What is the d/(dt)(te^{-2t}) ?

The d/(dt)(te^{-2t}) is e^{-2t}-2e^{-2t}t

What is the first d/(dt)(te^{-2t}) ?

The first d/(dt)(te^{-2t}) is e^{-2t}-2e^{-2t}t

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\frac{d}{d t} (t e^{- 2 t})

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

Solution steps

\frac{d}{d t} (t e^{- 2 t})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t, g = e^{- 2 t}

= \frac{d t}{d t} e^{- 2 t} + \frac{d}{d t} (e^{- 2 t}) t

\frac{d t}{d t} = 1

\frac{d}{d t} (e^{- 2 t}) = - 2 e^{- 2 t}

= 1 \cdot e^{- 2 t} + (- 2 e^{- 2 t}) t

Simplify

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

\frac{\partial}{\partial x} (cos (x^{2} + y^{2}))

\int (2400 x) 3 + 2 x d x

\frac{d}{d x} (4 + 2 x^{\frac{3}{2}})

\frac{d}{d x} (ln ((x + 3)^{2}))

What is the d/(dt)(te^{-2t}) ?
The d/(dt)(te^{-2t}) is e^{-2t}-2e^{-2t}t
What is the first d/(dt)(te^{-2t}) ?
The first d/(dt)(te^{-2t}) is e^{-2t}-2e^{-2t}t