What is the d/(dt)((4)(7+e^{-t^2})) ?

The d/(dt)((4)(7+e^{-t^2})) is -8e^{-t^2}t

What is the first d/(dt)((4)(7+e^{-t^2})) ?

The first d/(dt)((4)(7+e^{-t^2})) is -8e^{-t^2}t

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\frac{d}{d t} ((4) (7 + e^{- t^{2}}))

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

Solution steps

\frac{d}{d t} ((4) (7 + e^{- t^{2}}))

Take the constant out:

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

= 4 \frac{d}{d t} (7 + e^{- t^{2}})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= 4 (\frac{d}{d t} (7) + \frac{d}{d t} (e^{- t^{2}}))

\frac{d}{d t} (7) = 0

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

= 4 (0 - 2 e^{- t^{2}} t)

Simplify

= - 8 e^{- t^{2}} t

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

d x (x^{4} + 2 x^{3} - 7)

\frac{\partial}{\partial r} ((x) (r + 3 s - t))

1 + \frac{0.732 ( z + 0.1 )}{( z - 1 ) ( z - 0.9512 )}

(10 x y^{3} + 12 x^{3} + 6) d x + (15 x^{2} y - 5) y d y

What is the d/(dt)((4)(7+e^{-t^2})) ?
The d/(dt)((4)(7+e^{-t^2})) is -8e^{-t^2}t
What is the first d/(dt)((4)(7+e^{-t^2})) ?
The first d/(dt)((4)(7+e^{-t^2})) is -8e^{-t^2}t