To find the $n$th derivative of this function.

Question

Let $f(x)$ be smooth and continuous for $|x|<1$. I am interested in the $n$th derivative of: $$g(x) = f(x) e^{af(x)}$$ for some $a>0$.

Is it possible to write this in a neat form?

Thanks.

Answer 1

This requires a kind of 'chain rule' for $n^{th}$ derivatives, and unfortunately that becomes quite complicated. This is dealt with in full generality by Faa Di Bruno's formula which states that $$\frac{d^n}{dx^n}h(f(x))=\sum_{k=0}^n h^{(k)}(f(x))\cdot B_{n,k}\left(f'(x),\dots,f^{(n-k+1)}(x)\right)$$ where $B_{n,k}$ are the Bell polynomials. You are asking about the specific case when $h(x)=xe^{ax}$, in which case $$\frac{d^n}{dx^n} xe^{ax}=a^{n-1}e^{ax}(ax+k),$$ and so $$\frac{d^n}{dx^n} f(x)e^{af(x)}=e^{af(x)}\sum_{k=0}^n a^{k-1}(af(x)+k)\cdot B_{n,k}\left(f'(x),\dots,f^{(n-k+1)}(x)\right).$$

Answer 2

Let $$h_n(z)=g^{(n)}(z)e^{-af(z)}$$

Then $$\begin{align} h_n'(z)&=g^{(n+1)}(z)e^{-af(z)} -a f'(z)g^{(n)}(z)e^{-af(z)}\\ &=h_{n+1}(z) - af'(z)h_n(z) \end{align}$$

So:

$$\begin{align} h_0(z)&=f(z)\\ h_{n+1}(z)&=h_n'(z) + af'(z)h_n(z) \end{align}$$

Don't think you can do much better than that recursion.

Then $$g^{(n)}(z)=h_n(z)e^{af(z)}.$$