Given the definition of the elasticity coefficient in terms of a partial derivative, it is possible, for example, to determine the elasticity of an arbitrary rate law by differentiating the rate law by the independent variable and scaling. For example, the elasticity coefficient for a mass-action rate law such as:
where  is the reaction rate,  the reaction rate constant,  is the ith chemical species involved in the reaction and  the ith reaction order, then the elasticity,  can be obtained by differentiating the rate law with respect to  and scaling:
That is, the elasticity for a mass-action rate law is equal to the order of reaction of the species.
For example the elasticity of A in the reaction  where the rate of reaction is given by: , the elasticity can be evaluated using:
Elasticities can also be derived for more complex rate laws such as the Michaelis–Menten rate law. If
then it can be easily shown than
This equation illustrates the idea that elasticities need not be constants (as with mass-action laws) but can be a function of the reactant concentration. In this case, the elasticity approaches unity at low reactant concentration (s) and zero at high reactant concentration.
For the reversible Michaelis–Menten rate law:
where  is the forward ,  the forward ,  the equilibrium constant  and  the reverse , two elasticity coefficients can be calculated, one with respect to substrate, S, and another with respect to product, P. Thus:
where  is the mass-action ratio, that is . Note that when p = 0, the equations reduce to the case for the irreversible Michaelis–Menten law.
As a final example, consider the Hill equation:
where n is the Hill coefficient and  is the half-saturation coefficient (cf. Michaelis–Menten rate law), then the elasticity coefficient is given by:
Note that at low  concentrations of S the elasticity approaches n. At high concentrations of S the elasticity approaches zero. This means the elasticity is bounded between zero and the Hill coefficient.