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

1 Answer
Oct 23, 2015

Answer:

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

Explanation:

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]#