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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = t, g = e^{t}

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

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

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

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

Simplify

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

\frac{d}{d x} (x (x + 1)^{2})

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

y^{^{'}} - y = 9 t

\int sec^{6} (3 y) tan^{2} (3 y) d y

\frac{\partial}{\partial x} ((2 x) (3 z))