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

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

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tangent of e^{-2x}

tangent of

e^{- 2 x}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = e^{- 2 x}

x = a_{0} : m = - 2 e^{- 2 a_{0}}

Find the line with slope m=

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

and passing through

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

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

y = (x + x^{- 1})^{3}

\frac{d}{d x} ((x + 2)^{\frac{1}{2}})

\frac{\partial}{\partial x} (ln (x^{2} + y - 5))

\frac{d y}{d x} + y = 5 sin (2 x)

\frac{d}{d x} (3 (x - 2)^{2})

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