Find y' for y = xe^-x
Find $y'$ for $y = xe^{-x}$.
Explanation of Method
A product of a linear factor and a decaying exponential.
Step 1: Identify the factors and apply the product rule
For $y = uv$ recall $y' = u'v + uv'$. Using the result of Problem 61 for the second factor: \[u = x,\quad u' = 1, \qquad v = e^{-x},\quad v' = -e^{-x}.\]
Step 2: Substitute and factor
\[ y' = 1\cdot e^{-x} + x\bigl(-e^{-x}\bigr) = e^{-x} - xe^{-x} = e^{-x}(1-x). \]
Step 3: Verification
\[x = 1:\quad y' = e^{-1}(1-1) = 0, \text{ a critical point} \;\;\checkmark\]
\[x = 0:\quad y' = 1\cdot 1 = 1; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = e^{-x}(1-x)\,}\]
Remark
Remark. Because $e^{-x} \gt 0$ always, the sign of $y'$ is the sign of $1-x$: the function increases on $(-\infty,1)$ and decreases on $(1,\infty)$, so $x = 1$ gives the global maximum $y(1) = e^{-1} \approx 0.368$. This curve is the standard shape of a first-order response in physics and engineering.