What is the d/(dt)(te^{at}) ?

The d/(dt)(te^{at}) is e^{at}+ae^{at}t

What is the first d/(dt)(te^{at}) ?

The first d/(dt)(te^{at}) is e^{at}+ae^{at}t

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

e^{a t} + a e^{a t} t

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{d t}{d t} = 1

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

= 1 \cdot e^{a t} + e^{a t} a t

Simplify

= e^{a t} + a e^{a t} t

\frac{\partial}{\partial y} (x^{2} y^{3} + 2 x^{4} y)

\frac{d}{d x} (7 - 10 x)

\int \frac{sin ( x )}{5 + cos ^{2} ( x )} d x

\frac{d}{d x} (1 - ∣ 1 - x ∣)

\int \frac{x ^{4}}{x} d x

What is the d/(dt)(te^{at}) ?
The d/(dt)(te^{at}) is e^{at}+ae^{at}t
What is the first d/(dt)(te^{at}) ?
The first d/(dt)(te^{at}) is e^{at}+ae^{at}t