What is the derivative of te^{2t} ?

The derivative of te^{2t} is e^{2t}+2e^{2t}t

What is the first derivative of te^{2t} ?

The first derivative of te^{2t} is e^{2t}+2e^{2t}t

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derivative of

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

What is the derivative of te^{2t} ?
The derivative of te^{2t} is e^{2t}+2e^{2t}t
What is the first derivative of te^{2t} ?
The first derivative of te^{2t} is e^{2t}+2e^{2t}t