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Let $m,n\in\mathbb{N}$.

I need to express the derivative $\displaystyle\frac{d^n}{dx^n}\left(f(x)^m\right)$ in terms of sums/products of the derivatives of the function $f$ itself.

Here are results for several small fixed values of $n$: $$\frac{d}{dx}\left(f(x)^m\right)=m \cdot f(x)^{m-1} \cdot f^{'}(x)$$

$$\frac{d^2}{dx^2}\left(f(x)^m\right)=m \cdot (m-1) \cdot f(x)^{m-2} \cdot f^{'}(x)^2+m \cdot f(x)^{m-1} \cdot f^{''}(x)$$

$$\frac{d^3}{dx^3}\left(f(x)^m\right)=m \cdot (m-1) \cdot (m-2) \cdot f(x)^{m-3} \cdot f^{'}(x)^3+3 \cdot m \cdot (m-1) \cdot f(x)^{m-2} \cdot f^{'}(x) \cdot f^{''}(x) + m \cdot f(x)^{m-1} \cdot f^{'''}(x)$$

I tried to find a common pattern for these expressions, but failed.

Question: Is it possible to find a general formula for the derivative of any given order $n$?

X.C.
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1 Answers1

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A related problem. The Faà di Bruno's formula is given by

$$ {d^n \over dx^n} g(f(x)) = \sum_{k=1}^n g^{(k)}(f(x))\cdot B_{n,k}\left(f'(x),f''(x),\dots,f^{(n-k+1)}(x)\right). $$

Now, in your case $g(x)=x^m$ which we can find an explicit formula for it as

$$ \frac{d^k}{dx^k} x^m = \frac{\Gamma(m+1)}{\Gamma(m-k+1 )}x^{m-k}. $$

So, you will have the formula for your problem

$$ (g(f(x)))^{(n)} = \Gamma(m+1)\sum_{k=1}^n \frac{x^{m-k}}{\Gamma(m-k+1 )} B_{n,k}\left(f'(x),f''(x),\dots,f^{(n-k+1)}(x)\right). $$

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Mhenni Benghorbal
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