What is the derivative of #y=ln[x/(2x+3)]^(1/2)#? Calculus Differentiating Logarithmic Functions Overview of Different Functions 1 Answer Sasha P. Oct 17, 2015 #y' = 3/(2x(2x+3))# Explanation: #y=ln[x/(2x+3)]^(1/2) = 1/2 ln[x/(2x+3)]# #y'=1/2 * 1/(x/(2x+3)) * (2x+3-2x)/(2x+3)^2 = 3/(2x(2x+3))# Answer link Related questions How do you find the derivative of #y=sin2x+cos2x+ln(ex)#? What is the derivative of #y= e^(3x/4)#? What is the derivative of #y= 2x^4 - 2x^3 - 8#? What is the derivative of # e^(x^3)+log_5(pi)#? What is the derivative of #ln(x^2)#? What is the derivative of #1/logx#? Question #b5198 Question #09fd4 Question #56c3f See all questions in Overview of Different Functions Impact of this question 3048 views around the world You can reuse this answer Creative Commons License