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So I'm solving some exercises in which they ask to find where $f$ is analytic and then finding $f'(z)$. So I know that if a function doesn't satisfy Cauchy-Riemann then it is not analytic. I also know that there can be functions that satisfy Cauchy-Riemann but still they are not analytic. So, given a function that satisfies Cauchy-Riemann in all the complex plane, for example $f(z)=e^{-y}(\sin(x)+i\cos(x)$, can I calculate simply $f'(z)=e^{-y}(\cos(x)+\sin(x)$?

Here I'm using that if $f(z)=u(z)+iv(z)$ then $f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}$

However this implies the existence of $f'(z)$.

Putting everything together: If $f$ satisfies Cauchy-Riemann how can I know if $f'$ exists?

Another User
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Joaquin Liniado
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    $f$ is differentiable if it satisfies Cauchy-Riemann and its components are continuously differentiable. See http://math.stackexchange.com/questions/15875/when-is-a-function-satisfying-the-cauchy-riemann-equations-holomorphic – Michael M Aug 04 '15 at 20:56

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Usual rules for derivatives (derivative of $z^{\alpha}$, product, sum, composition) hold if you express everything in terms of $z$ instead of its components $x$ and $y$. In this example $f(z)=e^{-y}e^{ix}=e^{i(x+iy)}=e^{iz}$. Therefore $f'(z)=ie^{iz}$.

As for existence of complex derivative given Cauchy-Riemann equations it is sufficient that $f$ is Frechet differentiable when considered function $\mathbb R \to \mathbb R$. Condition that is easier to check and more useful in practice is that partial derivatives are continous.

Blazej
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