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How would I find the derivative with respect to $x$ of $$y = \left(x^{\sqrt{x}}\right)'.$$ I can find the correct answer using the method of logarithmic differentiation that my book mysteriously suggest to use for some reason, BUT how can I find the derivative without using that method?

Furthermore how would I know when I should use logarithmic differentiation? And why can I not get the correct answer using the Chain rule only?

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First of all logarithmic differentiation is not mysterious as it might seem, rather it makes the calculations look visibly simpler (in fact they are simpler to type also while using $\mathrm\LaTeX$). The idea is that a complicated expression of type $\{f(x)\}^{g(x)}$ can be handled easily by first taking its logarithm.

Thus if $$y = f(x) = x^{\sqrt{x}}\tag{1}$$ then we have $$\log y = \sqrt{x}\log x\tag{2}$$ The next step is to differentiate the above equation with respect to $x$ On the LHS we don't see the variable $x$ written explicitly, but that is not a problem because the LHS is actually $\log y = \log f(x)$. So differentiating it via chain rule we get $$\frac{d}{dx}\,\log y = \frac{d}{dx}\,\log f(x) = \frac{1}{f(x)}\cdot f'(x) = \frac{f'(x)}{f(x)}\tag{3}$$ Differentiating the RHS of equation $(2)$ via product rule we get its derivative as $$\frac{\sqrt{x}}{x} + \frac{\log x}{2\sqrt{x}} = \frac{2 + \log x}{2\sqrt{x}}$$ It now follows from $(3)$ that $$\frac{f'(x)}{f(x)} = \frac{2 + \log x}{2\sqrt{x}}$$ and hence the final answer is $$f'(x) = f(x)\cdot\frac{2 + \log x}{2\sqrt{x}} = \frac{x^{\sqrt{x}}(2 + \log x)}{2\sqrt{x}}$$

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You may write, for $x>0$, $$ x^{\sqrt{x}}=e^{\sqrt{x}\ln x} $$ then, you get $$ \left(x^{\sqrt{x}}\right)'=(\sqrt{x}\ln x)'e^{\sqrt{x}\ln x} $$ using $$ (e^u)'=u'e^u. $$ Finally, we have

$$ \left(x^{\sqrt{x}}\right)'=\left(1+\frac{\ln x}2\right)x^{\sqrt{x}-\frac12},\qquad x>0. $$

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Alternatively, using a logarithmic derivative as was suggested to you, from $$y=x^{\sqrt{x}}=e^{\sqrt{x}\ln x}$$ we have $$ (\ln y)'=\frac{y'}y=(\sqrt{x}\ln x)'=\frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}x $$ that is $$ \frac{(x^{\sqrt{x}})'}{x^{\sqrt{x}}}=\frac{\ln x}{2\sqrt{x}}+\frac1{\sqrt{x}} $$ giving

$$ \left(x^{\sqrt{x}}\right)'=\left(\frac{\ln x}2+1\right)x^{\sqrt{x}-\frac12},\qquad x>0. $$

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