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

The d/(dt)(3cos(3t)) is -9sin(3t)

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

The first d/(dt)(3cos(3t)) is -9sin(3t)

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

- 9 sin (3 t)

Solution steps

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

Take the constant out:

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

= 3 \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

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

Simplify

= - 9 sin (3 t)

\int (16 x + 3) d x

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

\int \frac{6}{10 x - x ^{2}} d x