What is the d/(dθ)(2sin(3θ)) ?

The d/(dθ)(2sin(3θ)) is 6cos(3θ)

What is the first d/(dθ)(2sin(3θ)) ?

The first d/(dθ)(2sin(3θ)) is 6cos(3θ)

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\frac{d}{d θ} (2 sin (3 θ))

6 cos (3 θ)

Solution steps

\frac{d}{d θ} (2 sin (3 θ))

Take the constant out:

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

= 2 \frac{d}{d θ} (sin (3 θ))

Apply the chain rule:

cos (3 θ) \frac{d}{d θ} (3 θ)

= cos (3 θ) \frac{d}{d θ} (3 θ)

\frac{d}{d θ} (3 θ) = 3

= 2 cos (3 θ) \cdot 3

Simplify

= 6 cos (3 θ)