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Math Help - Second derivative

  1. #1
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    Second derivative

    g(t) =-\frac{4}{(t+2)^2}

    Need some assistance finding the second derivartive of this equation. If you can show me in a step by step fashion how to solve this that'd be great.
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  2. #2
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    g(t)=-4(t+2)^{-2}
    g'(t)=-4(-2)(t+2)^{-3}=8(t+2)^{-3}
    g''(t)=8(-3)(t+2)^{-4}=-24(t+2)^{-4}=\ -\frac{24}{(t+2)^4}
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by hemi View Post
    g(t) =-\frac{4}{(t+2)^2}

    Need some assistance finding the second derivartive of this equation. If you can show me in a step by step fashion how to solve this that'd be great.

    Sure buddy.

    \frac{d}{dt}[-\frac{4}{(t+2)^2}]=\frac{d}{dt}[-4(t+2)^{-2}]

    Now by the product and chain rules, we defferentiate once to get:

    \frac{d}{dt}[-4(t+2)^{-2}]=-4(-2)(t+2)^{-3}(1)+0(t+2)^{-2}=8(t+2)^{-3}

    Now to find the second derivative, we just defferentiate again:

    \frac{d^2}{dt^2}[-\frac{4}{(t+2)^2}]=\frac{d}{dt}[8(t+2)^{-3}]=8(-3)(t+2)^{-4}(1)+0(t+2)^{-3}=-24(t+2)^{-4}

    therefore \frac{d^2g}{dt^2}=\frac{-24}{(t+2)^4}

    and there you have it. any questions?

    The product rule was unnecessary here. I only did it to illustrate that the thing would stillbe valid.
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