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calculus

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Can someone help me understand how to find the derivative of these two problems:

(t^2-1/t^2+2)^3=y and y=sin^32x?

I think with the first one, I could just foil it maybe and cancel but I don't know what to do with the ^3...and then the second one I haven't gotten the hang of how to eliminate those kinds of problems yet...:(

  • calculus -

    1)

    y'=d/dt (uv^-1) where u= t^2-1 du=2dt
    and v=(t^2+2)^3 dv= 3(t^2+2)^2 * (2t)=6t(t^2+2)^2

    y'= v^-1 du -uv^-2 dv

    y'=du/v -udv/v^2
    then put in u,v du, dv and you have it.

  • calculus -

    I guess it's the format but I am little confused about what you did....

  • calculus -

    1)
    use the quotient rule:
    y = u/v
    where u = t^2-1 and v = t^2+2

    y' = (u'v - uv')/v^2
    = [(2t)(t^2+2) - (t^2-1)(2t)]/(t^2+2)^2
    = (2t^3+4t-2t^3+2t)/(t^2+2)^2
    = 6t/(t^2+2)^2

    2) use the chain rule:
    y = u^3 where u = sin2x
    y' = 3u^2 u'
    = 3sin^2(2x) 2cos2x
    = 6sin^2(2x) cos(2x)
    or
    3sin4x sin2x

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