How do you find the derivative of #x^(2 lnx)#?

Question

2520 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-15T12:07:29

#y'=4x^(2lnx)lnx/x#

#y=x^(2lnx)#

by taking the natural logarithm to both sides

#lny=lnx^(2lnx)#

using the properties of the logarithmic functions
#color(green) (lnu^v=vlnu)#

#lny=2lnx*lnx#

#lny# =#2(lnx)^2#

Differentiate

#(y')/y=4lnx/x#

#y'=4ylnx/x#

Substitute for #y#

#y'=4x^(2lnx)lnx/x#