Using Logarithmic Differentiation

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 \(y^\prime \): \(y^\prime =y( 1+\ln x) =x^{x}( 1+\ln x).\)