How do you find the derivative of xsqrt(2x - 3)x√2x−3? Calculus Basic Differentiation Rules Chain Rule 1 Answer Sasha P. Apr 1, 2016 f'(x)= (3(x-1))/sqrt(2x-3)f'(x)=3(x−1)√2x−3 Explanation: Using product rule f=uv => f'=u'v+uv'f=uv⇒f'=u'v+uv', we have: f'(x)= (xsqrt(2x-3))'=1*sqrt(2x-3)+x*1/(2sqrt(2x-3))*2f'(x)=(x√2x−3)'=1⋅√2x−3+x⋅12√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 1728 views around the world You can reuse this answer Creative Commons License