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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

x \to 1 lim (ln (x) - 5 x)

f (b) = (4 - a - b - c) z

x \to + (- 0) lim (e^{\frac{1}{x}})

\frac{d}{d t} (5 cos (t))

\int_{0}^{1} \frac{1}{x ^{2} + 16} d x

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