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Math Help - Quotient Derivative

  1. #1
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    Quotient Derivative

    I'm stuck on this derivative:

    y= \frac{x^2 + 2ax -a}{(x+a)^2}

    Here is what I have so far:

    \frac{(x^2+2ax+a^2)(2x+2a) - (x^2+2ax-a)(2x+2a)}{(x+a)^4}

    \frac{2a^3+4a^2-2ax}{(x+a)^4}

    \frac{2a(a^2+2a-x)}{(x+a)^4}



    I know I made a mistake somewhere but I can't find it??
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by dbakeg00 View Post
    I'm stuck on this derivative:

    y= \frac{x^2 + 2ax -a}{(x+a)^2}

    Here is what I have so far:

    \frac{(x^2+2ax+a^2)(2x+2a) - (x^2+2ax-a)(2x+2a)}{(x+a)^4}

    \frac{2a^3+4a^2-2ax}{(x+a)^4}

    \frac{2a(a^2+2a-x)}{(x+a)^4}



    I know I made a mistake somewhere but I can't find it??

    The expansion of

    (x+a)^2 is x^2+2ax+a^2

    ,rather than

    (x^2+2ax-a).
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  3. #3
    MHF Contributor chisigma's Avatar
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    May be that this identity allows You to save time and efforts...

    \frac{x^{2} + 2 a x - a}{(x+a)^{2}} = 1- \frac{a (a+1)}{(x+a)^{2}} (1)

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by apcalculus View Post
    The expansion of

    (x+a)^2 is x^2+2ax+a^2

    ,rather than

    (x^2+2ax-a).
    That is how I have it on the left side of the numerator (of the first step)

    <br /> <br />
\frac{(x^2+2ax+a^2)(2x+2a) - (x^2+2ax-a)(2x+2a)}{(x+a)^4}<br />
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  5. #5
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    Quote Originally Posted by chisigma View Post
    May be that this identity allows You to save time and efforts...

    \frac{x^{2} + 2 a x - a}{(x+a)^{2}} = 1- \frac{a (a+1)}{(x+a)^{2}} (1)

    Kind regards

    \chi \sigma
    Can you show me how <br /> <br />
\frac{x^{2} + 2 a x - a}{(x+a)^{2}} = 1- \frac{a (a+1)}{(x+a)^{2}}<br />
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  6. #6
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by dbakeg00 View Post
    I'm stuck on this derivative:

    y= \frac{x^2 + 2ax -a}{(x+a)^2}

    Here is what I have so far:

    \frac{(x^2+2ax+a^2)(2x+2a) - (x^2+2ax-a)(2x+2a)}{(x+a)^4}

    \frac{2a^3+4a^2-2ax}{(x+a)^4}

    \frac{2a(a^2+2a-x)}{(x+a)^4}



    I know I made a mistake somewhere but I can't find it??
    \frac{2(x+a)[(x^2+2ax+a^2) - (x^2+2ax-a)]}{(x+a)^4}=

    \frac{2 [x^2-x^2+2ax-2ax+a^2+a]}{(x+a)^3}

    \frac{2a(a+1)}{(x+a)^3}
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  7. #7
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by dbakeg00 View Post
    Can you show me how <br /> <br />
\frac{x^{2} + 2 a x - a}{(x+a)^{2}} = 1- \frac{a (a+1)}{(x+a)^{2}}<br />
    \frac{x^{2} + 2 a x - a}{(x+a)^{2}} = \frac{x^{2} + 2 a x +a^{2} -a^{2} - a}{(x+a)^{2}} = 1 - \frac{a^{2} + a}{(x+a)^{2}} (1)

    Kind regards

    \chi \sigma
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  8. #8
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    Quote Originally Posted by apcalculus View Post
    \frac{2(x+a)[(x^2+2ax+a^2) - (x^2+2ax-a)]}{(x+a)^4}=

    \frac{2 [x^2-x^2+2ax-2ax+a^2+a]}{(x+a)^3}

    \frac{2a(a+1)}{(x+a)^3}
    Thank you very much. I don't know why I didn't see that earlier. I appreciate the help!
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  9. #9
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    Quote Originally Posted by chisigma View Post
    \frac{x^{2} + 2 a x - a}{(x+a)^{2}} = \frac{x^{2} + 2 a x +a^{2} -a^{2} - a}{(x+a)^{2}} = 1 - \frac{a^{2} + a}{(x+a)^{2}} (1)

    Kind regards

    \chi \sigma
    Now I see! Is that just an identity you had memorized?
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