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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

f (t) = t^{9} sin (t)

\int_{1}^{e} 4 x^{3} ln (x) d x

- \frac{1}{2} x

\frac{\partial}{\partial x} (x^{3} e^{x})

n = 0 \sum \infty \frac{9 ^{n}}{n !}