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