We can use the quotient rule, here, which states that, for #y=f(x)/g(x)#,
#(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/(g(x))^2#
Thus, as #f(x)=x# and #g(x)=2+e^x#,
#(dy)/(dx)=(1*(2+e^x)-x(e^x))/(2+e^x)^2=color(green)((2+e^x-xe^x)/(2+e^x)^2)#
Now, we can rewrite this first derivative as a product: #(2+e^x-xe^x)(2+e^x)^-2#
Applying the product rule, which states that when #y=f(x)g(x)#,
#(dy)/(dx)=f'(x)g(x)+f(x)g'(x)#
It'll be easier for us to find #f'(x)# and #g'(x)# before proceeding to the whole derivation, so we do not get lost.
#f'(x)=0+e^x+color(red)(1*e^x+x*e^x)#, thus #color(green)(f'(x)=2e^x+xe^x)#
Note that the red part was the demanded chain rule for #xe^x#.
Now, #g'(x)# demands chain rule, which states that #(dy)/(dx)=(dy)/(du)(du)/(dx)#.
Renaming #u=2+e^x#, we have #g(u)=u^-2#
So,
#(dy)/(du)=-2u^-3# and #(du)/(dx)=e^x#
Aggregating them in a multiplication as stated by chain rule:
#(dy)/(dx)=(-2u^-3)(e^x)=(-2(2+e^x)^-3)(e^x)#
#g'(x)=color(green)((-2(e^x))/(2+e^x)^3)#
Now that we have #f(x)#, #f'(x)#, #g(x)# and #g'(x)#, we can go to product rule:
#(d^2y)/(dx^2)=(e^x+e^x+xe^x)(e^x+2)^-2+(2+e^x+xe^x)(-2e^x(e^x+2)^-3)#
#(d^2y)/(dx^2)=(2e^x+xe^x)/(e^x+2)^2+((2+e^x+xe^x)(-2e^x))/(e^x+2)^3#
To operate this sum, we have that the lowest common denominator (l.c.d.) is #(e^x+2)^3#. Thus,
#(d^2y)/(dx^2)=((e^x+2)(2e^x+xe^x)+(-2e^x)(2+e^x+xe^x))/(e^x+2)^3#
Distributing the factors, we get the following, which allows us to operate some cancelling, as I will remark in collors:
#(d^2y)/(dx^2)=(color(red)(2e^(2x))color(green)(+xe^(2x))+color(blue)(4e^x)+2xe^xcolor(blue)(-4e^x)color(red)(-2e^(2x))color(green)(-2)xe^(2x))/(e^x+2)^3#
Final simplification:
#(d^2y)/(dx^2)=(2xe^x-xe^(2x))/(e^x+2)^3=color(green)((e^x(2-e^x)x)/(e^x+2)^3)#
