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

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

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

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

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

6 cos (3 t)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

cos (3 t) \frac{d}{d t} (3 t)

= cos (3 t) \frac{d}{d t} (3 t)

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

= 2 cos (3 t) \cdot 3

Simplify

= 6 cos (3 t)

\int 3 x^{2} - x + 3 d x

\frac{d}{d x} (\frac{8 x ^{2} + 2 x + 2}{x})

(80, 2000), (260, 5600)

x \to 1 lim (e^{x})