Question

One of the Leibniz Rules One of several Leibniz Rules in calculus deals with higher-order derivatives of products. Let \((f g)^{(n)}\) denote the \(n\) th derivative of the product \(f g,\) for \(n \geq 1\) a. Prove that \((f g)^{(2)}=f^{\prime \prime} g+2 f^{\prime} g^{\prime}+f g^{\prime \prime}\) b. Prove that, in general, $$(f g)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) f^{(k)} g^{(n-k)}$$ where \(\left(\begin{array}{l}n \\ k\end{array}\right)=\frac{n !}{k !(n-k) !}\) are the binomial coefficients. c. Compare the result of (b) to the expansion of \((a+b)^{n}\)

Short answer

Answer: The second derivative formula for the product of two functions is \((fg)^{(2)} = f''g + 2f'g' + fg''\), which can be derived using the Product Rule for differentiation. The general Leibniz Rule can be proved by mathematical induction: $$(fg)^{(n)} = \sum_{k=0}^{n}\binom{n}{k}f^{(k)}g^{(n-k)}$$ Comparing the general Leibniz Rule and the binomial expansion, we can see they share structural similarities. Both have a summation from k=0 to n and involve binomial coefficients in the sum. They also both have terms with exponents related to k and (n-k). The difference is that the Leibniz Rule deals with derivatives of functions, whereas the binomial expansion deals with powers of constants. The general Leibniz Rule can be viewed as a generalization of the binomial expansion applied to derivatives of product functions instead of constants.

Step-by-step solution

Proving the Second Derivative Formula

To prove that \((fg)^{(2)} = f''g + 2f'g' + fg''\), we will first compute the first and second derivatives of the product by applying the Product Rule. The Product Rule states that: $$(fg)' = f'g + fg'$$ Now, apply the Product Rule again to find the second derivative: $$(fg)'' = (fg')' + (f'g)'$$ Now, applying the Product Rule to both terms: $$(fg)'' = (f''g + fg'') + (f''g' + f'g'')$$ Combining terms, we get: $$(fg)^{(2)} = f''g + 2f'g' + fg''$$

Proving the General Leibniz Rule

To prove the general Leibniz Rule, we use mathematical induction. The base case \(n = 1\) is the Product Rule itself, which we know is true. Now assume that the Leibniz Rule is true for some arbitrary \(n\). We want to prove that it also holds for \(n+1\). Then we have: $$(fg)^{(n+1)} = \frac{d}{dx}(fg)^{(n)}$$ Using the induction hypothesis as the formula for \((fg)^{(n)}\), we get: $$(fg)^{(n+1)} = \frac{d}{dx}\left(\sum_{k=0}^{n}\binom{n}{k}f^{(k)}g^{(n-k)}\right)$$ Now differentiate term by term using the linearity of differentiation: $$(fg)^{(n+1)} = \sum_{k=0}^{n}\binom{n}{k}\left(f^{(k+1)}g^{(n-k)} + f^{(k)}g^{(n-k+1)}\right)$$ Now, we regroup terms with the same value of k: $$(fg)^{(n+1)} = \sum_{k=0}^{n}\binom{n}{k}f^{(k+1)}g^{(n-k)} + \sum_{k=0}^{n}\binom{n}{k}f^{(k)}g^{(n-k+1)}$$ To get the general Leibniz Rule for \(n + 1\), we need to adjust the sums' indices: $$\sum_{k=0}^{n+1}\binom{n+1}{k}f^{(k)}g^{((n+1)-k)}$$ To adjust the indices in the sum, we observe that when \(k = n + 1\), the binomial coefficient is 1. Therefore, we can write: $$(fg)^{(n+1)} = \sum_{k=0}^{n+1}\binom{n+1}{k}f^{(k)}g^{((n+1)-k)}$$ Which is the general Leibniz Rule for \(n + 1\). This completes the proof by induction.

