The Leibniz rule is a powerful formula for taking the nth derivative of a product of two functions. It generalizes the product rule and is particularly useful when differentiating products involving powers or sums.

Leibniz Rule Formula:

For two functions $f(x)$ and $g(x)$, the nth derivative of their product is given by:

[ \frac{d^n}{dx^n} \left[ f(x) \cdot g(x) \right] = \sum_{k=0}^{n} \binom{n}{k} \frac{d^k}{dx^k} \left[ f(x) \right] \cdot \frac{d^{n-k}}{dx^{n-k}} \left[ g(x) \right] ]

Where:

• $\binom{n}{k}$ is the binomial coefficient, which is $\frac{n!}{k!(n-k)!}$.
• The terms $\frac{d^k}{dx^k} \left[ f(x) \right]$ and $\frac{d^{n-k}}{dx^{n-k}} \left[ g(x) \right]$ are the kth and (n-k)th derivatives of $f(x)$ and $g(x)$, respectively.

Example:

Let’s say we want to compute the nth derivative of $f(x) = x^m$ and $g(x) = e^x$.

Using the Leibniz rule, we would have:

[ \frac{d^n}{dx^n} \left[ x^m \cdot e^x \right] = \sum_{k=0}^{n} \binom{n}{k} \frac{d^k}{dx^k} \left[ x^m \right] \cdot \frac{d^{n-k}}{dx^{n-k}} \left[ e^x \right] ]

1. $\frac{d^k}{dx^k} \left[ x^m \right] = \frac{m!}{(m-k)!} x^{m-k}$, for $k \leq m$, and 0 for $k > m$.
2. $\frac{d^{n-k}}{dx^{n-k}} \left[ e^x \right] = e^x$, since all derivatives of $e^x$ are just $e^x$.

Thus, the Leibniz rule gives us:

$\frac{d^n}{dx^n} \left[ x^m \cdot e^x \right] = \sum_{k=0}^{n} \binom{n}{k} \frac{m!}{(m-k)!} x^{m-k} e^x$

This expression simplifies depending on the values of $m$ and $n$.

Would you like to see a worked-out example with specific values for $m$ and $n$, or is there another problem you'd like to explore?