What is the d/(dz)(e^{iz}) ?

The d/(dz)(e^{iz}) is -sin(z)+icos(z)

What is the first d/(dz)(e^{iz}) ?

The first d/(dz)(e^{iz}) is -sin(z)+icos(z)

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

- sin (z) + i cos (z)

Solution steps

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

Simplify

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

\frac{d y}{d t} = 1.7 y

f (x) = (x - 5) (x + 1)

\int 4 e^{3 x + e^{3 x}} d x

\frac{d y}{d x}, 8 x + 3 y = 7

\int tan^{\frac{1}{2}} (x) sec^{4} (x) d x