What is the d/(dt)((x)(t^2e^t)) ?

The d/(dt)((x)(t^2e^t)) is x(2te^t+e^tt^2)

What is the first d/(dt)((x)(t^2e^t)) ?

The first d/(dt)((x)(t^2e^t)) is x(2te^t+e^tt^2)

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

(x + y)^{3} + 3 x y (1 - x - y) - 1

3 q^{2} + (5 q + 4)

\frac{6 - 4 3}{3}

x^{2} - 3 - 10

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