Find y' for y = x^x, x > 0

Find $y'$ for $y = x^{x}$, $x \gt 0$.

Step 1: Why neither standard rule works

The power rule $\bigl(x^{n}\bigr)' = nx^{n-1}$ needs a constant exponent; the exponential rule $\bigl(a^{x}\bigr)' = a^{x}\ln a$ needs a constant base. Here $x$ occupies both positions, so we take logarithms first.

Step 2: Take the natural logarithm

\[ \ln y = x\ln x. \]

Step 3: Differentiate both sides

The left side gives $\dfrac{y'}{y}$ by the chain rule; the right side needs the product rule: \[\frac{y'}{y} = \frac{d}{dx}\bigl(x\ln x\bigr) = \ln x + x\cdot\frac1x = \ln x + 1.\]

Step 4: Solve for $y'$

\[ y' = y\bigl(1+\ln x\bigr) = x^{x}\bigl(1+\ln x\bigr). \]

Step 5: Verification

\[x = 1:\quad y = 1,\quad y' = 1\cdot(1+0) = 1; \qquad \frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]

\[y' = 0 \;\Longleftrightarrow\; \ln x = -1 \;\Longleftrightarrow\; x = e^{-1} \approx 0.3679, \text{ the minimum } y = e^{-1/e} \approx 0.6922 \;\;\checkmark\]