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

The derivative of 2te^{-t} is 2(e^{-t}-e^{-t}t)

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

The first derivative of 2te^{-t} is 2(e^{-t}-e^{-t}t)

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derivative of

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)

(1, - 4), (4, - 5)

\int cos (π)

x \to 3 + lim (\frac{1}{x ^{2} - 9})

\frac{d}{d x} (\frac{3 x}{( x ^{2} + 1 ) ^{\frac{3}{5}}})

(x^{2} - 2 x + 4), 2

What is the derivative of 2te^{-t} ?
The derivative of 2te^{-t} is 2(e^{-t}-e^{-t}t)
What is the first derivative of 2te^{-t} ?
The first derivative of 2te^{-t} is 2(e^{-t}-e^{-t}t)