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

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

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

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

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

- 6 sin (3 t)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= 2 (- sin (3 t) \cdot 3)

Simplify

= - 6 sin (3 t)

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