Is the derivative of $x^2 + C$?

Question

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 $2x$?

Answer 1

We have, for any constant $C$, as $h \to 0$, $$ \frac{f(x+h)-f(x)}{h}=\frac{(x+h)^2+C-x^2-C}{h}=2x+h \to 2x. $$

Answer 2

its simply $2x$. It doesn't matter what particular constant $C$ is.

Answer 3

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).

Answer 4

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.