How do you find the derivative of #xsqrt(2x - 3)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Sasha P. Apr 1, 2016 #f'(x)= (3(x-1))/sqrt(2x-3)# Explanation: Using product rule #f=uv => f'=u'v+uv'#, we have: #f'(x)= (xsqrt(2x-3))'=1*sqrt(2x-3)+x*1/(2sqrt(2x-3))*2# #f'(x)= sqrt(2x-3)+x/sqrt(2x-3)# #f'(x)= (3x-3)/sqrt(2x-3)# #f'(x)= (3(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 1628 views around the world You can reuse this answer Creative Commons License