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

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Answer 1 · 2018-04-14T20:47:02

#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} ]#

#{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} #