What is the d/(dθ)(e^{-iθ}) ?

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

What is the first d/(dθ)(e^{-iθ}) ?

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

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\frac{d}{d θ} (e^{- i θ})

- sin (θ) - i cos (θ)

Solution steps

\frac{d}{d θ} (e^{- i θ})

Simplify

e^{- i θ} : cos (θ) - i sin (θ)

= \frac{d}{d θ} (cos (θ) - i sin (θ))

Apply the Sum/Difference Rule:

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

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

\frac{d}{d θ} (cos (θ)) = - sin (θ)

\frac{d}{d θ} (i sin (θ)) = i cos (θ)

= - sin (θ) - i cos (θ)

\frac{\partial}{\partial x} (ln (tan (\frac{x}{2})))

\int \frac{1 - x ^{2}}{x} d x

x \to 0 + lim (cos (x))

y^{^{''}} + 0.2 x^{2} - 0.35 x = 0

n = 1 \sum \infty \frac{1}{2 ^{3 n}}

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