What is the d/(dt)(3sin(2t)) ?

The d/(dt)(3sin(2t)) is 6cos(2t)

What is the first d/(dt)(3sin(2t)) ?

The first d/(dt)(3sin(2t)) is 6cos(2t)

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\frac{d}{d t} (3 sin (2 t))

6 cos (2 t)

Solution steps

\frac{d}{d t} (3 sin (2 t))

Take the constant out:

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

= 3 \frac{d}{d t} (sin (2 t))

Apply the chain rule:

cos (2 t) \frac{d}{d t} (2 t)

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

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

= 3 cos (2 t) \cdot 2

Simplify

= 6 cos (2 t)

\int_{3}^{5} x^{2} d x

\int \frac{t ^{3}}{t ^{3} + 1} d t

\frac{1}{( s ^{2} + 5 s + 4 )}

n = 0 \sum \infty \frac{2 n}{n + 1}

t e^{-, at}