Como sabemos que <span class="math-container">$Q*P=const.$</span> para las preferencias de Cobb-Douglas, podemos concluir que <span class="math-container">$\frac{dQ/Q}{dP/P}$</span> siempre es <span class="math-container">$-1$</span>:
<span class="math-container">$$ QP=const. \implica 0=d(PQ)=Q\ dP+P\ dQ \implica \frac{dQ}{Q}=-\frac{dP}{P} $$</span>