Single Variable Calculus, Chapter 7, 7.4-2, Section 7.4-2, Problem 38

Differentiate $y = \sqrt{x^2}$


$
\begin{equation}
\begin{aligned}

\sqrt{x^x} =& (e^{\ln x})^{\frac{x}{2}}
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\sqrt{x^x} =& e^{\frac{x}{2} \ln x}
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y =& e^{\frac{x}{2} \ln x}
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y' =& \frac{d}{dx} (e^{\frac{x}{2} \ln x})
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y' =& e^{\frac{x}{2} \ln x} \frac{d}{dx} \left( \frac{x}{2} \ln x \right)
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y' =& e^{\frac{x}{2} \ln x} \left[ \frac{x}{2} \frac{d}{dx} (\ln x) + \ln x \frac{d}{dx} \left( \frac{x}{2} \right) \right]
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y' =& e^{\frac{x}{2} \ln x} \left[ \frac{\cancel{x}}{2} \cdot \frac{1}{\cancel{x}} + \ln x \cdot \frac{1}{2 } \right]
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y' =& e^{\frac{x}{2} \ln x} \left( \frac{1}{2} + \frac{1}{2} \ln x \right)
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y' =& \frac{1}{2} e^{\frac{x}{2} \ln x} (1 + \ln x)
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& \text{ or }
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y' =& \frac{1}{2} \sqrt{x^x} (1 + \ln x)

\end{aligned}
\end{equation}
$