How do I find the derivative of #ln(ln(2x))#? Calculus Basic Differentiation Rules Chain Rule 1 Answer 1s2s2p May 11, 2018 #dy/dx=1/(xln(2x))# Explanation: #y=ln(ln(2x))# #dy/dx=d/dx[ln(ln(2x))]# #dy/dx=(d/dx[ln(2x)])/ln(2x)# #dy/dx=(((d/dx[2x])/(2x)))/ln(2x)# #dy/dx=((2/(2x)))/ln(2x)# #dy/dx=((1/x))/ln(2x)# #dy/dx=1/(xln(2x))# 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 11022 views around the world You can reuse this answer Creative Commons License