Comparing the General Rule to Binomial Expansion

The general Leibniz Rule is given by: $$(fg)^{(n)} = \sum_{k=0}^{n}\binom{n}{k}f^{(k)}g^{(n-k)}$$ The binomial expansion of \((a+b)^n\) is given by: $$(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{k}b^{(n-k)}$$ Comparing the general Leibniz Rule and the binomial expansion, we can see that they have the same structure: 1. Both involve a summation from \(k=0\) to \(n\). 2. Both use binomial coefficients \(\binom{n}{k}\) in their sum. 3. Both have terms with exponents that depend on \(k\) and \((n-k)\). The difference between the two is that, in the Leibniz Rule, the terms are derivatives of functions (\(f^{(k)}\) and \(g^{(n-k)}\)), whereas in the binomial expansion, the terms are powers of constants (\(a^{k}\) and \(b^{(n-k)}\)). Both expressions represent a weighted sum of products with coefficients given by the binomial coefficients. The general Leibniz Rule can be seen as a generalization of the binomial expansion, applying it to derivatives of product functions instead of constants.

Key concepts

Higher-Order Derivatives

Higher-order derivatives are essentially the repeated application of the differentiation process to a function. For instance, taking the first derivative of a function gives us the rate of change of that function. By taking the derivative of the derivative, we get the second-order derivative, which can provide us with information about the curvature of the function's graph, or acceleration if thinking in terms of motion.

Continuing this process, the third-order derivative, fourth-order derivative, and so on are obtained, collectively known as higher-order derivatives. These are crucial in various fields, such as physics, where higher-order derivatives describe phenomena like jerk and jounce in kinematics. In the exercise, the second-order derivative of a product of functions showcases a more complex application of differentiation involving higher-order terms.

Product Rule

The product rule is a key calculation method used when finding the derivative of the product of two functions. The rule states that the derivative of a product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function. Symbolically, it can be expressed as \( (fg)' = f'g + fg' \).

This concept is foundational in calculus, especially when dealing with more complex cases that involve higher-order derivatives, as seen in the exercise where the product rule is applied repeatedly to derive the required formula.

Binomial Coefficients

Binomial coefficients are the numbers that appear as coefficients in the binomial expansion. They can also be viewed as the number of ways to choose a subset of a certain size from a larger set, known as combinations in combinatorics. They are denoted by \( \binom{n}{k} \) and can be calculated using the formula \( \frac{n!}{k!(n-k)!} \), where \( ! \) denotes the factorial operation.

These coefficients play a crucial role in the general Leibniz rule for differentiation of products, since they determine the weight of each term in the expansion. Understanding how these coefficients work is vital for students to make sense of how the formula for the nth derivative of a product is derived.

Mathematical Induction

Mathematical induction is a powerful proof technique often utilized to establish the truth of an infinite sequence of statements. The process involves two steps: first, proving the base case (the first statement in the sequence is true), and second, proving the inductive step (if any one statement in the sequence is true, then so is the next one).

Induction mirrors the 'domino effect'. Once the first domino falls (the base case), and provided that each domino is set up to knock down the next (the inductive step), you can conclude all dominoes will fall. In the context of the exercise, mathematical induction is used to prove the general form of the Leibniz rule, which confirms the pattern of differentiation holds true for all natural numbers.

Binomial Expansion

Binomial expansion describes the algebraic expansion of powers of a binomial. According to the binomial theorem, the expansion of \( (a+b)^n \) can be expressed as a sum of terms involving powers of \( a \) and \( b \) and the binomial coefficients. The formula is \( (a+b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{k}b^{n-k} \).

The resemblance between the binomial expansion and the pattern obtained through the Leibniz rule reveals the deep connection between algebra and calculus. As highlighted in the general Leibniz rule, instead of simple exponents, we have derivatives of functions. Grasping this analogy assists students in understanding the structure and behavior of derivative products in a familiar algebraic context.