Through the product rule:
#y'=(x^2+5^x)color(blue)(d/dx[e^(2x)])+e^(2x)color(magenta)(d/dx[x^2+5^x])#
Find each derivative individually (both will require the chain rule):
#color(blue)(d/dx[e^(2x)])=e^(2x)d/dx[2x]=color(blue)(2e^(2x))#
#color(magenta)(d/dx[x^2+5^x])=2x+color(green)(d/dx[5^x]#
Finding #d/dx[5^x]# will be much simpler if we recognize that #5^x=e^(ln5^x)=e^(xln5)#.
#color(green)(d/dx[e^(xln5)])=overbrace(e^(xln5))^("still"=5^x)d/dx[xln5]=color(green)(5^xln5#
So, plug this back into the derivative we were finding earlier:
#d/dx[x^2+5^x]=color(magenta)(2x+5^xln5#
Plug everything we know back into our original equation for #y'#:
#y'=color(blue)(2e^(2x))(x^2+5^x)+e^(2x)(color(magenta)(2x+5^xln5))#
#y'=e^(2x)(2x^2+2(5^x)+2x+5^xln5)#
