Is the derivative of $x^2 + C$? What is the derivative of $x^2+C$, except if $C$ was set to the imaginary unit $i$?  It wouldn't be possible to take, or would it simply be $2x$?
 A: its simply $2x$. It doesn't matter what particular constant $C$ is.
A: We have, for any constant $C$, as $h \to 0$,
$$
\frac{f(x+h)-f(x)}{h}=\frac{(x+h)^2+C-x^2-C}{h}=2x+h \to 2x.
$$
A: As others have said, if $x \in \mathbb{R}$, then $\frac{d}{dx} (x^2 + C) = 2x$, even if $C \in \mathbb{C}$.
However, this is not generally true if $x$ is complex-valued. Differentiability of complex functions is different than for real-valued functions.
As it turns out, if $x = a+bi$, then we have $x^2+C = (a^2-b^2) + 2abi + C$.
Let $u$ denote the real part and $v$ denote the imaginary parts of this function. Then,
$$\frac{du}{da} = 2a = \frac{dv}{db}$$
and
$$\frac{du}{db} = -2b = -\frac{dv}{da}$$
Therefore, by coincidence, the complex-valued version of the function is differentiable, as well. But this is not necessarily the case for an arbitrary polynomial (and in fact is almost never the case).
A: The derivative of a sum is the sum of the derivatives, so your first term is immaterial and you are essentially asking about the derivative of $i$, a constant.
As shown by Olivier Oloa, this is null for any constant.
