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  1. #1
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    finding derivative

    I need help finding the derivative of this. I am a bit confused. I have y = (x^2)/(x^2+3). the derivative is 6x/(x^2+3)^2. I understand the quotient rule, and how they got the squared on the bottom. but how did they get the 6x on the top?
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    Re: finding derivative

    Quote Originally Posted by droidus View Post
    I need help finding the derivative of this. I am a bit confused. I have y = (x^2)/(x^2+3). the derivative is 6x/(x^2+3)^2. I understand the quotient rule, and how they got the squared on the bottom. but how did they get the 6x on the top?
    Next time show your work. You probably made a mistake somewhere...

    y = \frac{x^2}{x^2+3}

    \frac{dy}{dx} = \frac{\left(x^2+3\right)(2x)-x^2(2x)}{\left(x^2+3\right)^2}

    = \frac{2x^3+6x-2x^3}{\left(x^2+3\right)^2}

    = \frac{6x}{\left(x^2+3\right)^2}
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  3. #3
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    Re: finding derivative

    ok - here is what i get for the second derivative then...

    [(x^2 + 3)^2 + 6 - 6x(4x)] / (x^2+3)^4
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  4. #4
    MHF Contributor Reckoner's Avatar
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    Re: finding derivative

    Quote Originally Posted by droidus View Post
    ok - here is what i get for the second derivative then...

    [(x^2 + 3)^2 + 6 - 6x(4x)] / (x^2+3)^4
    The derivative of \left(x^2+3\right)^2 is not 4x. And shouldn't you be multiplying that 6 instead of adding it? Let's be more careful:

    \frac{dy}{dx} = \frac{6x}{\left(x^2+3\right)^2}

    \frac{d^2y}{dx^2} = \frac{\left(x^2+3\right)^2\cdot\frac d{dx}[6x] - 6x\cdot\frac d{dx}\left[\left(x^2+3\right)^2\right]}{\left(x^2+3\right)^4}

    = \frac{6\left(x^2+3\right)^2 - 6x\cdot2\left(x^2+3\right)\cdot\frac d{dx}\left[x^2+3\right]}{\left(x^2+3\right)^4}

    = \frac{6\left(x^2+3\right)^2 - 24x^2\left(x^2+3\right)}{\left(x^2+3\right)^4}

    = \frac{6\left(x^2+3\right) - 24x^2}{\left(x^2+3\right)^3}

    = \frac{-18\left(x^2-1\right)}{\left(x^2+3\right)^3}

    = \frac{-18(x-1)(x+1)}{\left(x^2+3\right)^3}


    Make sure that you're using the chain rule properly. We must remember to differentiate the "inner" function after we differentiate the "outer" function of a composite function.
    Last edited by Reckoner; June 13th 2012 at 02:32 PM.
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