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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](http://en.wikipedia.org/wiki/Bell_polynomials).2012-10-25
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    Do you have a link perhaps? I don't find it2012-10-25
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    You can check you formula by induction (replacing $n$ by $k$ in the product).2012-10-25
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    @user43418: I found the link in Mhenni Benghorbal's comment worked. The more specific Wikipedia page is [here](http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula)2012-10-25
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    My formula is correct2012-10-25
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    @user43418, I don't think it correct as written (need to replace n-1 with k and start the product at k=0. Which is what I think you meant, anyway). See my answer for why.2012-10-25
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    @user43418, $f^{(n-1)}$ should be $f^{(k)}$2012-10-25
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    $\prod_{k=0}^{n-1} f'(f^{(k)}(x))$2012-10-25
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    So now I show that the formula works by induction.2012-10-25

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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 $$\frac{d}{dx} f^{(n)}(x) = f'(x) \cdot f'(f(x)) \cdot f'(f^{(2)}(x)) \cdots f'(f^{(n-1)}(x)).$$