How do you find the derivative of #x^ln x#?

Question

2005 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-04T18:55:45

#f'(x)=2x^(ln(x))*(ln(x)/x)#

Taking the logarithm on both sides we get

#ln(f(x))=ln(x)*ln(x)#
differentiating with respect to #x#:
#1/f(x)*f'(x)=ln(x)/x+ln(x)/x)#
so we get

#f'(x)=2x^(ln(x))*(ln(x)/x)#

Answer 2 · 2018-07-04T18:59:17

#\frac{d}{dx}(x^{\ln x})=x^{\lnx}(\frac{1}{x^2}+\ln x)#

For a positive real number #x#, applying chain rule of differentiation as follows

#\frac{d}{dx}(x^{\ln x})#

#=x^{\lnx-1}\frac{d}{dx}(\ln x)+x^{\lnx}\ln x\frac{d}{dx}(x)#

#=x^{\lnx-1}(\frac{1}{x})+x^{\lnx}\ln x#

#=x^{\lnx}\frac{1}{x^2}+x^{\lnx}\ln x#

#=x^{\lnx}(\frac{1}{x^2}+\ln x)#