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Use logarithmic differentiation to find the derivative of y = x √ x.
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\[ y = x^1\cdot x^{1/2} = x^{3/2} \]
the x is being raised to sqrt of X
\[ \ln(y) = \ln(x^{3/2}) \]
This is a good start.
\[ x= x^1, \sqrt{x} = x^{1/2}, x\sqrt{x} = x^{1+1/2} = x^{3/2} \]
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sorry the x is being raised to the power of sqrtX
\[X^\sqrt{X}\]
Okay then \[ \ln(x^{\sqrt{x}}) = \sqrt{x}\ln(x) \]
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