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Math Help - Find the derivative of teh algebraic function.

  1. #1
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    Find the derivative of teh algebraic function.

    f(x) = (3-2x-x^2) / (x^2-1)

    I know the product rule and can get too

    (x^2-1)(-2-2x) - (3-2x-x^2)(2x) / (x^2-1)^2

    The answer in the back of the book says its 2/(x+1)^2

    I feel like a dumbass, math used to be my strongest subject, but now I can't even come close the answer... If someone could give me a STEP by STEP approach I would greatly appreciate it. I think my algebra / simplifying skills are in need of a serious tune up. Thanks again
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  2. #2
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    I also need help with x(1-(4/x+3)) without using the chain rule. b/c the chain rule hasn't been introduced yet...thanks again
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  3. #3
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    bump... anyone?

    should i rewrite the functions to make it easier to read?
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  4. #4
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    Quote Originally Posted by action812 View Post
    bump... anyone?

    should i rewrite the functions to make it easier to read?
    do not bump.


    f(x) = \frac{3-2x-x^2}{x^2-1} = \frac{(3+x)(1-x)}{(x+1)(x-1)} = -\frac{3+x}{x+1}

    f'(x) = - \frac{(x+1) - (3+x)}{(x+1)^2}

    clean up the algebra.
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  5. #5
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    Ok sorry about bumping, I have another question... how would I know whether to factor like you did, or just try and use the quotient rule without factoring?
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  6. #6
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    Quote Originally Posted by action812 View Post
    Ok sorry about bumping, I have another question... how would I know whether to factor like you did, or just try and use the quotient rule without factoring?
    you do not have to factor ... however, it makes taking the derivative a bit easier.
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  7. #7
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    Well when i don't factor I don't get the correct answer... is my algebra to blame...?
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  8. #8
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    Quote Originally Posted by action812 View Post
    Well when i don't factor I don't get the correct answer... is my algebra to blame...?
    I can't make that determination w/o seeing your algebra.
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  9. #9
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    If we were to do it without factoring first:

    f(x)=\frac{3-2x-x^2}{x^2-1}


    f'(x)=\frac{(x^2-1)(-2-2x)-(3-2x-x^2)(2x)}{(x^2-1)^2}

    =\frac{-2x^2+2-2x^3+2x-(6x-4x^2-2x^3)}{(x-1)^2(x+1)^2}

    =\frac{2x^2-4x+2}{(x-1)^2(x+1)^2}

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

    =\frac{2}{(x+1)^2}
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  10. #10
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    Ok thanks a lot fellas, it makes sense now. My factoring is just a little rusty.

    If you all could help me on this problem as well i'd greatly appreciate it.

    x(1-(4/x+3))
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  11. #11
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    Quote Originally Posted by action812 View Post
    Ok thanks a lot fellas, it makes sense now. My factoring is just a little rusty.

    If you all could help me on this problem as well i'd greatly appreciate it.

    x(1-(4/x+3))
    start a new problem w/ a new thread.

    show what you have tried, also.
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