Find $y^\prime$ if $y=x^{x}$, $x>0$.

Solution Notice that $x^{x}$ is neither $x$ raised to a fixed power $a$, nor a fixed base $a$ raised to a variable power. We follow the steps for logarithmic differentiation:

Step 1 Take the natural logarithm of each side of $y=x^{x}$, and simplify: $$\ln y=\ln x^{x}=x\ln x$$

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Step 2 Differentiate implicitly. $$\begin{eqnarray*} \dfrac{d}{dx}\ln y&=&\dfrac{d}{dx}(x\ln x)\\ \dfrac{y^\prime }{y} &=& x\cdot \dfrac{d}{dx}\ln x+\left( \dfrac{d}{dx}x\right) \cdot \ln x=x\left(\dfrac{1}{x}\right) +1\cdot \ln x=1+\ln x \end{eqnarray*}$$

Step 3 Solve for $y^\prime$: $y^\prime =y( 1+\ln x) =x^{x}( 1+\ln x).$