Derivatives! Help!

Well, we have the following information:

$PV^{1.4} = C$

$\frac{dp}{dt} = -15 \frac{kPa}{min}$

$\frac{dv}{dt} = \frac{dv}{dp}\frac{dp}{dt}$

Well, looks like we need to find $\frac{dv}{dp}$

Let's take a look at it, if we implicitly differentiate the first equation, we are left with a $\frac{dv}{dp}$ are we not?

$\frac{dv}{dp}[PV^{1.4} = C] = V + 1.4V^{0.4}P\frac{dv}{dp} = 0$

$1.4V^{0.4}P\frac{dv}{dp} = -V$

$\frac{dv}{dp} = -\frac{V^{0.6}}{1.4P}$

Looks like it's something we can work with, now we plug it back in to the equation given for $\frac{dv}{dt}$

$\frac{dv}{dt} = \frac{15V^{0.6}}{1.4P}$

Now we plug in the values:

$V = 500 cm^3$

$P = 93 kPa$

$\frac{dv}{dt} = \frac{15(500)^{0.6}}{1.4(93)}$

$\frac{dv}{dt} = 4.78 \frac{cm^3}{min}$

Notice that it's positive, as the question states, it is increasing.

Just to see how the units work here's the equation without the numbers:

$\frac{dv}{dt} = \frac{dv}{dp}\frac{dp}{dt} = \frac{cm^3}{kPa}\frac{kPa}{min} = \frac{cm^3}{min}$