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Leibnitz Theorem

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    • Thread Starter

    Hi, Im finding trouble trying to understand the Leibnitz theorem, can anyone help and just try simplifying it for me, Im just not getting it!:mad:

    I got these 2 questions:
    Use Leibnitz theorem to compute the 5th derivative of:
    a) x^3cos(x)
    b) x^2ln(x)


    This looks not fun. Is it the same as what this page is describing http://www.math.ohio-state.edu/~neva...formula_H6.pdf

    If so letting u(x) = x^3 and v(x) = cos(x), I'd then find the 1-5 th derivatives of each of them, then just stick it in that big sum formula.
    • Thread Starter

    Yeah we have to use the formula given at the top of the page but I dont really understand it.

    I'll try write out the first few stages of the first one then.

    You have  x^3 \cos x so split this into two (easily differentiable) functions  u(x) = x^ and  v(x) = \cos x . Find their 1st, 2nd, ..., 5th derivatives and just keep them somewhere handy.

    Now do you understand the sum notation? This is what you'd do using the formula (I'm not going to write it out, I don't know the symbols)

    Firstly, in the big sum, r = 0, and n = 5 because you want the 5th derivative.
    So you get (5C0 is 5 choose 0, the binomial thingy)
     5C0 \ x^3 \frac{d^5 (cos x)}{dx} = - x^3 \sin x
    And now r = 1 so
     5C1 \frac{dx^3}{dx} \frac{d^4 (cos x)}{dx^4} = 15x^2 \cos x
    And now do this for r= 2, 3, 4, and 5.
    Finally add up all those bits because its a sum. Hopefully some stuff cancels out too.

    It's just the same as using the binomial theorem but substituting in derivatives, what part of it don;t you understand
    • Thread Starter

    Yeahh I think I get it now, I jus wasnt sure about the 'r' and 'n' term things. Thanks alot though
    • Thread Starter

    Sebbie I think I get it now, thanks

    if you were to find the odd derivatives how would you do that?
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