What is the d/(dt)(4te^{-t}) ?

The d/(dt)(4te^{-t}) is 4(e^{-t}-e^{-t}t)

What is the first d/(dt)(4te^{-t}) ?

The first d/(dt)(4te^{-t}) is 4(e^{-t}-e^{-t}t)

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\frac{d}{d t} (4 t e^{- t})

4 (e^{- t} - e^{- t} t)

Solution steps

\frac{d}{d t} (4 t e^{- t})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d}{d t} (t e^{- t})

Apply the Product Rule:

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

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

= 4 (\frac{d t}{d t} e^{- t} + \frac{d}{d t} (e^{- t}) t)

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

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

= 4 (1 \cdot e^{- t} + (- e^{- t}) t)

Simplify

= 4 (e^{- t} - e^{- t} t)

f (x) = \frac{x ^{2}}{x ^{2} - 16}

\int 5 cos^{4} (3 x) d x

\frac{d}{d x} (2 π \frac{x}{9.8})

9 x e^{x}

\frac{d}{d x} (y^{2} x)

What is the d/(dt)(4te^{-t}) ?
The d/(dt)(4te^{-t}) is 4(e^{-t}-e^{-t}t)
What is the first d/(dt)(4te^{-t}) ?
The first d/(dt)(4te^{-t}) is 4(e^{-t}-e^{-t}t)