Final answer:
To show that f'(z) does not exist anywhere for the given function f(z) = 2x + xy^2i, we need to verify if the function satisfies the Cauchy-Riemann equations.
Step-by-step explanation:
To show that $$f'(z)$$ does not exist anywhere for the given function $$f(z) = 2x + xy^2i$$, we need to verify if the function satisfies the Cauchy-Riemann equations. The Cauchy-Riemann equations state that for a complex-valued function $$f(z) = u(x,y) + iv(x,y)$$ to be differentiable at a point $$z = x + yi$$, the partial derivatives of $$u$$ and $$v$$ with respect to $$x$$ and $$y$$ must satisfy certain conditions.
In this case, the partial derivatives of $$u$$ and $$v$$ are:
$$\frac{\partial u}{\partial x} = 2$$
$$\frac{\partial u}{\partial y} = 0$$
$$\frac{\partial v}{\partial x} = y^2$$
$$\frac{\partial v}{\partial y} = 2xy$$
Since $$\frac{\partial u}{\partial x}$$ is not equal to $$\frac{\partial v}{\partial y}$$ and $$\frac{\partial u}{\partial y}$$ is not equal to $$-\frac{\partial v}{\partial x}$$, the Cauchy-Riemann equations are not satisfied. Therefore, $$f'(z)$$ does not exist anywhere for the given function.