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I need to express the derivative of $f^{(n)}(x)$ in terms of $f'(x)$, meaning $f$ composed with itself n times.

I was able to express $f(f(x))=f'(f(x))f'(x)$

Is the derivative of the composition $\prod_{k=0}^{n-1} f'(f^{(k)}(x))$ ?

Please help

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Check Faà di Bruno's formula. – Mhenni Benghorbal Oct 25 '12 at 13:08
Do you have a link perhaps? I don't find it – user43418 Oct 25 '12 at 13:09
You can check you formula by induction (replacing $n$ by $k$ in the product). – Davide Giraudo Oct 25 '12 at 13:13
My formula is correct – user43418 Oct 25 '12 at 13:22
$\prod_{k=0}^{n-1} f'(f^{(k)}(x))$ – user43418 Oct 25 '12 at 13:28
up vote 0 down vote accepted

Note: Sometimes $f^{(n)}$ means the $n$th derivative. In this context it means the $n$th iterate.

Put $g(x) = f^{(n)}(x)$ in $\frac{d}{dx} f(g(x)) = g'(x) f'(g(x))$ to get $$\frac{d}{dx} f^{(n+1)}(x) = [\frac{d}{dx} f^{(n)}(x)] f'(f^{(n)}(x)).$$

Using $f^{(0)}(x) = x$ we find, by induction, that $$\frac{d}{dx} f^{(n)}(x) = f'(x) \cdot f'(f(x)) \cdot f'(f^{(2)}(x)) \cdots f'(f^{(n-1)}(x)).$$

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