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

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

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

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

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

e^{2 t} + 2 e^{2 t} t

Solution steps

\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}

= \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

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

Simplify

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

\int (x + 1) e^{5 x^{2} + 10 x} d x

(1 + 9 x^{2}) (x - x^{2})

\frac{d}{d x} (7 x (sin (x) + cos (x)))

\frac{d x}{d t} = 2 e^{t} - 2

(- 1.2) (4)

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