The slope of #f(x)# at #x=-1# is determined by differentiating the function at #x=-1# that is by computing #color(red)(f'(-1))#

Differentiate #f(x)# is determined by using the quotient rule differentiation

Let #u(x)=x+2 and v(x)=e^(x-x^2)#

Then #f(x)=(u(x))/(v(x))#

#color(red)(f'(x)=(color(blue)(u'(x))*v(x)-color(green)(v'(x))*u(x))/(v^2(x))#

Let us determine #color(blue)(u'(x))# and #color(green)(v'(x))#

#u(x)=x+2 rArr color(blue)(u'(x)=1)#

# v(x)=e^(x-x^2)#

is a composite of two function ,the exponential and polynomial so its derivative is determined by using chain rule.

Let #color(brown)(g(x)=e^x and h(x)=x-x^2)#

#color(brown)(v(x)=g(h(x)))#

Then the derivative of the composite function is#color(green)(v'(x)=g'(h(x))*h'(x))#

#g(x)=e^xrArrg'(x)=e^x#

#g'(h(x))=e^(h(x))#

#color(green)(g'(h(x))=e^(x-x^2))#

# h(x)=x-x^2rArrcolor(green)(h'(x)=1-2x)#

#color(green)(v'(x)=e^(x-x^2)*(1-2x))#

#color(red)(f'(x)=(color(blue)(u'(x))*v(x)-color(green)(v'(x))*u(x))/(v^2(x))#

#color(red)(f'(x))=(color(blue) 1(e^(x-x^2))-color(green)(e^(x-x^2)(1-2x)) *(x+2))/ (e^(x-x^2))^2#

#color(red)(f'(x))=((e^(x-x^2))-(e^(x-x^2))(1-2x) *(x+2))/ (e^(x-x^2))^2#

#color(red)(f'(x))=(e^(x-x^2)[1-(1-2x) *(x+2)])/ (e^(x-x^2))^2#

Simplifying the quotient by #e^(x-x^2)#

#color(red)(f'(x))=(1-(1-2x) *(x+2))/ (e^(x-x^2))#

Then,

#color(red)(f'(-1))=(1-(1-2(color(red)(-1))) *(color(red)(-1)+2))/ (e^(color(red)(-1)-color(red)((-1))^2))#

#color(red)(f'(x))=(1-(1+2) *(+1))/ (e^(-2))#

#color(red)(f'(x))=(-2)/ (e^(-2))#

#color(red)(f'(x))=-2e^2#