How do you differentiate #f(x)=ln(x)^x#?

Question

1219 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-20T01:04:36

#f'(x) = ln^x(x)(1/ln(x)+ln(ln(x)))#

Let #y = ln^x(x)#

#=>ln(y) = ln(ln^x(x)) = xln(ln(x))#

#=>d/dxln(y) = d/dxxln(ln(x))#

#=>1/ydy/dx = x(d/dxln(ln(x))) + ln(ln(x))(d/dxx)#

#=x(1/(ln(x))(d/dxln(x))) + ln(ln(x))#

#=x(1/ln(x)(1/x)) + ln(ln(x))#

#= 1/ln(x) + ln(ln(x))#

#=> dy/dx = y(1/ln(x) + ln(ln(x)))#

#= ln^x(x)(1/ln(x) + ln(ln(x)))#