What is the first (e^{-2t})^' ?

The first (e^{-2t})^' is -2e^{-2t}

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(e^{-2t})^'

(e^{- 2 t})^{^{'}}

- 2 e^{- 2 t}

Solution steps

(e^{- 2 t})^{'}

Apply the chain rule:

e^{- 2 t} (- 2 t)^{'}

= e^{- 2 t} (- 2 t)^{'}

(- 2 t)^{'} = - 2

= e^{- 2 t} (- 2)

Simplify

= - 2 e^{- 2 t}

\int \frac{- 6 x ^{4} + 6}{x ^{2} + 1} d x

\int \frac{1}{e ^{10 x}} d x

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

d er i v a t i v e f (x) = \frac{4 x + 3}{4 x - 3}

s l o p e in t erce pt (- 2, 7), (0, 5)