Find the n -th derivative of y = 1/(1+x)
Find the $n$-th derivative of $y = \dfrac{1}{1+x}$.
Explanation of Method
Signed factorials again, now over a rising power.
Step 1: The first few derivatives
Write $y = \bigl(1+x\bigr)^{-1}$; the inner derivative is $1$, so only the power rule acts: \[y' = -\bigl(1+x\bigr)^{-2}, \quad y'' = 2\bigl(1+x\bigr)^{-3}, \quad y''' = -6\bigl(1+x\bigr)^{-4}, \quad y^{\mathrm{IV}} = 24\bigl(1+x\bigr)^{-5}.\]
Step 2: The pattern
Each step contributes one more negative factor, and the product $\bigl(-1\bigr)\bigl(-2\bigr)\cdots\bigl(-k\bigr) = \bigl(-1\bigr)^{k}k!$: \[y^{(k)} = \frac{\bigl(-1\bigr)^{k}\,k!}{\bigl(1+x\bigr)^{k+1}}. \tag{1}\]
Step 3: Induction
Assuming (1) and differentiating: \[y^{(k+1)} = \bigl(-1\bigr)^{k}k!\cdot\bigl(-(k+1)\bigr)\bigl(1+x\bigr)^{-(k+2)} = \frac{\bigl(-1\bigr)^{k+1}\bigl(k+1\bigr)!}{\bigl(1+x\bigr)^{k+2}}.\]
Step 4: Verification
\[y^{(n)}(0) = \bigl(-1\bigr)^{n}n!, \quad\text{so the Maclaurin coefficients are } \bigl(-1\bigr)^{n} \;\;\checkmark\]
\[\frac{1}{1+x} = 1-x+x^{2}-x^{3}+\cdots \quad (|x|\lt 1) \;\;\checkmark\]
\[\text{Since } \frac{1}{1+x} = \bigl(\ln(1+x)\bigr)', \text{ Problem 288 with } k = n+1 \text{ gives the same thing} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y^{(n)} = \dfrac{\bigl(-1\bigr)^{n}\,n!}{\bigl(1+x\bigr)^{n+1}}\,}\]
Remark
Remark. This is Problem 288 differentiated once: since $\dfrac{1}{1+x} = \dfrac{d}{dx}\ln(1+x)$, the $n$-th derivative here is the $(n+1)$-st there, and indeed $\bigl(-1\bigr)^{n}n!$ is $\bigl(-1\bigr)^{(n+1)-1}\bigl((n+1)-1\bigr)!$. The geometric series and the logarithmic series are two faces of the same computation.