#d/dx((x^2+2x+3)e^-x)#
#=\underbrace(d/dx(x^2+2x+3))_((1)) * e^-x+(x^2+2x+3) * \underbrace(d/dx(e^-x))_((2)#
#\color(white)(...... ..............................)#(By the product rule)
Lets calculate #(1)# first:
#d/dx(x^2+2x+3)=d/dx(x^2)+d/dx(2x)+d/dx(3) = 2x+2+0=2x+2#
Now, let's calculate #(2)#:
#d/dx(e^-x)=d/(d(-x))e^-xd/dx(-x)#
#\color(white)(...... ..............................)#(By the chain rule)
#= e^-x*-1=-e^-x#
Combining the two, we get:
#(2x+2)* e^(-x)+(x^2+2x+3)*(-e^(-x))#
#=\cancel(2xe^-x)+2e^-x -x^2e^-x\cancel(-2xe^-x)-3e^-x#
#=-x^2e^-x-e^-x#
