How do you differentiate #g(x) = xsqrt(x^2-x)# using the product rule? Calculus Basic Differentiation Rules Product Rule 1 Answer Topscooter Jan 16, 2016 #g'(x) = sqrt(x^2 - x) + (2x^2 - x)/(2sqrt(x^2 - x))# Explanation: By the product rule, #(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)#. Here, #u(x) = x# so #u'(x) = 1# and #v(x) = sqrt(x^2 - x)# so #v'(x) = (2x-1)/(2sqrt(x^2 - x))#, hence the result. Answer link Related questions What is the Product Rule for derivatives? How do you apply the product rule repeatedly to find the derivative of #f(x) = (x - 3)(2 - 3x)(5 - x)# ? How do you use the product rule to find the derivative of #y=x^2*sin(x)# ? How do you use the product rule to differentiate #y=cos(x)*sin(x)# ? How do you apply the product rule repeatedly to find the derivative of #f(x) = (x^4 +x)*e^x*tan(x)# ? How do you use the product rule to find the derivative of #y=(x^3+2x)*e^x# ? How do you use the product rule to find the derivative of #y=sqrt(x)*cos(x)# ? How do you use the product rule to find the derivative of #y=(1/x^2-3/x^4)*(x+5x^3)# ? How do you use the product rule to find the derivative of #y=sqrt(x)*e^x# ? How do you use the product rule to find the derivative of #y=x*ln(x)# ? See all questions in Product Rule Impact of this question 1437 views around the world You can reuse this answer Creative Commons License