What is the d/(dt)(6cos(2t)) ?

The d/(dt)(6cos(2t)) is -12sin(2t)

What is the first d/(dt)(6cos(2t)) ?

The first d/(dt)(6cos(2t)) is -12sin(2t)

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\frac{d}{d t} (6 cos (2 t))

- 12 sin (2 t)

Solution steps

\frac{d}{d t} (6 cos (2 t))

Take the constant out:

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

= 6 \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

= 6 (- sin (2 t) \cdot 2)

Simplify

= - 12 sin (2 t)

y = x^{3}, y = x^{2} - 9 x, [- 1, 1]

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\frac{d}{d x} (tan (x^{2} - 1))

\int 2 x y z^{3} d y

\frac{d y}{d x} = 15 - 3 y