What is the d/(dθ)(2sin(2θ)) ?

The d/(dθ)(2sin(2θ)) is 4cos(2θ)

What is the first d/(dθ)(2sin(2θ)) ?

The first d/(dθ)(2sin(2θ)) is 4cos(2θ)

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\frac{d}{d θ} (2 sin (2 θ))

4 cos (2 θ)

Solution steps

\frac{d}{d θ} (2 sin (2 θ))

Take the constant out:

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

= 2 \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

= 2 cos (2 θ) \cdot 2

Simplify

= 4 cos (2 θ)

f (x) = 6 + 5 x^{2} - 2 x^{3}, at x = 1

\frac{d}{d x} (x^{2} + 2 x y)

n = 0 \sum \infty (\frac{e}{4 π})^{n}

\int \frac{x ^{2} + 5 x + 4}{x ^{2} - 2 x + 1} d x

\frac{d}{d x} (tan (sec (cos (x))))