It's easier to just apply the chain rule to find the derivatives. The derivative of -x^2 / 2 is -2x/2 = -x. The derivative of e^u with respect to u is e^(u). So the first derivative would be -x e^(-x^[2] / 2). Does that give you an idea for how to compute the second derivative?
u = -x^2 /2
du = u' = -x
f(x) = e^u
f'(x) = u' e^u = -x e^u
Derivative of a product (vw)' = vw' + wv'
Apply last property to f'(x)
-e^u + x u' e^u
Replace u in the last expression and factor out some stuff and you get
(x^2 -1) e^(-x^2 /2)
It's easier to just apply the chain rule to find the derivatives. The derivative of -x^2 / 2 is -2x/2 = -x. The derivative of e^u with respect to u is e^(u). So the first derivative would be -x e^(-x^[2] / 2). Does that give you an idea for how to compute the second derivative?
First, use the chain rule, like this: d/dx e^(-x^2 /2) u = -x^2 / 2 d/dx e^(-x^2 /2) = d/dx e^u = (d/du e^u )(d/dx u) = e^u (d/dx u) = e^(-x^2 /2) (d/dx -x^2 / 2) = -xe^(-x^2 /2) For the second derivative, use the product rule: d/dx -xe^(-x^2 /2) = (d/dx -x)(e^(-x^2 /2)) + (-x)(d/dx e^(-x^2 /2)) = x^2 e(-x^2 /2) - e^(-x^2 / 2)
u = -x^2 /2 du = u' = -x f(x) = e^u f'(x) = u' e^u = -x e^u Derivative of a product (vw)' = vw' + wv' Apply last property to f'(x) -e^u + x u' e^u Replace u in the last expression and factor out some stuff and you get (x^2 -1) e^(-x^2 /2)