What is the tangent of e^{-8x} ?

The tangent of e^{-8x} is y=-8e^{-8a_{0}}x+e^{-8a_{0}}+8a_{0}e^{-8a_{0}}

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tangent of e^{-8x}

tangent of

e^{- 8 x}

y = - 8 e^{- 8 a_{0}} x + e^{- 8 a_{0}} + 8 a_{0} e^{- 8 a_{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, e^{- 8 a_{0}})

Find the slope of

y = e^{- 8 x} : \frac{d y}{d x} = - 8 e^{- 8 x}

Compute the slope of

y = e^{- 8 x}

x = a_{0} : m = - 8 e^{- 8 a_{0}}

Find the line with slope m=

- 8 e^{- 8 a_{0}}

and passing through

(a_{0}, e^{- 8 a_{0}}) : y = - 8 e^{- 8 a_{0}} x + e^{- 8 a_{0}} + 8 a_{0} e^{- 8 a_{0}}

y = - 8 e^{- 8 a_{0}} x + e^{- 8 a_{0}} + 8 a_{0} e^{- 8 a_{0}}

y^{^{'}} = - \frac{4 x}{y}, y (2) = 3

x \to - 4 lim (- 2)

\int 6 x^{5} e^{x^{6}} d x

\frac{d}{d x} (x^{3} + 8 x)

f (x) = e^{2 y} (1 - y)

What is the tangent of e^{-8x} ?
The tangent of e^{-8x} is y=-8e^{-8a_{0}}x+e^{-8a_{0}}+8a_{0}e^{-8a_{0}}