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

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

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

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

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

2 (e^{2 t} + 2 e^{2 t} t)

Solution steps

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

Take the constant out:

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

= 2 \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}

= 2 (\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}) = e^{2 t} \cdot 2

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

Simplify

= 2 (e^{2 t} + 2 e^{2 t} t)

\frac{d}{d x} (\frac{4}{x} - \frac{1}{6 x ^{3}} + \frac{8}{x ^{5}})

\int x^{3} 121 - x^{2} d x

6 x e^{x}

10 (x^{3} - x)^{4}

\frac{d}{d x} (sin (9 cos (x)))

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