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Math Help - Second Derrivative

  1. #1
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    Second Derrivative

    Can someone help me take the 2nd Derivative of this function. I can't seem to get it right. y=\frac{x}{(x+1)^{2}}

    Thanks.
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    Quote Originally Posted by johntuan View Post
    Can someone help me take the 2nd Derivative of this function. I can't seem to get it right. y=\frac{x}{(x+1)^{2}}
    Before to contemplate a derivative, try to do a make-up first:

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

    So from here first and second derivative are practically done.

    f'(x) =  - (x + 1)^{ - 2}  + 2(x + 1)^{ - 3}  \implies f''(x) = 2(x + 1)^{ - 3}  - 6(x + 1)^{ - 4} .

    Now it remains to do the algebra make-up:

    f''(x) = 2(x + 1)^{ - 4} \Big[ {(x + 1) - 3} \Big] = \frac{{2(x - 2)}}<br />
{{(x + 1)^4 }}.

    (Observe that we've applied the Chain Rule always, it's just that in this case is pretty trivial, since (x+1)'=1.)
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    Thanks for the reply but, I don't quite understand your first step, What's the Make-up?
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  4. #4
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    Quote Originally Posted by johntuan View Post
    Thanks for the reply but, I don't quite understand your first step, What's the Make-up?
    It means try to re-arrange the given expression into something that's easier to work with.
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  5. #5
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    Quote Originally Posted by johntuan View Post
    Thanks for the reply but, I don't quite understand your first step, What's the Make-up?
    He means he's putting it into a nicer form for taking the derivative. You can also do this by the brute force method:
    y=\frac{x}{(x+1)^{2}}

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

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

    So
    y^{\prime \prime} = -\frac{1 \cdot (x + 1)^3 - (x - 1) \cdot 3(x + 1)^2}{(x + 1)^6}

    y^{\prime \prime} = 2 \frac{x - 2}{(x + 1)^4}
    after some work.

    -Dan
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    In this step y^{\prime} = \frac{-x^2 + 1}{(x + 1)^4} = -\frac{x - 1}{(x + 1)^3} how did u get -x^2+1 to cancel out one part of the denominator, in order to cancel out shouldn't it equal x+1^{x}
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  7. #7
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    Quote Originally Posted by johntuan View Post
    In this step y^{\prime} = \frac{-x^2 + 1}{(x + 1)^4} = -\frac{x - 1}{(x + 1)^3} how did u get -x^2+1 to cancel out one part of the denominator, in order to cancel out shouldn't it equal x+1^{x}
    -x^2 + 1 = 1 - x^2 = (1 - x)(1 + x) = -(x - 1)(x + 1) and the (x + 1) is clearly seen to be a common factor of numerator and denominator.
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