What is (\partial)/(\partial x)(e^{ikx}) ?

The answer to (\partial)/(\partial x)(e^{ikx}) is -sin(kx)k+icos(kx)k

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(\partial)/(\partial x)(e^{ikx})

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

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

Solution steps

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

Simplify

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

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

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (cos (k x)) = - sin (k x) k

\frac{\partial}{\partial x} (i sin (k x)) = i cos (k x) k

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

\int (1 + sin^{2} (x)) d x

\frac{d}{d x} (ln ((x - 4)^{3}))

\int x \cdot x^{2} + 5 d x

ln (4 - x)

\frac{\partial}{\partial z} (ln (1 + x^{2} + e^{y} + z))

What is (\partial)/(\partial x)(e^{ikx}) ?
The answer to (\partial)/(\partial x)(e^{ikx}) is -sin(kx)k+icos(kx)k