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

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$5) to do some experiments. Thanks a lot ^^ • wolframalpha.com/input/?i=nth+derivative+e%5E%7Bf%28x%7D%7D. You were doing great. Just needed to find the general pattern and then prove the result using induction or some such mechanism. Regards Commented Jul 11, 2013 at 19:24 • A related problem. In general, Check Faà di Bruno's formula. Commented Jul 11, 2013 at 19:32 • Another related problem. Commented Jul 11, 2013 at 19:39 • How about just writing$f'(x)$, for clarity? Commented Jul 11, 2013 at 19:45 • Note that, the Wolfram formula requires the computing the$n$derivative of the powers of the function$f(x)$which is another hard job. Commented Jul 11, 2013 at 19:53 ## 1 Answer Faà di Bruno's formula has been mentioned in the comments. Since the derivative of the exponential is the exponential itself, the (possibly) easiest closed form expression is the one in terms of complete Bell polynomials in the derivatives of$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

• (1) Feng Qi, Da-Wei Niu, Dongkyu Lim, and Yong-Hong Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, Journal of Mathematical Analysis and Applications vol. 491 (2020), no. 2, Paper No. 124382, 31 pages; available online at doi.org/10.1016/j.jmaa.2020.124382. (2) Feng Qi, Da-Wei Niu, Dongkyu Lim, and Bai-Ni Guo, Closed formulas and identities for the Bell polynomials and falling factorials, Contributions to Discrete Mathematics vol. 15 (2020), no. 1, 163--174; available online at doi.org/10.11575/cdm.v15i1.68111. Commented Sep 27, 2021 at 12:09