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

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

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

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

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

- sin (x) - i cos (x)

Solution steps

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

Simplify

e^{- i x} : cos (x) - i sin (x)

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

= - sin (x) - i cos (x)

x \to 0 lim ((1 + x) \frac{1}{x})

\int \frac{15}{1 + 9 x ^{2}} d x

- 4 x + x^{2} + 10, 9 - 2 x, 2 x + 1

y = x - 2 3 x

\frac{d P}{d t} = P - P^{2}

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