What is the d/(dt)(4te^{3t}) ?

The d/(dt)(4te^{3t}) is 4(e^{3t}+3e^{3t}t)

What is the first d/(dt)(4te^{3t}) ?

The first d/(dt)(4te^{3t}) is 4(e^{3t}+3e^{3t}t)

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\frac{d}{d t} (4 t e^{3 t})

4 (e^{3 t} + 3 e^{3 t} t)

Solution steps

\frac{d}{d t} (4 t e^{3 t})

Take the constant out:

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

= 4 \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}

= 4 (\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}) = e^{3 t} \cdot 3

= 4 (1 \cdot e^{3 t} + e^{3 t} \cdot 3 t)

Simplify

= 4 (e^{3 t} + 3 e^{3 t} t)

y = x^{2} - 3 x - 5

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\int \frac{x}{- 7 - 8 x - x ^{2}} d x

\int_{0}^{5} 5 x d x

What is the d/(dt)(4te^{3t}) ?
The d/(dt)(4te^{3t}) is 4(e^{3t}+3e^{3t}t)
What is the first d/(dt)(4te^{3t}) ?
The first d/(dt)(4te^{3t}) is 4(e^{3t}+3e^{3t}t)