Complex derivative of $\cos(x) \cosh(y)-i \sin(x) \sinh(y)$ 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:
$f'(z)=-\sin(z)$
 A: 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.
A: Use Cauchy-Riemann (with the obvious $u,v,u_x,\dots$)
$$u_x = -\sin x \cosh y =  v_y$$
$$u_y = \cos x \sinh y  = - v_x$$
This shows that f is complex differentiable, and then you compute
$$f'(z) = u_x+iv_x= \sin x \cosh y - i \cos x \sinh y.$$
A: Observe that the form of your function recalls the addition formula for trigonometric functions; 
\begin{align*}
\cos z
&=\cos(x+iy)\\
&=\cos x\cos iy-\sin x\sin iy\\
&=\cos x\cosh y -i\sin x\sinh y\\
\end{align*}
Thus the expression given is simply a reformulation of the complex cosine, i.e.
$$
f(z)=\cos z
$$
thus the complex derivative of $f$ is $f'(z)=-\sin z$.
A: Let $f(z)=\cos x \cosh y - \sin x \sinh y$ 
Now recall that, $$ \sinh z = \frac{e^{y}-e^{-y}}{2} ~~~and~~~ \cosh z = \frac{e^{y}+e^{-y}}{2}$$
$$ \Longrightarrow \cos x \cosh y - \sin x \sinh y = \cos x \left(\frac{e^{-y}+e^{y}}{2}\right) -i \sin x\left(\frac{e^{y}-e^{-y}}{2}\right) $$
$$= \frac{ e^{-y}(\cos x +i \sin x) + e^{y}(\cos x -i \sin x)}{2}$$
$$= \frac{ e^{-y+ix} + e^{y-ix}}{2}  = \frac{ e^{i(x+iy)} + e^{-i(x+iy)}}{2}$$ $$ =  \frac{ e^{iz} + e^{-iz}}{2} = \cos z$$
where $z=x+iy$.
$$\therefore f'(z) =  \frac{d}{dz}(\cos z) = \frac{d}{dz} \left(\frac{ e^{iz} + e^{-iz}}{2}\right) = \frac{i( e^{iz} - e^{-iz})}{2} = - \frac{( e^{iz} - e^{-iz})}{2i} = 
- \sin z$$
Hence, $f'(z) = -\sin z$
