What is the d/(dθ)((e^{iθ}+e^{-iθ})/2) ?

The d/(dθ)((e^{iθ}+e^{-iθ})/2) is -sin(θ)

What is the first d/(dθ)((e^{iθ}+e^{-iθ})/2) ?

The first d/(dθ)((e^{iθ}+e^{-iθ})/2) is -sin(θ)

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\frac{d}{d θ} (\frac{e ^{i θ} + e ^{- i θ}}{2})

- sin (θ)

Solution steps

\frac{d}{d θ} (\frac{e ^{i θ} + e ^{- i θ}}{2})

Simplify

\frac{e ^{i θ} + e ^{- i θ}}{2} : cos (θ)

= \frac{d}{d θ} (cos (θ))

Apply the common derivative:

\frac{d}{d θ} (cos (θ)) = - sin (θ)

= - sin (θ)

y = \frac{t - t}{t ^{\frac{1}{3}}}

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

\frac{d}{d x} (x^{2} \cdot e^{4 x})

\int 3 x - 6 x d x

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

What is the d/(dθ)((e^{iθ}+e^{-iθ})/2) ?
The d/(dθ)((e^{iθ}+e^{-iθ})/2) is -sin(θ)
What is the first d/(dθ)((e^{iθ}+e^{-iθ})/2) ?
The first d/(dθ)((e^{iθ}+e^{-iθ})/2) is -sin(θ)