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

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

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

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

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

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