Obtaining a Previously Constructed Function as Desired Output

Question

I was playing around with some Generalized Trigonometric Functions defined as follows;

 w[n_] := E^((2*\[Pi]*I)/n)

 Ncos[n_, x_] := Expand[(1/n)*Sum[E^(x*w[2 n]^(2 k + 1)), {k, 0, n - 1}]]

 Nsin[n_, j_, x_] := Expand[(1/(n*w[2 n]^j))*Sum[w[n]^(-j*k) E^(x*w[2 n]^(2 k + 1)), {k, 0, n - 1}]]

These functions act like the familiar trigonometric functions and have the property that

$$\frac{d^{2n}}{dx^{2n}}f(x)=f(x)$$ or $$f^{(n)}(x)+f(x)=0$$

As I was testing with paper and pencil I wrote up the above functions when I got home to my computer and observed that these relations hold. However, when I employ the derivative function $D$ in Mathematica, it will give the derivative in terms of the expanded function. For example, if I type in

 D[Ncos[3,x],x]

It returns $$-\frac{e^{-x}}{3}+\frac{1}{3}e^{{\frac{-i\pi}{3}-e^{\frac{-i\pi}{3}}}x}+\frac{1}{3}e^{{\frac{i\pi}{3}-e^{\frac{i\pi}{3}}}x}$$

The derivative should equal -Nsin[3,2,x], and when you input

 -Nsin[3,2,x]

it returns

$$-\frac{e^{-x}}{3}-\frac{1}{3}e^{{\frac{-4i\pi}{3}-e^{\frac{-i\pi}{3}}}x}-\frac{1}{3}e^{{\frac{-2i\pi}{3}-e^{\frac{i\pi}{3}}}x}$$

While Mathematica doesn't output the same form, subtracting the two yields 0, proving their equality...

My question is this. Is there a way to get mathematica, when I input

 D[Ncos[3,x],x]

to output

 -Nsin[3,2,x]

Or for other derivative, for example, if I input

 D[D[D[Ncos[3,x],x],x],x]

I would like to see

 -Ncos[3,x]

Is this possible?