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

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

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

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

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

2 e^{- t} 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}

Simplify

= 2 e^{- t} t - e^{- t} t^{2}

x \cdot y^{2} \cdot y^{^{'}} = x^{3} + y^{3}

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

11 x - 6 y x^{2} + 1 \frac{d y}{d x} = 0

\int \frac{x - 1}{x ^{2} + 4} d x

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

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