What is the tangent of f(x)=e^{-x^2},\at x=5 ?

The tangent of f(x)=e^{-x^2},\at x=5 is y=-(10)/(e^{25)}x+(51)/(e^{25)}

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tangent of f(x)=e^{-x^2},\at x=5

tangent of

f (x) = e^{- x^{2}}, at x = 5

y = - \frac{10}{e ^{25}} x + \frac{51}{e ^{25}}

Solution steps

Find the tangent point:

(5, \frac{1}{e ^{25}})

Find the slope of

f (x) = e^{- x^{2}} : \frac{df ( x )}{d x} = - 2 e^{- x^{2}} x

Compute the slope of

f (x) = e^{- x^{2}}

x = 5 : m = - \frac{10}{e ^{25}}

Find the line with slope m=

- \frac{10}{e ^{25}}

and passing through

(5, \frac{1}{e ^{25}}) : y = - \frac{10}{e ^{25}} x + \frac{51}{e ^{25}}

y = - \frac{10}{e ^{25}} x + \frac{51}{e ^{25}}

x \to 0 lim (- \frac{3}{1 - 3 x})

f (x) = cot^{2} (sin (θ))

\frac{d}{d x} (sec (x) y)

\frac{d y}{d t} = y - 5

\int (5 x^{2} + 4)^{3} d x

What is the tangent of f(x)=e^{-x^2},\at x=5 ?
The tangent of f(x)=e^{-x^2},\at x=5 is y=-(10)/(e^{25)}x+(51)/(e^{25)}