Si $e^x = \frac{\sqrt{1+t} - \sqrt{1-t}}{\sqrt{1+t} + \sqrt{1-t}}$ y $\tan \frac{y}{2} = \sqrt{\frac{1-t}{1+t}}$, determinar $\frac{dy}{dx}$ en $t = \frac{1}{2}$.

• [a.] $-\frac{1}{2}$
• [b.] $\frac{1}{2}$
• [c.] $0$
• [d.] none of these