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Math Help - derivitives

  1. #1
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    derivitives

    I have to find y'' for rad(t^3+1)

    I got y' = 3t^2/(2rad(t^3+1)

    i need help with y'' I cannot get to the answer
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  2. #2
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    y = \sqrt{t^{3} + 1}

    You calculated y' correctly. To calculate y'', you can use the quotient rule:
    y' = \frac{3}{2} \cdot \frac{t^{2}}{(t^{3}+1)^{\frac{1}{2}}}
    y'' = \frac{3}{2} \cdot  \frac{d}{dt}\left( \frac{t^{2}}{(t^{3}+1)^{\frac{1}{2}}}\right) = \frac{3}{2} \cdot \underbrace{\frac{\left(t^2\right)'(t^{3}+1)^{\fra  c{1}{2}} \: \: - \: \: t^{2}\left[\left(t^3 + 1\right)^{\frac{1}{2}}\right]'}{\left[\left(t^3+1\right)^{\frac{1}{2}}\right]^{2}}}_{\text{Quotient rule}}

    In case you forgot: \left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - f(x)g'(x)}{\left[g(x)\right]^2}
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by weezie23 View Post
    I have to find y'' for rad(t^3+1)

    I got y' = 3t^2/(2rad(t^3+1)

    i need help with y'' I cannot get to the answer
    rad? If you mean

    \frac{d}{dx}\bigg[\sqrt{x^3+1}\bigg]

    Then let y=\sqrt{x^3+1}\Rightarrow{y^2=x^3+1}

    Differentiating we get

    2y\cdot{y'}=3x^2

    Or y'=\frac{3x^2}{2y}

    Remembering our original equation is y we get

    y'=\frac{3x^2}{2\sqrt{x^3+1}}

    So now we take the second derivative first rewriting this as

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

    So y''=\frac{3}{2}\bigg[2x(x^3+1)^{\frac{-1}{2}}+\frac{-x^2}{2}(x^3+1)^{\frac{-3}{2}}\bigg]

    I think you can go from there
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  4. #4
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    Thanks!

    I included the 2/3 while calculating q. rule.

    Thanks!
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