How do you find the derivative of #sqrt(2x-3)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Mauricio M. Apr 18, 2018 #f'(x) = 1/(sqrt(2x-3))# Explanation: #f(x) = sqrt(2x-3)# #f'(x) = 1/(2sqrt(2x-3))*2# #f'(x) = 1/(cancel2sqrt(2x-3))*cancel2# #f'(x) = 1/(sqrt(2x-3))# 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 4749 views around the world You can reuse this answer Creative Commons License