What is the tangent of e^{-3t},\at t=0 ?

The tangent of e^{-3t},\at t=0 is y=-3t+1

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tangent of e^{-3t},\at t=0

tangent of

e^{- 3 t}, at t = 0

y = - 3 t + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

Compute the slope of

y = e^{- 3 t}

t = 0 : m = - 3

Find the line with slope m=

- 3

and passing through

(0, 1) : y = - 3 t + 1

y = - 3 t + 1

\frac{d}{d x} (- 5 cos (x))

π^{2}

\frac{d}{d x} (36 - x^{2})

\int y csc^{2} (y) cot (y) d y

2 e^{3 t} - e^{2 t}

What is the tangent of e^{-3t},\at t=0 ?
The tangent of e^{-3t},\at t=0 is y=-3t+1