What is the derivative of e^{-(t-3)^2} ?

The derivative of e^{-(t-3)^2} is e^{-t^2+6t-9}(-2t+6)

What is the first derivative of e^{-(t-3)^2} ?

The first derivative of e^{-(t-3)^2} is e^{-t^2+6t-9}(-2t+6)

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derivative of

e^{- (t - 3)^{2}}

e^{- t^{2} + 6 t - 9} (- 2 t + 6)

Solution steps

\frac{d}{d t} (e^{- (t - 3)^{2}})

Simplify

e^{- (t - 3)^{2}} : e^{- t^{2} + 6 t - 9}

= \frac{d}{d t} (e^{- t^{2} + 6 t - 9})

Apply the chain rule:

e^{- t^{2} + 6 t - 9} \frac{d}{d t} (- t^{2} + 6 t - 9)

= e^{- t^{2} + 6 t - 9} \frac{d}{d t} (- t^{2} + 6 t - 9)

\frac{d}{d t} (- t^{2} + 6 t - 9) = - 2 t + 6

= e^{- t^{2} + 6 t - 9} (- 2 t + 6)

\frac{d}{d x} (arctan (sin (x)))

\frac{\partial}{\partial y} (x^{3} y^{4})

\frac{d}{d y} (cos (x^{2}) - y)

\int_{- 4}^{6} 6 d x

f (x) = - 4 x^{2} - 4 x + 4, at x = 1

What is the derivative of e^{-(t-3)^2} ?
The derivative of e^{-(t-3)^2} is e^{-t^2+6t-9}(-2t+6)
What is the first derivative of e^{-(t-3)^2} ?
The first derivative of e^{-(t-3)^2} is e^{-t^2+6t-9}(-2t+6)