Question

\(\quad(d / d x)\left[x^{-n} j_{n}(x)\right]=-x^{-n} j_{n-1}(x)\)

Short answer

The derivative is \( \frac{d}{dx} \bigg[ x^{-n} j_{n}(x) \bigg] = -x^{-n} j_{n-1}(x) \)

Step-by-step solution

Identify the Differential Expression

We need to compute the derivative of the expression: \( \frac{d}{dx} \bigg[ x^{-n} j_{n}(x) \bigg] \)

Apply the Product Rule

The expression inside the derivative is a product of two functions: \( x^{-n} \) and \( j_{n}(x) \). According to the product rule, \[ \frac{d}{dx} [u(x) v(x)] = u'(x)v(x) + u(x)v'(x) \], where \( u(x) = x^{-n} \) and \( v(x) = j_{n}(x) \)

Differentiate Each Function

First, find the derivative of \( u(x) = x^{-n} \): \[ u'(x) = -n x^{-n-1} \]. Next, find the derivative of \( v(x) = j_{n}(x) \), which we'll denote as \( v'(x) = j'_{n}(x) \). The exact form of \( j'_{n}(x) \) isn't required for this problem.

Combine Using the Product Rule

Apply the product rule using our derivatives: \[ \frac{d}{dx} [ x^{-n}j_{n}(x) ] = -n x^{-n-1} j_{n}(x) + x^{-n} j'_{n}(x) \]

Simplify and Use Function Properties

Combine terms and note any simplifications. Specifically, observe the given identity: \[ \frac{d}{dx} \bigg[ x^{-n} j_{n}(x) \bigg] = -x^{-n} j_{n-1}(x) \] We verify that our derivative satisfies this relation.

Key concepts

Bessel functions

Bessel functions are essential in many fields, including physics and engineering. They appear in problems involving wave propagation and static potential. The primary reason for their usefulness is that they are solutions to Bessel's differential equation: They can be divided into two kinds: the first kind, denoted as \(J_{n}(x)\), and the second kind, denoted as \(Y_{n}(x)\). In this problem, we are particularly interested in the first kind, \(J_{n}(x)\). To calculate derivatives involving Bessel functions, you typically use known identities, which can simplify complex expressions.

Product rule

The product rule is a fundamental technique in differential calculus. It allows us to differentiate functions that are products of two simpler functions. The rule states:

Derivative computation

Computing derivatives is a foundational skill in calculus. Derivatives measure how functions change with respect to changes in their inputs. In the context of this problem, we first applied the product rule, and then focused on differentiating each individual component. To break it further down:

- For the term \({x^{-n}}\), we used the power rule which states: $$\frac{d}{dx} {x^{k}} = kx^{k-1}$$
- For the Bessel function \({j_{n}(x)}\), we noted that its derivative is denoted as \({j'_n(x)}\). Knowing these individual derivatives lets us apply the product rule effectively, and arrive at the desired result. Understanding the steps of differentiation is crucial to solving more complex calculus problems.

Differential equations

Differential equations play a significant role in physics and engineering. They help describe various phenomena, from heat transfer to electrical circuits. A differential equation involves functions and their derivatives, and solving these can be complex, depending on the scenario. In this exercise, we see how understanding specific derivatives contributes to solving more comprehensive differential equations. The given Bessel function \({j_{n}(x)}\) is an example of solutions to differential equations that model physical scenarios. Gaining a good grasp of these foundational techniques will serve students well, as such principles recur in many advanced topics in scientific studies.