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 $(\genfrac{}{}{0ex}{}{n}{k})$ 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 $u^{(k)}=\frac{d^k}{d{x}^k}x=\{\begin{array}{ll}1& \text{if }k=1\\ 0& \text{if }k>1\end{array}$
• Let $v=\log x$. The derivatives of $v$ are given by:
  • First derivative: $v^{\prime }=\frac{1}{x}$
  • 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 $k\ge 1$.

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

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!