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

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

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

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

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

e^{4 t} + 4 e^{4 t} t

Solution steps

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

Apply the Product Rule:

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

f = t, g = e^{4 t}

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

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

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

= 1 \cdot e^{4 t} + e^{4 t} \cdot 4 t

Simplify

= e^{4 t} + 4 e^{4 t} t

\int - 3 tan (3 y) d y

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\frac{\partial}{\partial y} (x^{2} sin^{2} (yz))

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\frac{1}{s}

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