What is the tangent of e^{-3x} ?

The tangent of e^{-3x} is y=-3e^{-3a_{0}}x+e^{-3a_{0}}+3a_{0}e^{-3a_{0}}

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tangent of e^{-3x}

tangent of

e^{- 3 x}

y = - 3 e^{- 3 a_{0}} x + e^{- 3 a_{0}} + 3 a_{0} e^{- 3 a_{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, e^{- 3 a_{0}})

Find the slope of

y = e^{- 3 x} : \frac{d y}{d x} = - 3 e^{- 3 x}

Compute the slope of

y = e^{- 3 x}

x = a_{0} : m = - 3 e^{- 3 a_{0}}

Find the line with slope m=

- 3 e^{- 3 a_{0}}

and passing through

(a_{0}, e^{- 3 a_{0}}) : y = - 3 e^{- 3 a_{0}} x + e^{- 3 a_{0}} + 3 a_{0} e^{- 3 a_{0}}

y = - 3 e^{- 3 a_{0}} x + e^{- 3 a_{0}} + 3 a_{0} e^{- 3 a_{0}}

\frac{d}{d x} (tan (x))

x \to - \infty lim (1)

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

x \to 8 lim (\frac{( x - 2 )}{x - 8})

What is the tangent of e^{-3x} ?
The tangent of e^{-3x} is y=-3e^{-3a_{0}}x+e^{-3a_{0}}+3a_{0}e^{-3a_{0}}