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

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

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

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

- sin (x) + i cos (x)

Solution steps

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

Simplify

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

y^{^{'}} + y = t

\frac{d}{d x} (\frac{x}{x + 3})

\int \frac{1}{x ^{2} 9 + x ^{2}} d x

(x + 2) sin (y) d x + x cos (y) d y = 0

\frac{d y}{d x} = e^{- x^{2}}, y (3) = 5

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