Using the difference to product identity, we find that
$x=\cos 36^{\circ} - \cos 72^{\circ}$ is equivalent to $$x=\text{-}2\sin{\frac{(36^{\circ}+72^{\circ})}{2}}\sin{\frac{(36^{\circ}-72^{\circ})}{2}} \implies$$
$$x=\text{-}2\sin54^{\circ}\sin(\text{-}18^{\circ}).$$
Since sine is an odd function, we find that $\sin{(\text{-}18^{\circ})}= \text{-} \sin{18^{\circ}}$, and thus $\text{-}2\sin54^{\circ}\sin(\text{-}18^{\circ})=2\sin54^{\circ}\sin18^{\circ}$. Using the property $\sin{(90^{\circ}-a)}=\cos{a}$, we find
$$x=2\cos(90^{\circ}-54^{\circ})\cos(90^{\circ}-18^{\circ}) \implies$$
$$x=2\cos36^{\circ}\cos72^{\circ}.$$
We multiply the entire expression by $\sin36^{\circ}$ and use the double angle identity of sine twice to find
$$x\sin36^{\circ}=2\sin36^{\circ}\cos36^{\circ}\cos72^{\circ} \implies$$
$$x\sin36^{\circ}=\sin72^{\circ}\cos72^{\circ} \implies$$
$$x\sin36^{\circ}=\frac{1}{2}\sin144^{\circ}.$$
Using the property $\sin(180^{\circ}-a)=\sin{a}$, we find $\sin144^{\circ}=\sin36^{\circ}.$ Substituting this back into the equation, we have
$$x\sin36^{\circ}=\frac{1}{2}\sin36^{\circ}.$$
Dividing both sides by $\sin36^{\circ}$, we have
$$x=\boxed{\textbf{(B)}\ \frac{1}{2}}$$

1. We start with the given expression $x = \cos 36^\circ - \cos 72^\circ$.

2. We use the double angle formulas for cosine:
$$\cos 36^\circ = 1 - 2 \sin^2 18^\circ$$
and
$$\cos 72^\circ = 2 \cos^2 36^\circ - 1.$$

3. We need to express $\cos 36^\circ$ and $\cos 72^\circ$ in a form that allows us to simplify $x$. First, we use the identity for $\cos 72^\circ$:
$$\cos 72^\circ = 2 \cos^2 36^\circ - 1.$$

4. Let $y = \cos 36^\circ$. Then:
$$\cos 72^\circ = 2y^2 - 1.$$

5. Substitute these into the expression for $x$:
$$x = y - (2y^2 - 1).$$

6. Simplify the expression:
$$x = y - 2y^2 + 1.$$

7. We need to find the value of $y = \cos 36^\circ$. Using the known value:
$$\cos 36^\circ = \frac{\sqrt{5} + 1}{4}.$$

8. Substitute $y = \frac{\sqrt{5} + 1}{4}$ into the expression for $x$:
$$x = \frac{\sqrt{5} + 1}{4} - 2 \left( \frac{\sqrt{5} + 1}{4} \right)^2 + 1.$$

9. Calculate $\left( \frac{\sqrt{5} + 1}{4} \right)^2$:
$$\left( \frac{\sqrt{5} + 1}{4} \right)^2 = \frac{(\sqrt{5} + 1)^2}{16} = \frac{5 + 2\sqrt{5} + 1}{16} = \frac{6 + 2\sqrt{5}}{16} = \frac{3 + \sqrt{5}}{8}.$$

10. Substitute back into the expression for $x$:
$$x = \frac{\sqrt{5} + 1}{4} - 2 \cdot \frac{3 + \sqrt{5}}{8} + 1.$$

11. Simplify the terms:
$$x = \frac{\sqrt{5} + 1}{4} - \frac{3 + \sqrt{5}}{4} + 1.$$

12. Combine the fractions:
$$x = \frac{\sqrt{5} + 1 - 3 - \sqrt{5}}{4} + 1 = \frac{-2}{4} + 1 = -\frac{1}{2} + 1 = \frac{1}{2}.$$

13. Therefore, the value of $x$ is:
$$x = \frac{1}{2}.$$

The final answer is $\boxed{\frac{1}{2}}$.