Given: #f(x) = e^(-x^2#
Required:#(d^2f(x))/(dx)^2 = (f'(x))'#
Definitions and Principles:
a) Derivative of - #f'(x)=(de^x)/(dx)=e^x#
b) Chain Rule - #(df(x))/dx= (df(x))/(du)*(du)/dx#
c) Product Rule - #(d[f(x)*r(x)])/(dx) = (df(x))/(dx)*r(x)+(dr(x))/(dx)*f(x)#
Solution Strategy:
1) Set up #(d^2f(x))/(dx)^2# using b) the chain rule, use a)
2) For 2nd derivative you will have to set up chain and product rule
Solution:
1) Let #u=-x^2# then we we can #g(u)=e^u#
#h(x)= (df(x))/dx= (dg(u))/(du)*(du)/dx#
# h(x)= (de^u)/(du)*(dx^2)/dx= e^u(-2x)|_(u=x^2)= -2xe^(x^2)#
2) Now we need to take the derivative of h(x).
Let #r(x)=2x and f(x)=-e^(-x^2#
#h'(x)=(dr(x))/(dx)*f(x)+(df(x))/(dx)*r(x)#
#h'(x)=-2e^(-x^2) - 2x*color(red)[(de^(-x^2))/(dx)]#
Now from 1) we know:
#f'(x)=color(red)[(de^(-x^2))/(dx) =-2x e^(x^2)]#
#h'(x)=-2e^(-x^2) - (2x)*color(red)[-2xe^(x^2)]#
Answer: #f''(x)=-2e^(-x^2) + (2x)^2e^(x^2)=2e^(x^2)[x^2-1]#
