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

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

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

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

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

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

Solution steps

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

Apply the Product Rule:

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

f = t^{2}, g = e^{t}

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

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

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

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

\frac{3}{( x ^{2} - x - 2 )}

\frac{d}{d x} ((2 + x)^{\frac{1}{3}} (1 - x)^{\frac{2}{3}})

\int \frac{9}{( x - 1 ) ( x + 8 )} d x

4 x^{4} + 3 x^{3} + 2 x^{2} + x + 1

\frac{d}{d t} (6 ln (cos (t)))