The slope of f(x)f(x) at x=-1x=−1 is determined by differentiating the function at x=-1x=−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 iscolor(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
