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

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

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

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

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

Solution steps

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

Simplify

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

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

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

= cos (i y) (\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)

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

\int ln (x) \cdot \frac{1}{x} d x

\int (1 - x^{2} - x) x d x

\int \frac{x ^{4} + x - 16}{x ^{2} + 4} d x

ln (\frac{a ^{2} - z ^{2}}{a ^{2} + z ^{2}})

y = 3 (4 x^{2} + 4)

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