What is the derivative of t^2e^t ?

The derivative of t^2e^t is 2te^t+e^tt^2

What is the first derivative of t^2e^t ?

The first derivative of t^2e^t is 2te^t+e^tt^2

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derivative of

t^{2} e^{t}

2 t e^{t} + e^{t} t^{2}

Solution steps

\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}

= \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}

= 2 t e^{t} + e^{t} t^{2}

\frac{\partial}{\partial x} (sin (4 y) e^{- 5 x})

f (x) = - x^{3} - 150 x^{2} + 60000 x - 800

\int e^{x} (sin (x) + cos (x)) d x

\frac{\partial}{\partial y} (x - 4 y - 9)

\int (4 + 3 x)^{4} d x

What is the derivative of t^2e^t ?
The derivative of t^2e^t is 2te^t+e^tt^2
What is the first derivative of t^2e^t ?
The first derivative of t^2e^t is 2te^t+e^tt^2