What is the tangent of e^{-2t},\at t=0 ?

The tangent of e^{-2t},\at t=0 is y=-2t+1

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tangent of e^{-2t},\at t=0

tangent of

e^{- 2 t}, at t = 0

y = - 2 t + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

Compute the slope of

y = e^{- 2 t}

t = 0 : m = - 2

Find the line with slope m=

- 2

and passing through

(0, 1) : y = - 2 t + 1

y = - 2 t + 1

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

x \to - 1 lim (2 x^{2} + 6 x - 1)

x \to 2 lim (\frac{5}{∣ 2 - x ∣})

\frac{d}{d x} (e^{- \frac{x}{a}})

f (x) = e^{x} arctan (x)

What is the tangent of e^{-2t},\at t=0 ?
The tangent of e^{-2t},\at t=0 is y=-2t+1