General formula or a pattern for the $n$th derivatives of $e^{f(x)}$?

Question

I want to find the $nth$ derivatives of the function $e^{f(x)}$ with respect to $x$, the first derivative is $$e^{f(x)}f^{\prime}(x).$$ The second derivative is $$\left( {\frac {d^{2}}{d{x}^{2}}}f \left( x \right) \right) {{\rm e}^ {f \left( x \right) }}+ \left( {\frac {d}{dx}}f \left( x \right) \right) ^{2}{{\rm e}^{f \left( x \right) }} .$$ The third derivative is $$\left( {\frac {d^{3}}{d{x}^{3}}}f \left( x \right) \right) {{\rm e}^ {f \left( x \right) }}+3\, \left( {\frac {d^{2}}{d{x}^{2}}}f \left( x \right) \right) \left( {\frac {d}{dx}}f \left( x \right) \right) { {\rm e}^{f \left( x \right) }}+ \left( {\frac {d}{dx}}f \left( x \right) \right) ^{3}{{\rm e}^{f \left( x \right) }} $$ The fourth derivative is $$ \left( {\frac {d^{4}}{d{x}^{4}}}f \left( x \right) \right) {{\rm e}^ {f \left( x \right) }}+4\, \left( {\frac {d^{3}}{d{x}^{3}}}f \left( x \right) \right) \left( {\frac {d}{dx}}f \left( x \right) \right) { {\rm e}^{f \left( x \right) }}+3\, \left( {\frac {d^{2}}{d{x}^{2}}}f \left( x \right) \right) ^{2}{{\rm e}^{f \left( x \right) }}+6\, \left( {\frac {d^{2}}{d{x}^{2}}}f \left( x \right) \right) \left( { \frac {d}{dx}}f \left( x \right) \right) ^{2}{{\rm e}^{f \left( x \right) }}+ \left( {\frac {d}{dx}}f \left( x \right) \right) ^{4}{ {\rm e}^{f \left( x \right) }} $$

My question is: Is there a general formula or a pattern for the nth derivative of $e^{f(x)}$. You may use the maple command diff(exp(f(x)), x$5) to do some experiments. Thanks a lot ^^

Answer 1

Faà di Bruno's formula has been mentioned in the comments. Since the derivative of the exponential is the exponential itself, the (possibly) easiest closed form expression is the one in terms of complete Bell polynomials in the derivatives of $f$:

$$ \begin{align*} \frac{d^n}{dx^n}\exp\big(f(x)\big) = \, & \sum_{k=1}^n \exp\big(f(x)\big) \cdot B_{n,k}\big(f'(x), f''(x), \dots, f^{(n-k+1)}(x)\big) \\ = \, & \exp\big(f(x)\big) \sum_{k=1}^n B_{n,k}\big(f'(x), f''(x), \dots, f^{(n-k+1)}(x)\big) \end{align*} $$

Of course, now you need to deal with the Bell polynomials, which are commonly defined recursively. Here is a thread on the sum involved, which unfortunately does not have an overwhelmingly informative answer: Sum of Bell Polynomials of the Second Kind