What is the derivative of #(e^(2x)sin(3x))#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Base e 1 Answer GiĆ³ Mar 29, 2015 I would use the Product Rule between functions #e# and #sin# and the Chain Rule to deal with the exponent and argument: #f'(x)=2e^(2x)sin(3x)+e^(2x)3cos(3x)=# #=e^(2x)[2sin(3x)+3cos(3x)]# Answer link Related questions What is the derivative of #y=3x^2e^(5x)# ? What is the derivative of #y=e^(3-2x)# ? What is the derivative of #f(theta)=e^(sin2theta)# ? What is the derivative of #f(x)=(e^(1/x))/x^2# ? What is the derivative of #f(x)=e^(pix)*cos(6x)# ? What is the derivative of #f(x)=x^4*e^sqrt(x)# ? What is the derivative of #f(x)=e^(-6x)+e# ? How do you find the derivative of #y=e^x#? How do you find the derivative of #y=e^(1/x)#? How do you find the derivative of #y=e^(2x)#? See all questions in Differentiating Exponential Functions with Base e Impact of this question 10272 views around the world You can reuse this answer Creative Commons License