Find the n -th derivative of y = e^(-x/2)
Find the $n$-th derivative of $y = e^{-x/2}$.
Explanation of Method
The chain rule contributes the same constant factor every time.
Step 1: The first few derivatives
Each differentiation brings down the inner derivative $-\tfrac12$: \[y' = -\frac12e^{-x/2}, \qquad y'' = \frac14e^{-x/2}, \qquad y''' = -\frac18e^{-x/2}.\]
Step 2: The pattern, and its proof
\[ y^{(k)} = \left(-\frac12\right)^{k}e^{-x/2}. \tag{1} \]
The base case $k = 1$ is the first line above. Assuming (1) for $k$ and differentiating once more multiplies by another $-\tfrac12$: \[y^{(k+1)} = \left(-\frac12\right)^{k}\cdot\left(-\frac12\right)e^{-x/2} = \left(-\frac12\right)^{k+1}e^{-x/2}.\]
Step 3: Verification
\[\text{General rule } \bigl(e^{ax}\bigr)^{(n)} = a^{n}e^{ax} \text{ with } a = -\tfrac12 \;\;\checkmark\]
\[y^{(n)}(0) = \left(-\frac12\right)^{n}, \quad\text{matching } e^{-x/2} = \sum_{n=0}^{\infty}\frac{\bigl(-1\bigr)^{n}}{2^{n}n!}x^{n} \;\;\checkmark\]
\[\text{The signs alternate and the magnitudes halve: } 1,\ \tfrac12,\ \tfrac14,\ \tfrac18,\ \ldots \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y^{(n)} = \left(-\dfrac12\right)^{n}e^{-x/2} = \dfrac{\bigl(-1\bigr)^{n}}{2^{n}}\,e^{-x/2}\,}\]
Remark
Remark. Because $|a| = \tfrac12\lt 1$, the derivatives shrink geometrically rather than growing — the opposite of the logarithm in Problem 288. A function whose successive derivatives stay bounded on an interval has a Taylor series converging on that whole interval, which is why the expansion of $e^{-x/2}$ is valid for every $x$.