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Math Help - The inverse of Leibniz's rule

  1. #1
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    The inverse of Leibniz's rule

    The product rule states, if f(x) and g(x) are two differentiable functions, then:

    \displaystyle [f(x)\cdot g(x)]' = f'(x)\cdot g(x)+g'(x)\cdotf(x).

    If f(x) and g(x) are n-times differentiable, then Liebniz's rule generalises this to:

    \displaystyle (f(x)\cdot g(x))^{(n)}=\sum_{k=0}^n {n \choose k} f^{(k)} {(x)g^{(n-k)}(x)

    Now, the quotient rule states that, if f(x) and g(x) are two differentiable functions, then:

    \displaystyle \left[\frac{f(x)}{g(x)}\right]' = \frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}

    Perhaps I've misused the word 'inverse' for what I want, but what does this generalise to?
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  2. #2
    MHF Contributor chisigma's Avatar
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    From the expression...

    \displaystyle \frac{d}{dx} \{f(x)\ g(x)}\} = f(x)\ g^{'} (x) + f^{'} (x)\ g(x) (1)

    ... setting \frac{1}{g(x)} instead of g(x) You obtain...

    \displaystyle \frac{d}{dx} \{\frac{f(x)}{g(x)}\} = \frac{f^{'}(x)}{g(x)} - f(x)\ \frac{g^{'}(x)}{g^{2} (x)} = \frac{f^{'} (x)\ g(x) - f(x)\ g^{'}(x)}{g^{2}(x)} (2)

    With a little of patience You can apply the now described procedure and find the expression for...

    \displaystyle \frac{d}{dx} \{\frac{ f_{1} (x)\ f_{2} (x) \dots f_{m} (x)}{g_{1}(x)\ g_{2} (x) \dots g_{n} (x)} \} (3)

    Kind regards

    \chi \sigma
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