By Leibniz' rule, write the formula for \(\left(d^{n} / d x^{n}\right)(u v) .\)

Understanding the nth derivative means we need to know how to differentiate a function multiple times. If you take the first derivative of a function, you get a new function that represents the slope or rate of change of the original function.
For example, if we have a function \( f(x) \), its first derivative is denoted as \( f'(x) \). If we take the derivative of \( f'(x) \), we get the second derivative, denoted as \( f''(x) \).
Building on this, if we differentiate \( f''(x) \) again, we get the third derivative, \( f'''(x) \), and so on. The nth derivative, written as \( f^{(n)}(x) \), is obtained by differentiating the function \( n \) times.

When we take the derivative of the product of two functions, we apply the product rule. The product rule states that the derivative of the product of two functions \( u(x) \) and \( v(x) \) is:
\[ \frac{d}{dx}(uv) = u'v + uv' \ \text{where } u' \text{ and } v' \text{ are the derivatives of } u \text{ and } v \text{ respectively}. \] However, for higher-order derivatives, we use Leibniz's rule, which is an extension of the product rule. Leibniz's rule allows us to find the nth derivative of a product by expressing it as a sum involving the derivatives of each function.
So the nth derivative of the product of two functions \( u(x) \) and \( v(x) \) is:
\[ \frac{d^n}{dx^n}(u \times v) = \binom{n}{0}u^{(n)}v + \binom{n}{1}u^{(n-1)}v^{(1)} + \binom{n}{2}u^{(n-2)}v^{(2)} + \text{...} + \binom{n}{n}uv^{(n)}. \]

In the formula for Leibniz's rule, one of the critical components is the binomial coefficient, denoted as \( \binom{n}{k} \). The binomial coefficient \( \binom{n}{k} \) represents the number of ways to choose \( k \) items from \( n \) items without regard to order.
It is calculated using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \ \text{where } n! \text{ (n factorial) is the product of all positive integers up to } n. \] For example, \( \binom{5}{2} \ = \frac{5!}{2!(5-2)!} \ = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 3 \times 2 \times 1} \ = 10 \). This means there are 10 ways to choose 2 items from a set of 5 items.
In Leibniz’s rule, the binomial coefficient helps in determining the weight of each term in the sum, ensuring that each combination of derivatives is counted appropriately.