Is voltage difference always proportional to its derivative? We write, because of Ohm's law:
$$V=RI(t),$$
but also we have
$$C\frac{dV}{dt}=I(t).$$
From the first equation we deduce that $V\propto I$ and from the second $\dot V\propto I$. So we can conclude 
$$V \propto \dot V.$$
Is this really true or I'm doing something wrong?
 A: This is true but strictly limited to RC circuits without external sources: that is, a resistor hooked up to a capacitor with nothing else in between.
In that case, $V$ is indeed proportional to $\dot V$, with a crucial minus sign in between:
$$
\dot V=-\frac1\tau V,
$$
where $\tau>0$ is some constant. This equation implies that $V(t)=V(0)e^{-t/\tau}$, which is the well-known transient behaviour of a capacitor discharging into a resistor.
However, this is about as far as you will go by blindly applying the formulae $V=IR$ and $Q=VC$ without thinking about what they mean. The former gives the potential difference between the leads of a resistor, and the latter describes that between the plates of a capacitor. They are only equal in the circuit described above.
In a more complicated circuit, you'll have a bunch of different voltages across a bunch of different circuit elements, and it is only sums over closed loops that are equal to zero. Moreover, many circuits include inductive elements for which the voltage depends on the rate of change of the current,
$$V_L=L\dot I,$$
which means that overall the currents and voltages will follow coupled second-order differential equations, with their corresponding oscillatory behaviour.
A: Generally, no.
You've written two equations.  The first relates voltage and current for an isolated resistor.  The second relates voltage and current for an isolated capacitor.
Given only those two expressions, there's no reason at all that they can or should be combined.  That is, you have no circuit, only isolated components.  There are principles for relating such expressions when the components are part of a circuit.  (Kirchoff's Laws)  But not without a circuit describing how things are connected.
Your proportionality clearly would not hold in a circuit with no resistors, or a circuit with no capacitors.
However, it is true that in a circuit that comprises only resistors and capacitors, and voltage sources and current sources, that $V$ is indeed proportional to $\dot{V}$.  Unfortunately, you've arrived at your expression by accident, not as a result of a correct analysis. 
A: Like any set of equations in physics, this is true when the conditions stated apply. Here, you are using two equations:
$$
V(t) = RI(t)
$$
when the current passes through an element that is a perfect resistor, and
$$
C \frac{dV}{dt} = I(t)
$$
when the current passes through an element that is a perfect capacitor. So, in a simple RC circuit consisting of a perfect resistor and a perfect capacitor, is it true that
$$
\frac{dV}{dt} = \frac{1}{RC} V(t) ?
$$
NO. These are two different circuit elements, and the voltage and current cannot simultaneously be the same for both, unless you are very lucky. Also remember that there is not a single "voltage," but that these describe the voltages across each element. So, the equations really are
$$
V_1(t) = RI_1(t)
$$
and
$$
C \frac{dV_2}{dt} = I_2(t)
$$
In a circuit, these might be connected either in series or in parallel to a voltage source (among other more complicated possibilities). In the former case
$$ V_1(t) + V_2(t) = V_S \\
I_1(t) = I_2(t).$$
In the second case,
$$ V_1(t) = V_2(t) = V_S \\
I_1(t) \neq I_2(t) $$
A: One equation is for resistive circuit and the other is for capacitive circuit. Two can not be merged together.
A: The identity
$$ V = K \frac{dV}{dt} $$
is only guaranteed with a constant $K$ if your assumptions actually hold. The first identity $V=RI$ only holds for a resistor, while the other holds for a capacitor. So in this sense, the letters $V,I$ in these equations mean something else. In one of them, it's the current through (or voltage on) a particular resistor, in the other, it's the current from (or voltage on) a particular capacitor.
You may, however, make the letters $V$ mean the same thing in both equations and similarly for $I$ if you connect a capacitor and a resistor to a simple "circular" circuit. Then indeed, $V$ will be proportional to $dV/dt$, and the solution will be that the voltage will exponentially decrease with time
$$ V(t) = V(0) \exp(-t/t_0) $$
as the initial charge held by the capacitor gets discharged through the resistor – where you may easily calculate the time constant $t_0$. I guess it's right to say that the answer to your question is that "it doesn't always hold, it holds for this particular simple resistor-capacitor circuit".
