Derivative of $y=x^{\ln x}$? I only know how to do one step:
$$
\ln\left(\,y\,\right) = \ln\left(\, x^{\ln\left(\, x\,\right)}\,\right)
$$
how do i do the derivative of
$\ln\left(\, x^{\ln\left(\, x\,\right)}\,\right)$ ?. I know the answer to that is $\ln\left(\,x\,\right)\ln\left(\,x\,\right)$, but how does the $x$ go away ?.
 A: Hint: $x^{\ln x} = e^{(\ln x)^2}$
A: We have $$\ln(y) = \ln^2(x) \implies \dfrac{dy}y = \dfrac{2\ln(x)dx}x \implies \dfrac{dy}{dx} = \dfrac{2y\ln(x)}x = 2x^{\ln(x)-1}\ln(x)$$
A: You have
$$
y(x) = x^{\log x} = \exp (\log x \log x) = \exp(\log^2 x)
$$
Now using the chain rule:
$$
y'(x) = \exp(\log^2 x) \times 2\frac{\log x} x = 2x^{\log x - 1}\log x
$$
A: If $$\ln{y}=\ln{x^{\ln{x}}}=\ln{x}\cdot\ln{x}=\ln^2{x}$$ then 
$$\frac{y'}{y}=2\ln{x}\cdot\frac{1}{x}\implies y'=2y\frac{\ln{x}}{x}=2x^{\ln{x}}\cdot\frac{\ln{x}}{x}=2x^{(\ln{x}-1)}\ln{x}$$
A: Here are the steps
$$ \frac{d}{dx}\left[x^{\ln x}\right]=\frac{d}{dx}\left[e^{\ln x^{\ln x}}\right] =\frac{d}{dx}\left[e^{(\ln x)(\ln x)}\right]=\frac{d}{dx}\left[e^{(\ln x)^2}\right]=e^{(\ln x)^2}\frac{d}{dx}\left[(\ln x)^2\right]=e^{(\ln x)^2}(2\ln x)\frac{d}{dx}\left[\ln x\right]=e^{(\ln x)^2}(2\ln x)\frac{1}{x}=\frac{x^{\ln x}}{x}(2\ln x)=2x^{\ln x - 1}\ln x $$
A: $$\Big(x^{\ln x}\Big)'~=~\Big(x^n\Big)'_{n~{\large=}~\ln x}+\Big(a^{\ln x}\Big)'_{a~{\large=}~x}~=~\Big(n\cdot x^{n-1}\Big)_{n~{\large=}~\ln x}+\bigg(a^{\ln x}\cdot\ln a\cdot\ln'x\bigg)_{a~{\large=}~x}~=~$$
$$~=~\ln x\cdot x^{\ln x-1}+x^{\ln x}\cdot\dfrac{\ln x}x~=~2\cdot x^{\ln x-1}\cdot\ln x.$$
