Find y' for y = x^(1/x), x > 0
Find $y'$ for $y = x^{1/x}$, $x \gt 0$.
Explanation of Method
The function whose maximum identifies the number e.
Step 1: Take the natural logarithm
\[ \ln y = \frac{1}{x}\ln x = \frac{\ln x}{x}. \]
Step 2: Differentiate both sides
Quotient rule on the right: $\frac{y'}{y} = \frac{\dfrac1x\cdot x - \ln x\cdot 1}{x^{2}} = \frac{1-\ln x}{x^{2}}$.
Step 3: Solve for $y'$
\[ y' = x^{1/x}\cdot\frac{1-\ln x}{x^{2}}. \]
Step 4: Verification
\[x = e:\quad \ln e = 1 \;\Longrightarrow\; y' = \frac{e^{1/e}\cdot 0}{e^{2}} = 0, \text{ a critical point} \;\;\checkmark\]
\[x = 1:\quad y = 1,\quad y' = \frac{1\cdot 1}{1} = 1; \qquad \frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]
Step 5: The nature of the critical point
Since $x^{1/x} \gt 0$ and $x^{2} \gt 0$, the sign of $y'$ is the sign of $1-\ln x$: $y' \gt 0 \ \text{ for } 0 \lt x \lt e, \qquad y' \lt 0 \ \text{ for } x \gt e$.
So $x = e$ is a global maximum, with value $e^{1/e} \approx 1.44467$.
Result
Final answer:
\[\boxed{\,y' = \dfrac{x^{1/x}\bigl(1-\ln x\bigr)}{x^{2}}, \qquad x > 0\,}\]
Remark
Remark. This is the classical proof that $e^{1/e}$ is the largest value of $x^{1/x}$ — equivalently, that among all positive integers $3^{1/3} \gt 2^{1/2}$, since $3$ is the integer nearest $e$ on the decreasing side. Cubing and squaring gives the familiar comparison $\sqrt[3]{3} \approx 1.4422 \gt \sqrt{2} \approx 1.4142$.