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)

\int 2 cos^{2} (θ) - \frac{1}{2} d θ

\frac{\partial}{\partial y} ((x^{2} - 2 x) cos (y))

\frac{\partial ^{2}}{\partial x ^{2}} (ln (2 + x^{2} y^{2}))

\int tan^{3} (2 x) d x

y = f (x), at x = 7