What is the d/(dt)(at^2e^t+5t) ?

The d/(dt)(at^2e^t+5t) is a(2te^t+e^tt^2)+5

What is the first d/(dt)(at^2e^t+5t) ?

The first d/(dt)(at^2e^t+5t) is a(2te^t+e^tt^2)+5

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\frac{d}{d t} (a t^{2} e^{t} + 5 t)

a (2 t e^{t} + e^{t} t^{2}) + 5

Solution steps

\frac{d}{d t} (a t^{2} e^{t} + 5 t)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (a t^{2} e^{t}) + \frac{d}{d t} (5 t)

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

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

= a (2 t e^{t} + e^{t} t^{2}) + 5

x^{2} - 6 - 6

5 - 2 x + 4 x - 3

\frac{x ^{2} + 6 x - 40}{2}

(X^{2} - 9) + 1

\frac{d}{d t} ((x) (3 x))

What is the d/(dt)(at^2e^t+5t) ?
The d/(dt)(at^2e^t+5t) is a(2te^t+e^tt^2)+5
What is the first d/(dt)(at^2e^t+5t) ?
The first d/(dt)(at^2e^t+5t) is a(2te^t+e^tt^2)+5