What is the derivative of e^{ix} ?

The derivative of e^{ix} is -sin(x)+icos(x)

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

The first derivative of e^{ix} is -sin(x)+icos(x)

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

- sin (x) + i cos (x)

Solution steps

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

Simplify

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

\int \frac{2 cos ( ln ( x ) )}{x} d x

\frac{\partial}{\partial y} (e^{x^{y}})

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

\int \frac{5}{cos ^{2} ( x )} d x

\frac{d ^{2} y}{d x ^{2}} + 14 \frac{d y}{d x} + 98 y = 0

What is the derivative of e^{ix} ?
The derivative of e^{ix} is -sin(x)+icos(x)
What is the first derivative of e^{ix} ?
The first derivative of e^{ix} is -sin(x)+icos(x)