What is the derivative of (e^{ix}+e^{-ix}/2) ?

The derivative of (e^{ix}+e^{-ix}/2) is -sin(x)

What is the first derivative of (e^{ix}+e^{-ix}/2) ?

The first derivative of (e^{ix}+e^{-ix}/2) is -sin(x)

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

- sin (x)

Solution steps

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

Simplify

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

= \frac{d}{d x} (cos (x))

Apply the common derivative:

\frac{d}{d x} (cos (x)) = - sin (x)

= - sin (x)

\frac{\partial}{\partial y} (\frac{x}{9 + y})

y^{^{'}} + y cos (x) = sin (x) cos (x)

f (x) = (ln (x))^{6}, at x = 4

x y^{^{'}} + 2 y = - 3 cos (x)

x^{2} + 6

What is the derivative of (e^{ix}+e^{-ix}/2) ?
The derivative of (e^{ix}+e^{-ix}/2) is -sin(x)
What is the first derivative of (e^{ix}+e^{-ix}/2) ?
The first derivative of (e^{ix}+e^{-ix}/2) is -sin(x)