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How do you find the first and second derivative of #y=e^(-x^2)#?

Answer 1

Luke Alvarez

#dy/dx = -2xe^{-x^2} #

#{d^2y}/{dx^2} = 4x^2e^{-x^2}-2e^{-x^2} #

#y=e^{-x^2}#

#dy/dx=d/dx[e^{-x^2}]#

#{d^2y}/{dx^2} = d/dx[d/dx[e^{-x^2}]]#

let #u=-x^2#

#d/dx[e^{-x^2}]=d/{du}[e^u]d/dx[-x^2]#

#d/dx[e^{-x^2}]=e^u times -2x#

#d/dx[e^{-x^2}]=-2xe^{-x^2} #

#{d^2y}/{dx^2} = d/dx[-2xe^{-x^2} ]#

Product rule:

#{d^2y}/{dx^2} = d/dx[-2x]e^{-x^2} + -2x d/dx[e^{-x^2}]#

From earlier: #d/dx[e^{-x^2}]=-2xe^{-x^2} #

#{d^2y}/{dx^2} = d/dx[-2x]e^{-x^2} + -2x(-2xe^{-x^2} )#

# d/dx[-2x] = -2#

#{d^2y}/{dx^2} = -2e^{-x^2} + 4x^2e^{-x^2} #

#{d^2y}/{dx^2} = 4x^2e^{-x^2}-2e^{-x^2} #

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Answer 2

Axel Alvarez

To find the first derivative, use the chain rule: dy/dx = -2x * e^(-x^2).

To find the second derivative, differentiate again using the product rule: d^2y/dx^2 = (4x^2 - 2) * e^(-x^2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.