How do you use logarithmic differentiation to find the derivative of the function #y=x^x#?

2 Answers
Mar 28, 2015

#y=x^x#

#lny=lnx^x#

#lny=xlnx#

Now differentiate implicitly:

#1/y (dy)/(dx)=(1)(lnx)+(x)(1/x)#

#1/y (dy)/(dx)=1+lnx#

Solve for #(dy)/(dx)#

#(dy)/(dx)=y(1+lnx)#

Recall that #y=x^x#, so

#(dy)/(dx)=x^x (1+lnx)#

Mar 28, 2015

It's easy if you remember the logarithmic rule:

#[f(x)]^g(x)=e^ln([f(x)]^g(x))=e^(g(x)*lnf(x)#,

so:

#y=x^x=e^(xlnx)rArr#

#y'=e^(xlnx)(1*lnx+x*1/x)=e^(xlnx)(lnx+1)#,

(or, if you want: #y'=x^x(lnx+1)#).