Why is the derivative of the capacitance of a capacitor not 0?

Question

So while trying to get the differential equation of Uc, which is the voltage across a capacitor. We'll have to use the following: dq/dt. Where q is the charge across the capacitor. Then we'll obtain this: d(CUc)/dt. Which then becomes C dUc/dt. I don't know if this question is too basic or something, but I can't understand why didn't C which is a constant not translate into 0 after deriving it.

Answer 1

Hint:

What is wrong with the following reasoning:

If $f(x)=x$, and $g(x)=10\cdot f(x)$, then $\frac{dg}{dx} = 0\cdot \frac{df}{dx}$, because $10$ is just a constant and the derivative of a constant is $0$.

Answer 2

Let's say that $f$ is a derivable function and $a\in \mathbb{R}$. Let's define $g(x) = a.f(x)$. $$g'(x) = \lim_{h\to 0} \frac{g(x+h)-g(x)}{h} = a\cdot \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = a \cdot f'(x)$$ A constant $C \in \mathbb{R}$ does not result in $0$ in your derivative, but give instead a multiplicative factor.