Let f,g: all real numbers ----> all real numbers have derivatives of all orders. Recall the product rule: D(fg)=D(f)g+fD(g), where D denotes differentiation. show that D^n(fg)= Sum from k=0 to n (n,k)D^k(f)D^(n-k)(g)

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.

Let f, g: all real numbers ----> all real numbers have derivatives of all orders. Recall the product rule: D(fg)=D(f)g+fD(g), where D denotes differentiation. Show that D^n(fg)= Sum from k=0 to n (n,k)D^k(f)D^(n-k)(g)

Explore the proof of the n-th derivative of the product of two functions f and g using the product rule, binomial theorem, and mathematical induction. This detailed explanation covers differentiation concepts and is suitable for advanced high school and undergraduate students.

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