By taking the first derivative, we have

`dy/dx = d/dx[e^-2^(x^2)]`

`=[e^-2^(x^2)]*d/dx(-2x^2)`

`=[e^-2^(x^2)]*(-2*2x^(2-1))`

`=[e^-2^(x^2)]*(-4x) or-4x[e^-2^(x^2)]`

So the second derivative will be

`(d^2y)/dx^2 = -4[x*d/dx[e^-2^(x^2)]+[e^-2^(x^2)]*d/dx(x)]`

`(d^2y)/dx^2 = -4[x*[e^-2^(x^2)]*d/dx(-2x^2)+[e^-2^(x^2)]*1]`

`(d^2y)/dx^2 = -4[x*[e^-2^(x^2)]*(-4x)+[e^-2^(x^2)]]`

`(d^2y)/dx^2 = -4[e^-2^(x^2)](-4x^2+1)`

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The second derivative of `y = e^(-2x^2)` has to be determined.

Apply chain rule twice to obtain the first derivative,

`(dy)/(dx)=e^(-2x^2)*(-2x^2)'`

`=e^(-2x^2)*(-2)*2x`

`=-4xe^(-2x^2) `

Now apply product rule as well as chain rule as above to obtain the second derivative,

`(d^2y)/(dx^2)=-4[x(e^(-2x^2))'+(x)'e^(-2x^2)]`

`=-4[x*(-4)xe^(-2x^2) +e^(-2x^2)]`

`=-4e^(-2x^2)[-4x^2+1]`

`=e^(-2x^2)[16x^2-4]`

`=e^(-2x^2)[(4x)^2-2^2]`

`=e^(-2x^2)(4x-2)(4x+2)`