#f(x) = e^(-x^2) - x^2e^x#
derive #e^(-x^2)# and #-x^2e^x# separately.
#f'(x) = (Deltay)/(Deltax) (e^(-x^2)) - (Deltay)/(Deltax)(x^2e^x)#
using the chain rule:
#(Deltay)/(Deltax) ( e^(-x^2)) = e^(-x^2) * (Deltay)/(Deltax) ( -x^2)#
#= e^(-x^2) * - (Deltay)/(Deltax)(x^2)#
#-2xe^(-x^2)#
using the product rule:
#(Deltay)/(Deltax) (-x^2e^x) = ((Deltay)/(Deltax)(-x^2) * e^x) + ((Deltay)/(Deltax)(e^x) + -x^2)#
#(Deltay)/(Deltax)(-x^2) = -2x#
#(Deltay)/(Deltax)(e^x) = e^x#
#((Deltay)/(Deltax)(-x^2) * e^x) + ((Deltay)/(Deltax)(e^x) + -x^2) = -2xe^x + -x^2e^x#
#(-2xe^(-x^2)) + (-2xe^x + -x^2e^x) = -2xe^(-x^2) - 2xe^x - x^2e^x#
#f'(x) = -2xe^(-x^2) - 2xe^x - x^2e^x#
to find the gradient of the tangent line, substitute #x# for #4:#
#-2xe^(-x^2) - 2xe^x - x^2e^x = -8e^(-16) - 8e^4 - 16e^4#
#-8e^(-16) - 8e^4 - 16e^4 = -8e^(-16) - 24e^4#
#= -1310.355# (3d.p.)
the tangent is a straight line, with equation #y = mx+c#.
at #x = 4, y = e^(-4^2) - 16e^4 = e^(-16) - 16e^4 = -873.570# (3d.p.)
at the tangent, #-1310.355x + c = -873.570#
#c = -873.570 - (-1310.355 * 4) = -873.570 - -5241.42#
#= 4367.85#
therefore the equation of the tangent is #y = -1310.355x + 4367.85# (figures to 3 d.p.)
