Calculating rate of voltage change through a circuit.

Here's the question:

Voltage across a circuit is given by Ohm's law, V=IR, where I=the current & R= resistance.

If we place two circuits with resistance R1 and R2 in parallel their combined resistance, R, is given as

$\displaystyle \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

Suppose the current is 2 amps and increasing at $\displaystyle 10^(-2)$ amp/sec and $\displaystyle R_1$ is 3 ohms and increasing at 0.5 ohm/sec, while $\displaystyle R_2$ is 5 ohms and decreasing at 0.1 ohm/sec.

Calculate the rate at which the voltage is changing.

This is what I've got:

Need to find $\displaystyle \frac{dV}{dt}$

Using the Chain Rule, I get this:

$\displaystyle \frac{dV}{dt}=I\frac{dR}{dt}+R\frac{dI}{dt}$

where

$\displaystyle R_1=3$ and $\displaystyle \frac{dR_1}{dt}=0.5$

$\displaystyle R_2=5$ and $\displaystyle \frac{dR_2}{dt}=-0.1$

$\displaystyle I=2$ and $\displaystyle \frac{dI}{dt}=0.01$

$\displaystyle \frac{1}{R}=\frac{1}{3}+\frac{1}{5} => R=\frac{15}{8}$

$\displaystyle \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} = \frac{R_1R_2}{R_1+R_2}$

Which all comes to:

$\displaystyle \frac{dV}{dt}=2\frac{dR}{dt}+\frac{15}{8}*\frac{1} {100}$

$\displaystyle \frac{dV}{dt}=2\frac{dR}{dt}+\frac{15}{800}$

and that's as far as I've got! How do I find the change in resistance? I've tried a few ways but none of them seem satisfactory. any ideas?