Question

Use mathematical induction to derive the Leibniz rule for differentiating a product: $$ \frac{d^{n}}{d x^{n}}(f \cdot g)=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) \frac{d^{k} f}{d x^{k}} \frac{d^{n-k} g}{d x^{n-k}} . $$

Short answer

By using mathematical induction, it is concluded that the Leibniz rule for differentiating a product is proved. The process involved validating the base case, making an induction hypothesis at n=k, and proving the hypothesis for n=k+1 using the product and binomial theorem.

Step-by-step solution

Base Case (n=0)

When n=0, both sides of the Leibniz rule simplifies to \(f \cdot g\). This is because the 0th derivative of any function is the function itself and \(\binom{0}{0}\) = 1.

Induction Hypothesis (Assume n=k)

Assume that the Leibniz rule is true for n=k. That is: \[\frac{d^{k}}{d x^{k}}(f \cdot g)=\sum_{i=0}^{k}\binom{k}{i}\frac{d^{i} f}{d x^{i}} \frac{d^{k-i} g}{d x^{k-i}}\]This represents the assumption that the formula holds for some integer \(k\).

Induction Step (Prove n=k+1)

It needs to be shown that the formula holds for n=k+1. As per the product rule, differentiate both sides of the equation obtained in the induction hypothesis with respect to x. This step involves the application of the binomial theorem to simplify the equation. The goal is to be able to express the left hand side as the (k+1)th derivative and the right hand side as the sum from 0 to k+1. After some simplification, this proves the Leibniz rule at n=k+1.

Conclusion

The base case for n = 0 is confirmed. It has also been demonstrated that if the rule holds for some value \(n = k\), then it also holds for \(n = k + 1\). By the principle of mathematical induction, the Leibniz rule for differentiating a product is proved.

Key concepts

Mathematical Induction

Mathematical induction is a powerful proof technique used to establish that a given statement is true for all natural numbers. It works in two critical steps: first, you show that the statement is true for an initial natural number, usually 0 or 1, which is known as the base case. Then, you assume that the statement is true for some arbitrary natural number k, this assumption is termed the induction hypothesis. Lastly, using the hypothesis, you prove that if the statement holds for k, it must also hold for k+1; this is known as the induction step. If both steps are successful, you've proved the statement for all natural numbers.

This process is somewhat analogous to falling dominos: if you know the first domino falls, and each domino will knock down the next one, you can conclude all dominos will fall without having to actually topple them sequentially.

Product Rule Differentiation

Product rule differentiation is a fundamental concept in differential calculus, used to find the derivative of the product of two functions. It states that the derivative of the product of two functions f(x) and g(x) is the product of the first function and the derivative of the second function plus the product of the second function and the derivative of the first. Formally, it is expressed as: \[ \frac{d}{dx}(f(x) \cdot g(x)) = f(x) \frac{d}{dx}g(x) + g(x) \frac{d}{dx}f(x) \] This rule is particularly useful when applying mathematical induction in the context of finding higher-order derivatives of product functions, which entails differentiating the product repeatedly.

Binomial Theorem

The binomial theorem provides a formula for expanding expressions that are raised to a power. Specifically, for a binomial (a + b) raised to the power n, where n is a non-negative integer, the expansion is given by: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \] In the formula, \( \binom{n}{k} \) represents the binomial coefficients, which tell us the number of ways to choose k elements from a set of n distinct elements. These coefficients are also the entries of Pascal's triangle. The theorem allows us to determine individual terms of the expansion directly without multiplying out the whole expression, which becomes particularly handy in various calculus problems, including the proof of Leibniz rule for higher-order derivatives.

Higher-Order Derivatives

Higher-order derivatives refer to the derivatives of a function taken successively more than once. The nth-order derivative of a function is essentially the derivative of the function's (n-1)th derivative. Higher-order derivatives come into play when we are investigating the curvature of graphs, acceleration in physics (which is the second derivative of the position), and in many other complex physical and mathematical problems.

They are notated by \( \frac{d^n}{dx^n}f(x) \) for the nth derivative. Calculating higher-order derivatives can often become challenging and hence the necessity for rules and methods such as the Leibniz rule, which simplifies the process when dealing with the product of two functions.

In applying the Leibniz rule, it's essential to understand how to differentiate terms of a product repeatedly, and how combinations of these derivatives interact, as shown in the mathematics of binomial expansion.