#### Answer

$$y'=\frac{2x^{\ln x}\ln x}{x}=2x^{\ln x-1}\ln x$$

#### Work Step by Step

$$y=x^{\ln x}$$
We can also use logarithmic differentiation to solve these types of problems. Take the natural logarithm of both sides: $$\ln y=\ln(x^{\ln x})$$ $$\ln y=\ln x\times\ln x=(\ln x)^2$$
Now use implicit differentiation to find $y'$ with respect to $x$: $$(\ln y)' =\Big(\ln x)^2\Big)'$$
$$\frac{y'}{y}=2\ln x(\ln x)'$$
$$\frac{y'}{y}=\frac{2\ln x}{x}$$
Find $y'$: $$y'=\frac{2\ln x}{x}\times y=\frac{2y\ln x}{x}$$
$$y'=\frac{2x^{\ln x}\ln x}{x}=2x^{\ln x-1}\ln x$$