To prove the given identity for the $n$-th derivative of the product of two functions $f$ and $g$, we can use the principle of mathematical induction. The identity to prove is:

$D^n(fg) = \sum_{k=0}^n \binom{n}{k} D^k(f) D^{n-k}(g)$

where $D^n$ denotes the $n$-th derivative and $\binom{n}{k}$ is the binomial coefficient.

Base Case: $n = 1$

For $n = 1$, the identity becomes the product rule for differentiation:

$D^1(fg) = D(fg) = D(f)g + fD(g)$

The right-hand side according to the formula is:

$\sum_{k=0}^1 \binom{1}{k} D^k(f) D^{1-k}(g)$

Let's compute each term:

• For $k = 0$: $\binom{1}{0} D^0(f) D^1(g) = 1 \cdot f \cdot D(g) = fD(g)$
• For $k = 1$: $\binom{1}{1} D^1(f) D^0(g) = 1 \cdot D(f) \cdot g = D(f)g$

Adding these, we get:

$D^1(fg) = D(f)g + fD(g)$

This shows that the formula holds for $n = 1$.

Inductive Step

Assume the identity holds for some $n = m$, i.e.,

$D^m(fg) = \sum_{k=0}^m \binom{m}{k} D^k(f) D^{m-k}(g)$

We need to show that it holds for $n = m+1$, i.e.,

$D^{m+1}(fg) = \sum_{k=0}^{m+1} \binom{m+1}{k} D^k(f) D^{m+1-k}(g)$

By the definition of the derivative,

$D^{m+1}(fg) = D\left(D^m(fg)\right)$

Using the inductive hypothesis:

$D^{m+1}(fg) = D\left(\sum_{k=0}^m \binom{m}{k} D^k(f) D^{m-k}(g)\right)$

Applying the derivative $D$ to each term in the sum, we get:

$D^{m+1}(fg) = \sum_{k=0}^m \binom{m}{k} \left( D^{k+1}(f) D^{m-k}(g) + D^k(f) D^{m-k+1}(g) \right)$

We can separate this into two sums:

$D^{m+1}(fg) = \sum_{k=0}^m \binom{m}{k} D^{k+1}(f) D^{m-k}(g) + \sum_{k=0}^m \binom{m}{k} D^k(f) D^{m-k+1}(g)$

Let's reindex the second sum by setting $j = k+1$. Then the second sum becomes:

$\sum_{k=0}^m \binom{m}{k} D^k(f) D^{m-k+1}(g) = \sum_{j=1}^{m+1} \binom{m}{j-1} D^{j-1}(f) D^{m+1-j}(g)$

Now, we combine the two sums:

$D^{m+1}(fg) = D^{m+1}(f)g + \sum_{j=1}^m \left( \binom{m}{j} + \binom{m}{j-1} \right) D^j(f) D^{m+1-j}(g) + fD^{m+1}(g)$

Using the identity for binomial coefficients:

$\binom{m}{j} + \binom{m}{j-1} = \binom{m+1}{j}$

We get:

$D^{m+1}(fg) = D^{m+1}(f)g + \sum_{j=1}^m \binom{m+1}{j} D^j(f) D^{m+1-j}(g) + fD^{m+1}(g)$

This simplifies to:

$D^{m+1}(fg) = \sum_{j=0}^{m+1} \binom{m+1}{j} D^j(f) D^{m+1-j}(g)$

This is exactly the formula we needed to prove for $n = m+1$.

Conclusion

By induction, the identity holds for all $n \geq 1$:

$D^n(fg) = \sum_{k=0}^n \binom{n}{k} D^k(f) D^{n-k}(g)$

Would you like further details on any part of this proof?

Here are some related questions:

1. How is the binomial coefficient defined and used in this context?
2. Can you provide an example applying this formula to specific functions $f(x)$ and $g(x)$?
3. How does the principle of induction work in proving identities like this one?
4. How does this formula relate to the Leibniz rule for higher-order derivatives?
5. What happens to the formula when $f$ or $g$ is a constant function?

Tip: The principle of mathematical induction is a powerful tool for proving statements that are true for all natural numbers. Always verify the base case and the inductive step carefully.