# Implicit derivatives and logarithmic derivatives: $\left(x^{\sqrt{x}}\right)'=?$

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?

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