What is the d/(dt)(e^{it}) ?

The d/(dt)(e^{it}) is -sin(t)+icos(t)

What is the first d/(dt)(e^{it}) ?

The first d/(dt)(e^{it}) is -sin(t)+icos(t)

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\frac{d}{d t} (e^{i t})

- sin (t) + i cos (t)

Solution steps

\frac{d}{d t} (e^{i t})

Simplify

e^{i t} : cos (t) + i sin (t)

= \frac{d}{d t} (cos (t) + i sin (t))

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (cos (t)) + \frac{d}{d t} (i sin (t))

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

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

= - sin (t) + i cos (t)

\frac{d x}{d t} = sin (2 π (t + 3))

1

n = 3 \sum \infty \frac{1}{3} (\frac{2}{5})^{n}

\int \frac{1}{5 ^{4 x}} d x