calculus

Find the second derivative of: h(t)= (t^2 + 5)^8

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  1. Use the chain rule, or substitute
    y=(t^2+5), and
    dy/dt = 2t in
    h(t)= (t^2 + 5)^8
    to give
    h(y)=y^8
    and
    h'(t)=dh/dy * dy/dt

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  2. I'm not following. For the first derivative I got 8(t^2 + 5)^7 x (2t). I get stuck after that.

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  3. What I did was for the first derivative.
    You can proceed in a similar way for the second derivative, but I guess substitution is out.
    Take the derivative of what you've got as a product.

    u=(t^2 + 5)^7
    v=16t
    d(uv)/dt = udv/dt + vdu/dt

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