 2\Delta x + 2

    \Delta y = 2x\Delta x + (\Delta x)^2 - 2\Delta x

    \frac{\Delta y}{\Delta x} = 2x + \Delta x - 2

    \frac{dy}{dx} = \lim_{\Delta x \rightarrow 0}(2x + \Delta x - 2) = 2x - 2
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  4. #4
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    Re: Differentiate by definition

    -
    Last edited by jpritch422; March 23rd 2013 at 01:53 PM.
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  5. #5
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    Re: Differentiate by definition

    Quote Originally Posted by MINOANMAN View Post
    Cal...

    (x+Δx)^2=Χ^2+2(Χ)(ΔΧ)+(ΔΧ)^2 ΑΝD -2(Χ+ΔΧ) = -2Χ-2ΔΧ
    CORRECT YOUR SECOND LINE (EXPANSION)
    Anyway the final result is ok.
    No, the final result is not correct.

    Minoas
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  6. #6
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    Re: Differentiate by definition

    Quote Originally Posted by HallsofIvy View Post
    No, the final result is not correct.
    So, then is Jpritch22 correct?
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  7. #7
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    Re: Differentiate by definition

    Halls of Ivy was responding to minoanman's response. My solution should be correct
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  8. #8
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    Re: Differentiate by definition

    Ok great. Thanks for your help!
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