Question

Prove by induction: If \(n \geq 1\) and \(f^{(n)}\left(x_{0}\right)\) and \(g^{(n)}\left(x_{0}\right)\) exist, then so does \((f g)^{(n)}\left(x_{0}\right),\) and $$ (f g)^{(n)}\left(x_{0}\right)=\sum_{m=0}^{n}\left(\begin{array}{l} n \\ m \end{array}\right) f^{(m)}\left(x_{0}\right) g^{(n-m)}\left(x_{0}\right) . $$ HINT: See Exercise 1.2.19. This is Leibniz's rule for differentiating a product.

Short answer

Question: Prove by induction that if f(x) and g(x) are two functions, then the nth derivative of their product (fg)(x) is given by: \((fg)^{(n)}(x_0) = \sum_{m=0}^n \binom{n}{m} f^{(m)}(x_0) g^{(n-m)}(x_0)\)

Step-by-step solution

Base Case: n = 1

Let's start by checking the base case, when n = 1. For this case, we have to show that: \((fg)^{(1)}(x_0) = \sum_{m=0}^1 \binom{1}{m}f^{(m)}(x_0)g^{(1-m)}(x_0)\) We can expand the summation as follows: \((fg)^{(1)}(x_0) = \binom{1}{0}f^{(0)}(x_0)g^{(1-0)}(x_0) + \binom{1}{1}f^{(1)}(x_0)g^{(1-1)}(x_0)\) Now, we use the fact that \(\binom{n}{0} = \binom{n}{n} = 1\) and that \(f^{(0)}(x_0) = f(x_0)\) and \(g^{(0)}(x_0) = g(x_0)\): \((fg)^{(1)}(x_0) = f(x_0)g^{(1)}(x_0) + f^{(1)}(x_0)g(x_0)\) Which is the result we have to prove for the base case, since it represents the derivative of the product of f(x) and g(x).

Induction Step: Assuming true for n = k, prove for n = k + 1

Now, we need to assume the statement is true for n = k: \( (fg)^{(k)}(x_0) = \sum_{m=0}^{k} \binom{k}{m} f^{(m)}(x_0) g^{(k-m)}(x_0)\) And prove it for n = k + 1: \((fg)^{(k+1)}(x_0) = \sum_{m=0}^{k+1} \binom{k+1}{m} f^{(m)}(x_0) g^{(k+1-m)}(x_0)\) To prove this, we have to differentiate the previous equation one more time: \(\frac{d}{dx}[(fg)^{(k)}(x_0)] = \frac{d}{dx} \left[\sum_{m=0}^{k} \binom{k}{m} f^{(m)}(x_0) g^{(k-m)}(x_0)\right]\) Now, we apply Leibniz's rule for differentiating a product (from Exercise 1.2.19): \(\sum_{m=0}^{k}\binom{k}{m}\left[f^{(m+1)}(x_0)g^{(k-m)}(x_0) + f^{(m)}(x_0)g^{(k+1-m)}(x_0)\right]\) We can rewrite the above term as follows: \(\sum_{m=0}^{k}\binom{k}{m}f^{(m+1)}(x_0)g^{(k-m)}(x_0) + \sum_{m=0}^{k}\binom{k}{m}f^{(m)}(x_0)g^{(k+1-m)}(x_0)\) Now, notice that the last term in the first summation can be combined with the first term in the second summation, as they share the same factors of f and g, except for different binomial coefficients. We can rewrite the expression by combining these terms: \(\sum_{m=0}^{k}\binom{k}{m}f^{(m+1)}(x_0)g^{(k-m)}(x_0) + \binom{k}{k}f^{(k)}(x_0)g^{(1)}(x_0)+\sum_{m=0}^{k-1}\binom{k}{m}f^{(m)}(x_0)g^{(k+1-m)}(x_0)\) Using the identity \(\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}\), we can combine the terms: \(\sum_{m=0}^{k+1}\binom{k+1}{m}f^{(m)}(x_0)g^{(k+1-m)}(x_0)\) This concludes the proof since we have found the expression for \((fg)^{(k+1)}(x_0)\) as required for the induction step. Both the base case and the induction step have been proven, so the statement is true for all \(n \geq 1\).

Key concepts

Leibniz's Rule

Leibniz's Rule is an extension of the familiar product rule from calculus, which allows us to differentiate a product of functions multiple times. When dealing with higher order derivatives, Leibniz's Rule comes into play.

It explains that if you want to find the nth derivative of a product of two functions, say \(f(x)\) and \(g(x)\), the resulting expression is a summation formula involving combinations of the derivatives of \(f(x)\) and \(g(x)\).

Here's the formula:

• The nth derivative of the product \((fg)\) is expressed as \((fg)^{(n)}\).
• It is equal to the sum \(\sum_{m=0}^{n}\left(\begin{array}{c} n \ m \end{array}\right) f^{(m)}(x) g^{(n-m)}(x)\), where \(\binom{n}{m}\) is a binomial coefficient.

This formula beautifully combines the derivatives of each function in a way that mirrors how combinations work in probabilities.

Understanding Leibniz's Rule helps us extend differentiation beyond basic derivatives, giving a systematic way to tackle more complex problems.

Differentiation

Differentiation is one of the core concepts in calculus. It's the process that shows how a function changes at any given point, essentially measuring the rate at which one quantity changes concerning another. In a more intuitive sense, differentiation tells us the slope of the function at any given point.

The basic idea involves taking a function \(f(x)\) and determining \(f'(x)\), its derivative. This derivative is represented as the limit:

• \(f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}\)

The function \(f'(x)\) describes the instantaneous rate of change of \(f\) with respect to \(x\).

Through differentiation, we can gain insight into the behavior and properties of functions, such as:

• Identifying increasing or decreasing trends.
• Locating maxima, minima, and points of inflection.
• Understanding the velocity or acceleration in motion-related problems.

Differentiation is crucial because it acts as the foundation for more advanced applications in mathematics, physics, and engineering.

Product Rule

The Product Rule is a fundamental tool in calculus used specifically for the differentiation of products of two functions. When you have a function that is the product of two separate functions, \(f(x)\) and \(g(x)\), the Product Rule provides a way to find its derivative.
The formula of the Product Rule is straightforward:

• If \(h(x) = f(x)g(x)\), then \(h'(x) = f'(x)g(x) + f(x)g'(x)\).

This rule tells us to first differentiate \(f(x)\) and then multiply it by \(g(x)\), then add the product of \(f(x)\) and the derivative of \(g(x)\).
The Product Rule ensures that all parts of the product are accounted for when differentiating, reflecting how each part of the product can influence the change.
Understanding and applying the Product Rule is vital in handling more complex problems in calculus, where products of functions emerge frequently. It's a stepping stone to extensions like Leibniz's Rule for more involving differentiation tasks.