How do you differentiate #g(x)=sqrtxe^x#? Calculus Basic Differentiation Rules Product Rule 1 Answer Monzur R. Dec 27, 2016 #g'(x)=e^x(1/(2sqrt(x))+sqrt(x))# Explanation: #g(x)=sqrt(x)e^x=(x)^(1/2)e^x# #u=(x)^(1/2), v=e^x# #g'(x)=vu'+uv'# #u'=1/2x^(-1/2)=1/(2sqrt(x)), v'=e^x# #g'(x)=e^x/(2sqrt(x))+sqrt(x)e^x=e^x(1/(2sqrt(x))+sqrt(x))# 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 2478 views around the world You can reuse this answer Creative Commons License