What is the tangent of y=e^{-x^2} ?

The tangent of y=e^{-x^2} is y=-2e^{-a_{0}^2}a_{0}x+e^{-a_{0}^2}+2a_{0}^2e^{-a_{0}^2}

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tangent of y=e^{-x^2}

tangent of

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

y = - 2 e^{- a_{0}^{2}} a_{0} x + e^{- a_{0}^{2}} + 2 a_{0}^{2} e^{- a_{0}^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, e^{- a_{0}^{2}})

Find the slope of

y = e^{- x^{2}} : \frac{d y}{d x} = - 2 e^{- x^{2}} x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 2 e^{- a_{0}^{2}} a_{0}

Find the line with slope m=

- 2 e^{- a_{0}^{2}} a_{0}

and passing through

(a_{0}, e^{- a_{0}^{2}}) : y = - 2 e^{- a_{0}^{2}} a_{0} x + e^{- a_{0}^{2}} + 2 a_{0}^{2} e^{- a_{0}^{2}}

y = - 2 e^{- a_{0}^{2}} a_{0} x + e^{- a_{0}^{2}} + 2 a_{0}^{2} e^{- a_{0}^{2}}

f (x) = \frac{x ^{2} + 1}{x}

\frac{\partial}{\partial y} (x e^{y} + y + 1)

\frac{d}{d t} (\frac{6 t + 28}{t + 7})

d er i v a t i v e f (x) = x^{3} (3 - x)^{4}

\frac{d}{d t} (\frac{1}{1 - t})

What is the tangent of y=e^{-x^2} ?
The tangent of y=e^{-x^2} is y=-2e^{-a_{0}^2}a_{0}x+e^{-a_{0}^2}+2a_{0}^2e^{-a_{0}^2}