find the nth derivative of y=xlogx,using the leibniz theorem

To find the nth derivative of the function \( y = x \log x \) using Leibniz's theorem, we first need to understand what Leibniz's theorem states. This theorem provides a formula for the nth derivative of a product of two functions. In our case, we can consider \( u = x \) and \( v = \log x \). The theorem is expressed as follows:

Leibniz's Theorem Overview

Leibniz's theorem states that if you have two functions \( u(x) \) and \( v(x) \), the nth derivative of their product is given by:

(uv)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} u^{(k)} v^{(n-k)}

Here, \( u^{(k)} \) is the k-th derivative of \( u \), and \( v^{(n-k)} \) is the (n-k)-th derivative of \( v \). The binomial coefficient \( \binom{n}{k} \) represents the number of ways to choose \( k \) elements from a set of \( n \) elements.

Identifying the Functions

For our function \( y = x \log x \):

• Let \( u = x \) and thus \( u^{(k)} = \frac{d^k}{dx^k} x = \begin{cases} 1 & \text{if } k = 1 \\ 0 & \text{if } k > 1 \end{cases} \)
• Let \( v = \log x \). The derivatives of \( v \) are given by:
  • First derivative: \( v' = \frac{1}{x} \)
  • Second derivative: \( v'' = -\frac{1}{x^2} \)
  • Third derivative: \( v''' = \frac{2}{x^3} \)
  • In general, the k-th derivative of \( v \) can be expressed as \( v^{(k)} = (-1)^{k-1} \frac{(k-1)!}{x^k} \) for \( k \geq 1 \).

Applying Leibniz's Theorem

Now, we can apply Leibniz's theorem to find the nth derivative of \( y = x \log x \):

y^{(n)} = \sum_{k=0}^{n} \binom{n}{k} u^{(k)} v^{(n-k)}

Since \( u^{(k)} = 0 \) for \( k > 1 \), we only need to consider \( k = 0 \) and \( k = 1 \):

• For \( k = 0 \):
  • \( u^{(0)} = x \) and \( v^{(n)} = (-1)^{n-1} \frac{(n-1)!}{x^n} \)
  • Contribution: \( \binom{n}{0} x \cdot (-1)^{n-1} \frac{(n-1)!}{x^n} = (-1)^{n-1} \frac{(n-1)!}{x^{n-1}} \)
• For \( k = 1 \):
  • \( u^{(1)} = 1 \) and \( v^{(n-1)} = (-1)^{n-2} \frac{(n-2)!}{x^{n-1}} \)
  • Contribution: \( \binom{n}{1} \cdot 1 \cdot (-1)^{n-2} \frac{(n-2)!}{x^{n-1}} = n \cdot (-1)^{n-2} \frac{(n-2)!}{x^{n-1}} \)

Combining the Contributions

Now, we can combine both contributions to find the nth derivative:

y^{(n)} = (-1)^{n-1} \frac{(n-1)!}{x^{n-1}} + n \cdot (-1)^{n-2} \frac{(n-2)!}{x^{n-1}}

By simplifying this expression, we can factor out common terms:

y^{(n)} = \frac{(-1)^{n-2}}{x^{n-1}} \left( (n-1)! - n(n-2)! \right)

Finally, we can express this as:

y^{(n)} = \frac{(-1)^{n-2}}{x^{n-1}} \left( (n-1)! - n(n-2)! \right) = \frac{(-1)^{n-2}}{x^{n-1}} \left( (n-1)! - n(n-2)! \right)

This gives us the nth derivative of \( y = x \log x \) using Leibniz's theorem. Each step logically builds on the previous one, leading us to a clear and concise result. If you have any further questions or need clarification on any part of this process, feel free to ask!