How do you differentiate #y = ln(x^(1/2))#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Jim H Apr 1, 2015 #y=ln(x^(1/2)) = 1/2 ln(x)# #(dy)/(dx)=1/2 d/(dx)(ln(x))=1/2*1/x=1/(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 7317 views around the world You can reuse this answer Creative Commons License