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Math Help - Derivatives Quotient Rule

  1. #1
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    Derivatives Quotient Rule

    In the problem f(x) = -1/x^2+1 finding the derivative using the quotient rule.

    the solution would be -x^2+2x-1/(x^2+1)^2?

    It seems right but im not sure.
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  2. #2
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    Quote Originally Posted by oregon88 View Post
    In the problem f(x) = -1/x^2+1 finding the derivative using the quotient rule.

    the solution would be -x^2+2x-1/(x^2+1)^2?

    It seems right but im not sure.
    I don't get that answer, perhaps you could post your working?

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

    Wolfram gets the same answer I do. If you are unsure about how to use the quotient rule here the chain rule works equally well

    Spoiler:
    By the chain Rule:

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

    f'(x) = (x^2+1)^{-2} \cdot 2x = \frac{2x}{(x^2+1)^2}
    Last edited by e^(i*pi); May 17th 2010 at 01:02 PM. Reason: more descriptive
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  3. #3
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    f(x)=-x/x^2+1

    f'(x)=(x^2+1)d/dx(-x)-(-x)d/dx(x^2+1)/(x^2+1)^2

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

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

    Is what I came up with
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by oregon88 View Post
    f(x)=-x/x^2+1

    f'(x)=(x^2+1)d/dx(-x)-(-x)d/dx(x^2+1)/(x^2+1)^2

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

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

    Is what I came up with
    Hold on, you said f(x) = -\frac{1}{x^2+1} in the OP. Which one is correct?

    Your differentiating is correct but your simplification on the third line is not - there should be no x term. Plus be more clear with your syntax using brackets

    u = -x \: \rightarrow \: u' = -1

    v = x^2+1 \: \rightarrow \: v' = 2x


    f'(x) = \frac{-x^2-1 + 2x^2}{(x^2+1)^2} = \frac{x^2-1}{(x^2+1)^2}
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  5. #5
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    The original problem is

    <br />
f(x) = \frac {-x}{x^2+1}<br />

    Sorry about that, I see where I messed up on the third line I did not factor the (-x)(2x) on my paper. Thanks for the help e^(i*pi)
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