What is the derivative of ie^{ix} ?

The derivative of ie^{ix} is -cos(x)-isin(x)

What is the first derivative of ie^{ix} ?

The first derivative of ie^{ix} is -cos(x)-isin(x)

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

- cos (x) - i sin (x)

Solution steps

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

Simplify

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

f^{^{'}} (t) = 0

\frac{d}{d x} ((x - 1)^{3})

\frac{d}{d x} (e^{(- 0.5 x)})

\int sin (2 x) sin (n x) d x

\frac{d y}{d t} + 0.7 t y = 9 t, y (0) = 5

What is the derivative of ie^{ix} ?
The derivative of ie^{ix} is -cos(x)-isin(x)
What is the first derivative of ie^{ix} ?
The first derivative of ie^{ix} is -cos(x)-isin(x)