What is the derivative of #ln(2x+1)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Henry W. Oct 18, 2016 #2/(2x+1)# Explanation: #y=ln(2x+1)# contains a function within a function, i.e. #2x+1# within #ln(u)#. Letting #u=2x+1#, we can apply chain rule. Chain rule: #(dy)/(dx)=(dy)/(du)*(du)/(dx)# #(dy)/(du)=d/(du)ln(u)=1/u# #(du)/(dx)=d/(dx)2x+1=2# #:.(dy)/(dx)=1/u*2=1/(2x+1)*2=2/(2x+1)# 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 47705 views around the world You can reuse this answer Creative Commons License