How do you differentiate #y=(4/ln2)(x^(lnx))(lnx+2ln(cosx))#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Sasha P. Oct 23, 2015 #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]# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 1996 views around the world You can reuse this answer Creative Commons License