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    By using Leibniz' theorem, show that the value of the nth derivative of at is given by .

    Hmmmm...
    I don't know how to do this.

    Can someone please give me a hint or something, please?


    Lets do it step by step using what we know already:

    Leibniz' thereom:

    \frac{d^n}{dx^n}fg= \frac{d^n}{dx^n}(f)g + \displaystyle \binom{n}{1} \frac{d^{n-1}}{dx^{n-1}}(f) \frac{d}{dx}(g) + \displaystyle \binom{n}{2} \frac{d^{n-2}}{dx^{n-2}}(f) \frac{d^2}{dx^2}(g) +...+ \displaystyle \binom{n}{{n-1}} \frac{d}{dx}(f) \frac{d^{n-1}}{dx^{n-1}}(g) + f \frac{d}{dx} (g)

    So in this example, we let f=x^3 and g=e^{-4x}

    But now what? How do we know how much we have to diffrentiate it to?
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    Evaluate the first few derivatives of  e^{4x} , and from there the pattern for the nth derivative of  e^{-4x} should be fairly obvious. Also, what is the nth derivative of  x^3 for n > 3?
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    (Original post by GHOSH-5)
    Evaluate the first few derivatives of , and from there the pattern for the nth derivative of should be fairly obvious. Also, what is the nth derivative of for n > 3?
    hmmm so the nth derivative for is:

    \frac{3!}{(3-n)!}x^{3-n} right?

    and also, the first few derivatives for are:
    g' = -4e^{-4x}
    g'' = 16e^{-4x}
    g''' = -64e^{-4x}
    g'''' = 256e^{-4x}

    The pattern... is that they get multiplied by -4?
    but i still dont understand. please could you explain a bit more?
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    What is (3-n)! if n is > 3?
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    Sorry to steal the thread momentarially, but is 0! = 1 or 0?
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    (Original post by Rubgish)
    Sorry to steal the thread momentarially, but is 0! = 1 or 0?
    0!=1 by definition
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    I'm sorry, this is totally irrelevant but when I read 'Leibniz' I immediately thought it was a theory about those REALLY yummy chocolate buscuits. yum. Clearly it isn't. It's also pretty clear I don't do maths. Bye.
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    (Original post by hahaha_dontknow)
    The pattern... is that they get multiplied by -4?
    Yes, and hence:

     \dfrac{d^n}{dx^n} (e^{-4x}) = (-4)^ne^{-4x} .
    but i still dont understand. please could you explain a bit more?
    After you realise what the even simpler form of the nth derivative for  x^3 is for n greater than 3, use Leibniz's theorem.
 
 
 
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