# Thread: The inverse of Leibniz's rule

1. ## 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?

2. 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

$\chi$ $\sigma$