What is the d/(dn)(e^{nipi}) ?

The d/(dn)(e^{nipi}) is -sin(pin)pi+icos(pin)pi

What is the first d/(dn)(e^{nipi}) ?

The first d/(dn)(e^{nipi}) is -sin(pin)pi+icos(pin)pi

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\frac{d}{d n} (e^{niπ})

- sin (πn) π + i cos (πn) π

Solution steps

\frac{d}{d n} (e^{niπ})

Simplify

e^{niπ} : cos (πn) + i sin (πn)

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d n} (cos (πn)) = - sin (πn) π

\frac{d}{d n} (i sin (πn)) = i cos (πn) π

= - sin (πn) π + i cos (πn) π

y^{^{''}} + y^{^{'}} + y = 0

\int (y^{5} + 5 y) d y

y^{^{''}} + y = 0, y (\frac{π}{3}) = 16, y^{^{'}} (\frac{π}{3}) = - 32

\frac{\partial}{\partial y} (7 e^{x^{2} l n (y)})

5 Y^{\frac{5}{8}} X^{\frac{3}{8}}

What is the d/(dn)(e^{nipi}) ?
The d/(dn)(e^{nipi}) is -sin(pin)pi+icos(pin)pi
What is the first d/(dn)(e^{nipi}) ?
The first d/(dn)(e^{nipi}) is -sin(pin)pi+icos(pin)pi