[Math] Complex derivative of $\cos(x) \cosh(y)-i \sin(x) \sinh(y)$

Question

The problem: Determine the derivative of the following function
$f(z)=\cos(x) \cosh(y)-i \sin(x) \sinh(y)$

The original exercise can be found at 2.17 (e) page 36

Should i try to rewrite the function in terms of $z=x+iy$ or is there some connection between the partial derivatives and the complex derivative that I am missing?

Edit:
Thanks @Claude Leibovici
The equations needed are:
$(1) \cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b)$
$(2) \cos(ic)=\cosh(c)$
$(3) \sin(id)=i\sinh(d)$

With (2) and (3) I can rewrite the cosh and sinh in terms of cos and sin.
$f(z)=\cos(x)\cos(iy)-\sin(x)\sin(iy)$

Then I can use (1) to combine it into one cos.
$f(z)=\cos(x+iy)=\cos(z)$

Therefore:
$f'(z)=-\sin(z)$

Answer 1

Hint

$$\cos(a+b)=\cos(a]\cos(b)-\sin(a)\sin(b)$$ $$\cos(i c)=\cosh(c)$$ $$\sin(id)=i\sinh(d)$$

All of that would make $f(z)$ very nice.

I am sure that you can take it from here.