How to differentiate #y=xe^x(sinx+cosx)# ?

#y=xe^x(sinx+cosx)#

1 Answer
Steve M
Jun 19, 2018

# dy/dx = e^x(sinx + (1+2x)cosx) #

Explanation:

We seek the differential of:

# y=xe^x(sinx+cosx) #

Using the triple product rule we have:

# dy/dx = (d/dxx)e^x(sinx+cosx) #
# \ \ \ \ \ \ \ \ \ \ + x(d/dxe^x)(sinx+cosx) #
# \ \ \ \ \ \ \ \ \ \ + xe^x(d/dx(sinx+cosx)) #

# \ \ \ \ \ = (1)e^x(sinx+cosx) #
# \ \ \ \ \ \ \ \ \ \ + x(e^x)(sinx+cosx) #
# \ \ \ \ \ \ \ \ \ \ + xe^x(cosx-sinx)) #

# \ \ \ \ \ = e^x(sinx+cosx) #
# \ \ \ \ \ \ \ \ \ \ + xe^x{sinx+cosx + cosx-sinx)} #

# \ \ \ \ \ = e^x(sinx+cosx) #
# \ \ \ \ \ \ \ \ \ \ + xe^x(2cosx) #

# \ \ \ \ \ = e^x(sinx+cosx + 2xcosx) #

# \ \ \ \ \ = e^x(sinx + (1+2x)cosx) #

Related questions
Impact of this question
2187 views around the world
You can reuse this answer
Creative Commons License