What is the d/(dt)(4cos(2t)) ?

The d/(dt)(4cos(2t)) is -8sin(2t)

What is the first d/(dt)(4cos(2t)) ?

The first d/(dt)(4cos(2t)) is -8sin(2t)

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\frac{d}{d t} (4 cos (2 t))

- 8 sin (2 t)

Solution steps

\frac{d}{d t} (4 cos (2 t))

Take the constant out:

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

= 4 \frac{d}{d t} (cos (2 t))

Apply the chain rule:

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

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

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

= 4 (- sin (2 t) \cdot 2)

Simplify

= - 8 sin (2 t)