What is the derivative of y=(t^2+3)e^t ?

The derivative of y=(t^2+3)e^t is 2te^t+e^t(t^2+3)

What is the first derivative of y=(t^2+3)e^t ?

The first derivative of y=(t^2+3)e^t is 2te^t+e^t(t^2+3)

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derivative of

y = (t^{2} + 3) e^{t}

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

Solution steps

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

Apply the Product Rule:

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

f = t^{2} + 3, g = e^{t}

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

\frac{d}{d t} (t^{2} + 3) = 2 t

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

= 2 t e^{t} + e^{t} (t^{2} + 3)

y = 3 + 4 x^{2} - 2 x^{3}

y^{^{''}} + 9 y = t sin (5 t)

\frac{d}{d x} (5 5 x - 4)

\int 5 e^{2 x + e^{2 x}} d x

\int x^{2} + y^{2} d x

What is the derivative of y=(t^2+3)e^t ?
The derivative of y=(t^2+3)e^t is 2te^t+e^t(t^2+3)
What is the first derivative of y=(t^2+3)e^t ?
The first derivative of y=(t^2+3)e^t is 2te^t+e^t(t^2+3)