How do you differentiate #y=(4/ln2)(x^(lnx))(lnx+2ln(cosx))#?

Question

2022 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-23T07:51:43

#y'=4/ln2 [2x^(lnx-1)lnx(lnx+2ln(cosx))+x^(lnx-1)-2x^(lnx)tanx]#

Lets find #(x^(lnx))'# :

#g=x^(lnx)#

#lng = ln x^(lnx)#

#lng = lnxlnx = ln^2x#

#1/g*g' = 2lnx*1/x#

#g' = 2lnx*1/x * g#

#g' = 2lnx*1/x * x^(lnx) = 2x^(lnx-1)lnx#

#y=(4/ln2)(x^(lnx))(lnx+2ln(cosx))#

#y'=4/ln2 [2x^(lnx-1)lnx(lnx+2ln(cosx))+x^(lnx)(1/x-2/cosxsinx)]#

#y'=4/ln2 [2x^(lnx-1)lnx(lnx+2ln(cosx))+x^(lnx-1)-2x^(lnx)tanx]#