What is the derivative of e^{ix}+e^{-ix} ?

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

What is the first derivative of e^{ix}+e^{-ix} ?

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

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

- 2 sin (x)

Solution steps

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

Simplify

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

= \frac{d}{d x} (2 cos (x))

Take the constant out:

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

= 2 \frac{d}{d x} (cos (x))

Apply the common derivative:

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

= 2 (- sin (x))

Simplify

= - 2 sin (x)

\int 6 e^{- 3 x} d x

\int arccot (15 y) d y

\frac{d y}{d x} = (y + 4 x)^{2}

\int_{0}^{1} π^{9}

\frac{d}{d x} ((3 x^{2} - x - 4) (- x^{2} + 2))

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