The second derivative of the following: y = e^(-2x^2)

2 Answers

kingattaskus12's profile pic

kingattaskus12 | (Level 3) Adjunct Educator

Posted on

By taking the first derivative, we have

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



    `=[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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llltkl's profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

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

Apply chain rule twice to obtain the first derivative,



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

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


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