How do you differentiate #g(x) = lnxe^(2x)# using the product rule? Calculus Basic Differentiation Rules Product Rule 1 Answer Alan N. Sep 8, 2016 Answer: #g'(x) = (2x+1)/x# Explanation: #g(x) = ln(xe^(2x))# #g'(x) = 1/(xe^(2x)) * d/dx(xe^(2x))# Standerd differential and chain rule #= 1/(xe^(2x)) * (x*e^(2x)*2 + e^(2x)*1)# Product rule, chain rule and standard differential #=(e^(2x) (2x+1))/(xe^(2x))# #=(cancel(e^(2x)) (2x+1))/(xcancel(e^(2x)))# #= (2x+1)/x# 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 - ... 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 207 views around the world You can reuse this answer Creative Commons License