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The answer is (I think) $x^{\sqrt{x}-0.5} (1+0.5\ln(x))$, but how?

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    $\begingroup$ Hint: See if you can follow this one Derivative of $x^{x}$. Can you emulate it? $\endgroup$ – Amzoti Dec 19 '12 at 21:05
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    $\begingroup$ Hint: Write $x^{x^{1/2}} = e^{\ln(x)\cdot x^{1/2}}$ $\endgroup$ – Andrew Maurer Dec 20 '12 at 2:55
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The way to differentiate most expressions of the form $x^{f(x)}$ is to rewrite them as $e^{f(x)\ln(x)},$ and then use the chain and product rules, and that works in this case too.

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I'll probably get stones thrown at me for suggesting this, but let's view $x^{f(x)}$ as a hybrid of a power function $x^k$ and a composite exponential function $b^{f(x)}$. It's actually legitimate to take derivatives of these two simpler functions and add them together to get the derivative. $$\begin{align}\frac{d}{dx}x^{f(x)}&=\overbrace{f(x)x^{f(x)-1}}^{\text{power rule}}+\overbrace{x^{f(x)}\ln(x)}^{\text{exponential rule}}\cdot\overbrace{f'(x)}^{\text{chain rule}}\\ &=x^{f(x)} \left(\frac{f(x)}{x}+\ln(x)f'(x)\right)\end{align}$$ which is completely consistent with the "correct" approach of rewriting the function as $e^{f(x)\ln(x)}$ and using the chain and product rules. I don't recommend that you think this way, I just find this curious. I'd love to see a rigorous explanation for this happy coincidence that does not essentially appeal to rewriting as $e^{f(x)\ln(x)}$.


I think I found one. Let $F(y,z)=y^z$ with $y=g(x)$ and $z=f(x)$, then actually $F(y(x),z(x))=g(x)^{f(x)}$. Now using the chain rule for multivariate functions, $$\begin{align}\frac{d}{dx}g(x)^{f(x)}=\frac{dF}{dx}&=\frac{\partial F}{\partial y}\frac{dy}{dx}+\frac{\partial F}{\partial z}\frac{dz}{dx}\\&=zy^{z-1}\frac{dy}{dx}+y^z\ln(y)\frac{dz}{dx}\\&=f(x)(g(x))^{f(x)-1}g'(x)+g(x)^{f(x)}\ln(g(x))f'(x)\\&=g(x)^{f(x)}\left(\frac{f(x)g'(x)}{g(x)}+\ln(g(x))f'(x)\right)\end{align}$$

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Let's write $E'$ to mean the derivative of the expression $E$.

Then in general: $$\begin{align} u^v & = e^{v\log u} \\ (u^v)' &= \left(e^{v\log u}\right)'\\ &=e^{v\log u}\cdot (v\log u)'\\ &=u^v \left(v'\log u + v\frac {u'}u\right) \\ &=u^vv'\log u + u^{v-1}vu' \end{align}$$

To check, try taking one of $u$ or $v$ to be the identity function and the other to be a constant.

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$$y=x^{\sqrt{x}}$$ $$\ln y=\sqrt{x}\ln x$$ now differentiate w.r.t. x using chain rule $$\dfrac {1}{y}\cdot\dfrac {dy}{dx}=\dfrac{\sqrt{x}}{x}+\dfrac{\ln x}{2\sqrt x}$$ $$\dfrac {dy}{dx}=x^{\sqrt{x}}\cdot\left(\dfrac{2+\ln x}{2\sqrt x}\right)$$

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