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  1. #1
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    find the second derivative of...

    Find the second derivative of f(x)= sq rt(x^2 + 1)

    I think the first derivative is x(x^2 + 1)^-1/2

    But, I don't know how to find the derivative of that?


    Possible answers are:

    A) x/sq rt (x^2 + 1)

    B) sq rt(x^2 + 1) - 1/2(x^2 + 1)^-1/2

    C) 2x(sq rt(x^2 + 1) + 1/2(x^2 + 1)^-1/2

    D) 1/sq rt((x^2 + 1)^3)
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by obsmith08 View Post
    Find the second derivative of f(x)= sq rt(x^2 + 1)

    I think the first derivative is x(x^2 + 1)^-1/2

    But, I don't know how to find the derivative of that?


    Possible answers are:

    A) x/sq rt (x^2 + 1)

    B) sq rt(x^2 + 1) - 1/2(x^2 + 1)^-1/2

    C) 2x(sq rt(x^2 + 1) + 1/2(x^2 + 1)^-1/2

    D) 1/sq rt((x^2 + 1)^3)
    Your first derivative is correct.

    Now apply the product rule, which states \frac{\,d}{\,dx}\left[f\!\left(x\right)g\!\left(x\right)\right]=f\!\left(x\right)\frac{\,dg}{\,dx}+g\!\left(x\rig  ht)\frac{\,df}{\,dx}. Treat x as f\!\left(x\right) and \left(x^2+1\right)^{-1} as g\!\left(x\right).

    Can you try this now and see how far you get? If you still have questions on how to continue, feel free to post back.
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