n-th derivative of exponential function $\;e^{-f(x)}$ [closed]

Solution 1:

See Faa di Bruno's identity generalizing the chain rule to higher derivatives:

$${d^n \over dx^n} f(g(x))=\sum \frac{n!}{m_1!\,1!^{m_1}\,m_2!\,2!^{m_2}\,\cdots\,m_n!\,n!^{m_n}}\cdot f^{(m_1+\cdots+m_n)}(g(x))\cdot \prod_{j=1}^n\left(g^{(j)}(x)\right)^{m_j}$$

where the sum is over all $n$-tuples of nonnegative integers $(m_1, …, m_n)$ satisfying the constraint $$1\cdot m_1+2\cdot m_2+3\cdot m_3+\cdots+n\cdot m_n=n.\,$$

If you're familiar with Bell polynomials $B_{n,k}(x_1,...,x_{n−k+1})$:, the identity can be simplified as follows: $${d^n \over dx^n} f(g(x)) = \sum_{k=1}^n f^{(k)}(g(x))\cdot B_{n,k}\left(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\right).$$

In the formulae above, of course, $f$ is the exponential function, and $g(x)$ serves as your $-f(x)$.

With $f(x) = e^x$, all of the derivatives of $f$ are the same, and are a factor common to every term.