Find the n -th derivative of y = ln x
Find the $n$-th derivative of $y = \ln x$.
Explanation of Method
After one step the logarithm becomes a power, and factorials appear.
Step 1: The first few derivatives
\[ y' = x^{-1}, \qquad y'' = -x^{-2}, \qquad y''' = 2x^{-3}, \qquad y^{\mathrm{IV}} = -6x^{-4}. \]
The coefficients $1,-1,2,-6$ are the signed factorials $0!,-1!,2!,-3!$.
Step 2: The conjecture
\[ y^{(k)} = \bigl(-1\bigr)^{k-1}\bigl(k-1\bigr)!\,x^{-k}, \qquad k\ge1. \tag{1} \]
Check: $k=1$ gives $x^{-1}$; $k=2$ gives $-x^{-2}$; $k=3$ gives $2x^{-3}$; $k=4$ gives $-6x^{-4}$.
Step 3: Induction
Assume (1) for some $k\ge1$. Differentiating, the power rule supplies a factor $-k$: \[y^{(k+1)} = \bigl(-1\bigr)^{k-1}\bigl(k-1\bigr)!\cdot\bigl(-k\bigr)x^{-k-1} = \bigl(-1\bigr)^{k}\,k!\,x^{-(k+1)},\]
which is (1) with $k+1$ in place of $k$.
Step 4: Verification
\[x = 1:\ y^{(n)}(1) = \bigl(-1\bigr)^{n-1}\bigl(n-1\bigr)!\]
\[\ln\bigl(1+x\bigr) = \sum_{n=1}^{\infty}\frac{\bigl(-1\bigr)^{n-1}}{n}x^{n}, \quad\text{coefficient } \frac{\bigl(-1\bigr)^{n-1}\bigl(n-1\bigr)!}{n!} = \frac{\bigl(-1\bigr)^{n-1}}{n} \;\;\checkmark\]
\[n = 5,\ x = 2:\ y^{(5)} = \frac{4!}{32} = 0.75 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y^{(n)} = \dfrac{\bigl(-1\bigr)^{n-1}\bigl(n-1\bigr)!}{x^{n}}, \qquad n\ge1\,}\]
Remark
Remark. The alternating signs and growing factorials are why the series for $\ln(1+x)$ converges only for $|x|\le1$: the derivatives grow like $(n-1)!$, which the division by $n!$ in the Taylor coefficients barely tames, leaving terms of size $1/n$. Compare the exponential, whose derivatives do not grow at all and whose series converges everywhere.