Question 1183844
{{{y = (x+2)^x = e^(ln(x+2)^x) = e^(xln(x+2))}}}


===> {{{dy/dx = e^(xln(x+2))*(1*ln(x+2) + x*(1/(x+2))) =(x+2)^x*(ln(x+2) + x/(x+2) )}}}  after applying the chain rule.


===> {{{dy(-1)/dx =(-1+2)^(-1)*(ln(-1+2) + -1/(-1+2) ) = -1}}}