1. The problem states: If $y = x^{n-1} \log x$, prove that the $n^{th}$ derivative $y^{(n)} = \frac{(n-1)!}{x}$. 2. We start with the function: $$y = x^{n-1} \log x$$ 3. Use the Leibniz rule for the $n^{th}$ derivative of a product of two functions $u(x)$ and $v(x)$: $$\frac{d^n}{dx^n} [u(x)v(x)] = \sum_{k=0}^n \binom{n}{k} u^{(k)}(x) v^{(n-k)}(x)$$ 4. Here, let: - $u(x) = x^{n-1}$ - $v(x) = \log x$ 5. Compute derivatives: - For $u^{(k)}(x)$: $$u^{(k)}(x) = \frac{d^k}{dx^k} x^{n-1} = \frac{(n-1)!}{(n-1-k)!} x^{n-1-k}$$ for $k \leq n-1$ (otherwise zero). - For $v^{(m)}(x)$: $$v(x) = \log x$$ $$v'(x) = \frac{1}{x}$$ $$v''(x) = -\frac{1}{x^2}$$ $$v^{(m)}(x) = (-1)^{m-1} (m-1)! \frac{1}{x^m} \quad \text{for } m \geq 1$$ 6. Because $u^{(k)}(x) = 0$ for $k \geq n$, the sum reduces to $k=0$ to $n-1$. 7. Compute $y^{(n)}$: $$y^{(n)} = \sum_{k=0}^{n-1} \binom{n}{k} u^{(k)}(x) v^{(n-k)}(x)$$ 8. Substitute: $$y^{(n)} = \sum_{k=0}^{n-1} \binom{n}{k} \frac{(n-1)!}{(n-1-k)!} x^{n-1-k} \times (-1)^{n-k-1} (n-k-1)! \frac{1}{x^{n-k}}$$ 9. Simplify the powers of $x$: $$x^{n-1-k} \times \frac{1}{x^{n-k}} = x^{-1}$$ 10. Factorials: $$(n-1)! (n-k-1)! / (n-1-k)! = (n-1)!$$ because $(n-k-1)! / (n-1-k)! = 1$. 11. So $$y^{(n)} = x^{-1} (n-1)! \sum_{k=0}^{n-1} \binom{n}{k} (-1)^{n-k-1}$$ 12. Rearranging powers of $-1$: $$(-1)^{n-1} \sum_{k=0}^{n-1} \binom{n}{k} (-1)^{-k} = (-1)^{n-1} \sum_{k=0}^{n-1} \binom{n}{k} (-1)^k$$ 13. Using the binomial theorem: $$\sum_{k=0}^n \binom{n}{k} (-1)^k = (1 - 1)^n = 0$$ 14. Therefore, $$\sum_{k=0}^{n-1} \binom{n}{k} (-1)^k = -\binom{n}{n} (-1)^n = -(-1)^n = (-1)^{n+1}$$ 15. Substitute back: $$y^{(n)} = x^{-1} (n-1)! (-1)^{n-1} (-1)^{n+1} = x^{-1} (n-1)!$$ 16. Final result: $$y^{(n)} = \frac{(n-1)!}{x}$$ This completes the proof.