Choosing an inductor value for a buck regulator comes directly from V = \$\frac{\text{L di} }{\text{dt}}\$ . Where V is the voltage across the inductor, and i is the current through it. First, you want to design for the case where the inductor is in continuous conduction mode (CCM). This means that energy in the inductor doesn't run out during the switching cycle. So, there are two states, one where the switch is on, and another where the switch is off (and the rectifier is on). Voltage across the inductor during each state is essentially a constant (although it is a different value for each state). Anyway since the voltage is a constant, the inductor equation can be linearized (and rearranged to give L).
L = \$\frac{V \text{$\Delta $t}}{\text{$\Delta $I}}\$ this is the basis for the equation you saw in the app-note.
\$\text {$\Delta $I}\$ is something you define, not determine.
You will want to maintain CCM operation, so define \$\text {$\Delta $I}\$ as some small fraction of inductor current (I). A good choice is 10% of I. So, for your case \$\text {$\Delta $I}\$ would be 0.24A. This will also define the ripple current in the output capacitors, and less ripple current means less ripple voltage on the output.
Now you can choose an optimal value of L using \$V_{\text{in}}\$ and \$V_o\$ (and hence the duty cycle D = \$\frac {V_o} {V_ {\text {in}}}\$). But you can also make a quick over estimate for the inductance where you don't consider \$V_{\text{in}}\$ using L ~ \$\frac{10 V_o}{I_o F_{\text{sw}}}\$ (for more on this look here How to choose a inductor for a buck regulator circuit? ). An over estimate can be worthwhile, especially if you are early in development or uncertain exactly how much the output current will be (output current tends to end up higher than expected usually).
Since you are looking at Linear Tech you should (as Anindo Ghosh pointed out) also look at using their CAD support.
Firstly, if you are worried about the transient response of your regulator when load current suddenly decreases, you need to use a synchronous buck converter - a non synchronous buck converter isn't going to stop the voltage rising when load is removed because the converter may be in the process of transferring a sizable "lump" of energy from the inductor into the capacitor and, without the previous low load, the output voltage is going to rise above expectations and possibly damage the devices you have connected.
Sanity check
Going on what you've said let's say your average load is 10A at 0.9V - this is a load power of 9W. Let's also assume that your mark-space ratio is 50% for this load. You've stated the inductance so it is possible to calculate your switching frequency. If you were switching at 1MHz your inductor is being charged for 500ns and in that time it will attain a peak current: -
\$V=L\dfrac{di}{dt}\therefore di = dt\times\dfrac{V}{L} = 500ns\times\dfrac{12V-0.9V}{0.47\times 10^{-6}} = 11.81A\$
This translates to an energy of \$\dfrac{L\times 11.81^2}{2} = 32.78\mu J\$.
This gets transferred 1 million times per second so the power it is delivering is 32.78W.
Clearly this is more than what the load power is (9W) so you are probably using a higher frequency. At 2MHz switching (50:50 duty) the inductor current is half and the energy is a quarter of what it is at 1MHz but the energy gets transferred twice as often so the power reduces to 16.38W and this is still too high for the "average" load and a little too high for the peak load (15A x 0.9V = 13.5W).
You don't see many switchers operating above 2MHz so I have to assume that you are operating at somewhat less than 50% duty cycle for the "average" load of 10A. A 200ns inductor charge time would give a peak current of 4.72A and an energy of 5.24\$\mu J\$ and this feeds 10.48W into your load at 2MHz (near enough).
So, if your switcher circuit isn't actually operating at about 2MHz, then you might want to consider the effectiveness of your circuit.
Transient current of 6A
Assuming you are using a synchronous converter, a transient current of +/-6A is going to leave an energy deficit (or excess) that your output capacitor has to mop-up without causing (say) more than a +/-0.1V change on the output: -
\$Q = C.V \therefore \dfrac{dQ}{dt} = C\dfrac{dV}{dt}\$ = current (6A)
This allows you to calculate the capacitor on the output: -
\$C = \dfrac{250ns}{0.1V}\times 6A = 15uF\$
Conclusion
I reiterate - if your switcher isn't switching at about 2MHz then these numbers are going to be a little off the mark. I'd also say that if it is switching below 1MHz (low duty cycle) then extreme changes in output voltage due to load changes are going to be an issue. If the converter isn't a synchronous type then sudden load reductions are going to cause even bigger problems because the converter will go into discontinuous mode and you could easily see a peak in voltage (on top of the 0.9V) of several hundred millivolts.
Best Answer
Increasing the value of the inductor will sooner or later turn your circuit into Continuous Conduction Mode (CCM), where there is a constant DC current flowing through the inductor, and the ripple current (the triangle waveform) being superimposed on this base DC current.
This means that the current of the inductor in this mode never goes down to zero. The energy stored in the inductor is indeed \$\dfrac{LI^2}{2}\$, however, only a part of this energy gets transferred in a cycle to the consumer: \${\dfrac{LI_{max}^2}{2}} - {\dfrac{LI_{min}^2}{2}} = \dfrac{L(I_{max}^2-I_{min}^2)}{2}\$, the rest stays in the inductor (more or less permanently). The amount of energy used by the consumer will be the sum of the energy transferred through storage in the inductor, and the energy constantly being transferred by the DC current flowing through the inductor.
The amount of the maximum possible ripple current is determined by the inductance value, the switching frequency, the ON-OFF duty ratio (which more or less equals the output to input voltage ratio), and the voltages connected to the inductor (the input voltage minus the output voltage in the ON phase, and the output voltage in the OFF phase). Increasing the inductance value decreases the maximum possible ripple current. As soon as the output current needed is more than half of the maximum possible ripple current, operation will turn into CCM.
What disadvantages are there for having a larger inductance?
More turns, therefore higher DC resistance in the inductor, meaning larger copper losses, or having to increase wire thickness/number of strands to compensate for this, thereby increasing inductor size & cost.
Having a larger inductance, the feedback control loop will be slower, meaning that the power supply will be less flexible adapting to quickly changing loads. This will probably show itself as larger overshoots in response to a unit-step load change.
When using an asynchronous buck converter, in OFF-state, the free-running diode will be conducting. In DCM, the diode is conducting only as long as the energy stored in the inductor gets fully removed. In CCM, the diode is conducting during the whole OFF-phase, as the inductor current never goes to zero. This means higher losses on the free-running diode - which may be a problem especially because the losses on the FET can be decreased by using a FET with a lower Rdson, but you cannot do the same with the diode. The smaller the output to input voltage ratio is, the more grave this problem could be.