AL physics-Capacitor

Your derivation is mathematically correct, but the sign is incorrect. Consider a circuit consisting of a cell with e.m.f. E and a resistor and capacitor connected in series, then by
Q=CV
dQ/dt
=C(dV/dt)
But note that for the notation V, it represents the voltage across the resistor, so V = IR is not valid. Instead, E = IR + V, so V = E - IR, and
dV/dt = dE/dt - d(IR)/ dt, but since E, the e.m.f. of the cell, is assumed to be constant,
dE/dt = 0 and
dV/dt = -d(IR)/dt), and
dQ/dt
= - C(dIR/dt)
I/(CR)= - dI/dt
-t/(CR)= [ln(I)] (upper=I and lower=Io)
=ln(I)-ln(Io)
=ln(I/Io)
I=(Io)e^[-t/(CR)], as required.
Also, if a resistor is connected in parallel to the capacitor, the circuit with capacitor has no resistance at all. The capacitor is charged up very quickly and the equation for I doesn't apply (substituting R = 0 in the equation and you will see why)

Yet in the example:
A charged capacitor of capacitance 47F is connected to a voltmeter. The voltmeter reading is found to drop from 8V to 2V in 7.8s Find the resistance of the voltmeter
Note that the capacitor is not being charged up, instead, it is discharging. For discharging capacitor with a resistor in the circuit, we have the equation:
V = V0(e^[-t/(CR)]),
where V is the voltage across the resistor. Applying the equation and you can get the answer.

Your equation is wrong. By Q=CV
dQ/dt=V(dC/dt)=C(dV/dt)
=C(dIR/dt) < ---- this is wrong, V is not equal to IR.It is because V is the p.d. across the capacitor, but IR is the p.d. across the resistor. They are not the same. When the resistor is connected in parallel with the capacitor, the capacitor will be charged almost immediately to its full capacity after connected to the battery. The relevant equation for series connection could not apply.