## University Calculus: Early Transcendentals (3rd Edition)

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