What is the d/(dt)(te^{-3t}) ?

The d/(dt)(te^{-3t}) is e^{-3t}-3e^{-3t}t

What is the first d/(dt)(te^{-3t}) ?

The first d/(dt)(te^{-3t}) is e^{-3t}-3e^{-3t}t

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\frac{d}{d t} (t e^{- 3 t})

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

Solution steps

\frac{d}{d t} (t e^{- 3 t})

Apply the Product Rule:

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

f = t, g = e^{- 3 t}

= \frac{d t}{d t} e^{- 3 t} + \frac{d}{d t} (e^{- 3 t}) t

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

\frac{d}{d t} (e^{- 3 t}) = - 3 e^{- 3 t}

= 1 \cdot e^{- 3 t} + (- 3 e^{- 3 t}) t

Simplify

= e^{- 3 t} - 3 e^{- 3 t} t

\frac{d}{d x} (\frac{6 ( x ^{2} + 1 )}{( x ^{2} - 3 ) ^{3}})

e^{x}, x e^{x^{2}}, 0, 1

\frac{2 s + 5}{s ^{2} + 6 s + 34}

\frac{d}{d x} (x^{\frac{1}{4}} - 2)

(y + x^{2} + y^{2}) d x = x d y

What is the d/(dt)(te^{-3t}) ?
The d/(dt)(te^{-3t}) is e^{-3t}-3e^{-3t}t
What is the first d/(dt)(te^{-3t}) ?
The first d/(dt)(te^{-3t}) is e^{-3t}-3e^{-3t}t