What is the d/(dθ)(3sin(2θ)) ?

The d/(dθ)(3sin(2θ)) is 6cos(2θ)

What is the first d/(dθ)(3sin(2θ)) ?

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

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\frac{d}{d θ} (3 sin (2 θ))

6 cos (2 θ)

Solution steps

\frac{d}{d θ} (3 sin (2 θ))

Take the constant out:

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

= 3 \frac{d}{d θ} (sin (2 θ))

Apply the chain rule:

cos (2 θ) \frac{d}{d θ} (2 θ)

= cos (2 θ) \frac{d}{d θ} (2 θ)

\frac{d}{d θ} (2 θ) = 2

= 3 cos (2 θ) \cdot 2

Simplify

= 6 cos (2 θ)

n = 1 \sum \infty cos (4 n)

\frac{d}{d x} (a e^{a x})

\frac{d}{d x} ((x^{3} + 2 x - 7) (3 + x - x^{2}))

\frac{d}{d x} ((x + \frac{1}{x})^{- 1})

\frac{d y}{d x} - 2 x y = 8 x