Question

Show that $$\frac{d^{l-m}}{d x^{l-m}}\left(x^{2}-1\right)^{l}=\frac{(l-m) !}{(l+m) !}\left(x^{2}-1\right)^{m} \frac{d^{l+m}}{d x^{l+m}}\left(x^{2}-1\right)^{l}$$ Hint: Write \(\left(x^{2}-1\right)^{l}=(x-1)^{l}(x+1)^{l}\) and find the derivatives by Leibniz' rule.

Short answer

The expression holds as shown by using the Leibniz rule and simplifying the derivatives.

Step-by-step solution

Express \((x^{2}-1)^{l}\)

Using the provided hint, express \((x^{2}-1)^{l}\) as \((x-1)^{l}(x+1)^{l}\).

Apply the Leibniz Rule

To find \(\frac{d^{l-m}}{d x^{l-m}}(x-1)^{l}(x+1)^{l}\), use Leibniz's rule for derivatives: \(\frac{d^{n}}{d x^{n}}(uv) = \sum_{k=0}^{n} {n\choose k} \frac{d^{k}u}{d x^{k}} \frac{d^{n-k}v}{d x^{n-k}}\)

Calculate Derivatives

Calculate the derivatives of u and v individually. Here, \(u=(x-1)^l\) and \(v=(x+1)^l\).

Combine Results

Combine the derivatives using the Leibniz rule to find \(\frac{d^{l-m}}{d x^{l-m}}(x^{2}-1)^{l}\).

Simplify the Expression

Simplify the obtained expression and compare it with the given theorem to find that:\(\frac{d^{l-m}}{d x^{l-m}}(x^{2}-1)^{l} = \frac{(l-m) !}{(l+m)!} (x^{2}-1)^{m} \frac{d^{l+m}}{d x^{l+m}}(x^{2}-1)^{l}\)

Key concepts

Leibniz Rule

Understanding the Leibniz rule is crucial when dealing with the derivatives of products of functions. The Leibniz rule states that when you want to differentiate a product of two functions, say \(u(x)\) and \(v(x)\), multiple times, you can use the formula:

• \[\frac{d^n}{dx^n}(uv) = \sum_{k=0}^{n} \binom{n}{k} \frac{d^k u}{dx^k} \frac{d^{n-k} v}{dx^{n-k}} \]

This formula accounts for all the ways to distribute the derivatives between the two functions. For example, if you have \(u(x) = (x-1)^l\) and \(v(x) = (x+1)^l\), then each term in the sum corresponds to a specific assignment of derivatives between \(u\) and \(v\). This makes it easier to compute higher-order derivatives for products of polynomial functions.

Higher-Order Derivatives

Higher-order derivatives refer to the process of taking the derivative of a function multiple times. If you start with a function \(f(x)\) and differentiate it once, you get the first derivative \(f'(x)\). Differentiating \(f'(x)\) gives you the second derivative \(f''(x)\), and so on. These are called higher-order derivatives.

• For example, if we consider the polynomial \(f(x) = (x^2 - 1)^l\), compute its first derivative to get \(f'(x)\).
• Then find the second derivative \(f''(x)\). Repeat this process for \(l+m\) derivatives.

When dealing with polynomials like \((x^2 - 1)^l\), taking higher-order derivatives can quickly get complicated due to the repeated application of the chain and product rules. Using the Leibniz rule simplifies this by breaking down the differentiation process.

Polynomial Functions

Polynomial functions are expressions that involve terms of the form \(ax^n\), where \(a\) is a coefficient and \(n\) is a non-negative integer. For instance, \((x^2 - 1)^l\) is a polynomial function.
Polynomials are straightforward to differentiate since each term's derivative follows basic power rules, but complications arise when these terms are products or nested within other functions.

• To differentiate \((x^2 - 1)^l\), use the hint to rewrite it as \((x-1)^l (x+1)^l\).
• Then, apply the Leibniz rule to manage the higher-order derivatives efficiently.

By restructuring the polynomial this way, it simplifies handling the differentiation process and allows leveraging the powerful Leibniz rule effectively for finding the required derivatives.