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1. Equation 1 says that regulation vs changes in \$V_\text{CC}\$ is better when \$V_\text{CC}\gg V_\text{LED}\$ and that increases in \$V_\text{CC}\$ will lead to increases in \$I_\text{LED}\$.
2. Equation 2 says that regulation vs changes in \$V_\text{LED}\$ is better when \$V_\text{CC}\gg V_\text{LED}\$ and that increases in \$V_\text{LED}\$ will lead to decreases in \$I_\text{LED}\$.
3. Equation 3 says that regulation vs changes in \$R\$ is fixed at 1:1. So a +1% change in the resistor value will correspond to a -1% change in the current. This is simply because \$R\$ is in the divisor (and that we are talking about small changes in \$R\$.)

In your case, \$\frac{\%\,I_\text{LED}}{\%\,V_\text{CC}}=1.58\$ and \$\frac{\%\,I_\text{LED}}{\%\,V_\text{LED}}=-0.58\$. Given the LED datasheet I'd provided, the LEDs are \$3.3\:\text{V}\pm 9\%\$ and so we can compute that a 9% change in \$V_\text{LED}\$ would lead to a \$-0.58\,\cdot\,\pm 9\%= \mp 5.22\,\%\$ change in the LED current. Which is very close to what was observed in earlier calculations above.

1. Equation 1 says that regulation vs changes in \$V_\text{CC}\$ is better when \$V_\text{CC}\gg V_\text{LED}\$ and that increases in \$V_\text{CC}\$ will lead to increases in \$I_\text{LED}\$.
2. Equation 2 says that regulation vs changes in \$V_\text{LED}\$ is better when \$V_\text{CC}\gg V_\text{LED}\$ and that increases in \$V_\text{LED}\$ will lead to decreases in \$I_\text{LED}\$.
3. Equation 3 says that regulation vs changes in \$R\$ is fixed at 1:1 (but with opposite sign.) So a +1% change in the resistor value will correspond to a -1% change in the current. This is simply because \$R\$ is in the divisor (and that we are talking about small changes in \$R\$.)

In your case, but using my datasheet's LED range (\$3.0\:\text{V} \le V_\text{LED}\le 3.6\:\text{V}\$) and therefore choosing the midpoint value of \$V_\text{LED}\approx 3.3\:\text{V}\$, I get \$\frac{\%\,I_\text{LED}}{\%\,V_\text{CC}}=1.58\$ and \$\frac{\%\,I_\text{LED}}{\%\,V_\text{LED}}=-0.58\$. Given the LED datasheet I'd provided, the LEDs are \$3.3\:\text{V}\pm 9\%\$ and so we can compute that a 9% change in \$V_\text{LED}\$ would lead to a \$-0.58\,\cdot\,\pm 9\%= \mp 5.22\,\%\$ change in the LED current. Which is very close to what was observed in earlier calculations above.

Also note that the sensitivity equations can be used without knowing the value of \$R\$. The only thing that matters is the ratio of \$V_\text{CC}\$ and \$V_\text{LED}\$. This is an important observation for resistor regulation: regulation is better when the supply voltage is very much larger than the required load voltage. (Better regulation implies wasting more power. One of the reasons why active linear regulators were designed, which can provide good regulation without requiring a lot of overhead voltage to get it.)

Also note that the sensitivity equations can be used without knowing the value of \$R\$. The only thing that matters is the ratio of \$V_\text{CC}\$ and \$V_\text{LED}\$. This is an important observation for resistor regulation: regulation is better when the supply voltage is very much larger than the required load voltage. (Better regulation implies wasting more power by increasing the voltage drop across \$R\$. One of the reasons why active linear regulators were designed, which can provide good regulation without requiring a lot of overhead voltage to get it.)

$$R=\frac{V_\text{CC}-V_\text{LED}}{I_\text{LED}}=\frac{9\:\text{V}-3\:\text{V}}{20\:\text{mA}}=300\:\Omega$$

$$R=\frac{V_\text{CC}-V_\text{LED}}{I_\text{LED}}=\frac{9\:\text{V}-3\:\text{V}}{20\:\text{mA}}=300\:\Omega\tag{0}$$

Also note that the sensitivity equations can be used without knowing the value of \$R\$. The only thing that matters is the ratio of \$V_\text{CC}\$ and \$V_\text{LED}\$. This is an important observation for resistor regulation: regulation is better when the supply voltage is very much larger than the required load voltage. (Better regulation implies wasting more power. One of the reasons why active linear regulators were designed, which can provide good regulation without requiring a lot of overhead voltage to get it.)

In your case, \$\frac{\%\,I_\text{LED}}{\%\,V_\text{CC}}=1.58\$ and \$\frac{\%\,I_\text{LED}}{\%\,V_\text{LED}}=-0.58\$. Given the LED datasheet I'd provided, the LEDs are \$3.3\:\text{V}\pm 9\%\$ and so we can compute that a 9% change in \$V_\text{LED}\$ would lead to a \$-0.58\,\cdot\,\pm 9\%= \mp 5.22\,\%\$ change in the LED current. Which is very close to what was observed in earlier calculations above.