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

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

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Answer 1 · 2018-04-14T20:47:02

$\frac{d y}{d x} = - 2 x e^{- x^{2}}$ $dy/dx = -2xe^{-x^2}$

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

Explanation:

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

$\frac{d y}{d x} = \frac{d}{d x} [e^{- x^{2}}]$ $dy/dx=d/dx[e^{-x^2}]$

$\frac{d^{2} y}{{d x}^{2}} = \frac{d}{d x} [\frac{d}{d x} [e^{- x^{2}}]]$ ${d^2y}/{dx^2} = d/dx[d/dx[e^{-x^2}]]$

let $u = - x^{2}$ $u=-x^2$

$\frac{d}{d x} [e^{- x^{2}}] = \frac{d}{d u} [e^{u}] \frac{d}{d x} [- x^{2}]$ $d/dx[e^{-x^2}]=d/{du}[e^u]d/dx[-x^2]$

$\frac{d}{d x} [e^{- x^{2}}] = e^{u} \times - 2 x$ $d/dx[e^{-x^2}]=e^u times -2x$

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

--

$\frac{d^{2} y}{{d x}^{2}} = \frac{d}{d x} [- 2 x e^{- x^{2}}]$ ${d^2y}/{dx^2} = d/dx[-2xe^{-x^2} ]$

Product rule:

$\frac{d^{2} y}{{d x}^{2}} = \frac{d}{d x} [- 2 x] e^{- x^{2}} + - 2 x \frac{d}{d x} [e^{- x^{2}}]$ ${d^2y}/{dx^2} = d/dx[-2x]e^{-x^2} + -2x d/dx[e^{-x^2}]$

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

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

$\frac{d}{d x} [- 2 x] = - 2$ $d/dx[-2x] = -2$

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

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

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

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