Given f(x)=-x/(e^(x-x^3))f(x)=−xex−x3 at x=-1x=−1
Determine the first derivative f' (x) first, then compute for f' (-1) to obtain the slope
We make use of the derivative of quotient for this type of function
use the formula d/dx(u/v)=(v*d/dx(u)-u*d/dx(v))/v^2
Let u=x and v=e^(x-x^3)
f' (x)=d/dx(f(x))=
d/dx(-x/(e^(x-x^3)))=(-1)*(e^(x-x^3)*d/dx(x)-x*d/dx(e^(x-x^3)))/(e^(x-x^3))^2
d/dx(-x/(e^(x-x^3)))=(-1)*((e^(x-x^3)*1-xe^(x-x^3)(1-3x^2)))/(e^(x-x^3))^2
There is no need for further simplification just use x=-1 right away
f' (-1)=(-1)*((e^(-1-(-1)^3)*1-(-1)e^(-1-(-1)^3)(1-3(-1)^2)))/(e^(-1-(-1)^3))^2
f' (-1)=(-1)*((e^(-1+1)-(-1)e^(-1+1)(1-3)))/(e^(-1+1))^2
f' (-1)=(-1)*((e^(0)-(-1)e^(0)(-2)))/(e^(0))^2
f' (-1)=(-1)*((1-2))/(1)^2
f' (-1)=+1
God bless....I hope the explanation is useful.
