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 2103 views around the world You can reuse this answer Creative Commons License