What is the d/(dθ)(1+cos(2θ)) ?

The d/(dθ)(1+cos(2θ)) is -2sin(2θ)

What is the first d/(dθ)(1+cos(2θ)) ?

The first d/(dθ)(1+cos(2θ)) is -2sin(2θ)

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\frac{d}{d θ} (1 + cos (2 θ))

- 2 sin (2 θ)

Solution steps

\frac{d}{d θ} (1 + cos (2 θ))

Simplify

1 + cos (2 θ) : 2 cos^{2} (θ)

= \frac{d}{d θ} (2 cos^{2} (θ))

Take the constant out:

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

= 2 \frac{d}{d θ} (cos^{2} (θ))

Apply the chain rule:

2 cos (θ) \frac{d}{d θ} (cos (θ))

= 2 cos (θ) \frac{d}{d θ} (cos (θ))

\frac{d}{d θ} (cos (θ)) = - sin (θ)

= 2 \cdot 2 cos (θ) (- sin (θ))

Simplify

2 \cdot 2 cos (θ) (- sin (θ)) : - 2 sin (2 θ)

= - 2 sin (2 θ)

x \to - 1 - lim (\frac{x}{( x + 1 ) ^{2}})

(2 cos (2 x))^{^{'}}

3 s + 2

x \to \infty lim (e^{x} - 3)

\frac{d}{d x} (\frac{z ( x )}{x})