What is the derivative of (t^2+9)e^t ?

The derivative of (t^2+9)e^t is 2te^t+e^t(t^2+9)

What is the first derivative of (t^2+9)e^t ?

The first derivative of (t^2+9)e^t is 2te^t+e^t(t^2+9)

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derivative of

(t^{2} + 9) e^{t}

2 t e^{t} + e^{t} (t^{2} + 9)

Solution steps

\frac{d}{d t} ((t^{2} + 9) e^{t})

Apply the Product Rule:

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

f = t^{2} + 9, g = e^{t}

= \frac{d}{d t} (t^{2} + 9) e^{t} + \frac{d}{d t} (e^{t}) (t^{2} + 9)

\frac{d}{d t} (t^{2} + 9) = 2 t

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

= 2 t e^{t} + e^{t} (t^{2} + 9)

\frac{- 4 e ^{(- s)}}{( s ^{2} - 4 s + 3 )}

y^{^{'}} = (t + y)^{2} - 1, y (2) = 4

\int x (x - 1)^{3} d x

\frac{d}{d x} (\frac{x}{81 + x ^{2}})

\frac{\partial}{\partial x} ((9 x + 2 y)^{6})

What is the derivative of (t^2+9)e^t ?
The derivative of (t^2+9)e^t is 2te^t+e^t(t^2+9)
What is the first derivative of (t^2+9)e^t ?
The first derivative of (t^2+9)e^t is 2te^t+e^t(t^2+9)