**Toolbox:**

- Differentiate using chain rule and quotient rule.

$ \large\frac{dy}{dx}=\bigg( \large\frac{1}{\sqrt{b^2-a^2}} \bigg)$x $ \bigg(\large \frac{\sqrt{b+a}+\sqrt{b-a}tan\large\frac{x}{2}}{\sqrt{b+a}+\sqrt{b-a}tan\large\frac{x}{2}} \bigg)$ x$ \large\frac{d}{dx} \bigg( \large\frac{(\sqrt{b+a}+\sqrt{b-a}tan\large\frac{x}{2})}{(\sqrt{b+a}-\sqrt{b-a}tan\large\frac{x}{2})} \bigg)$

Simplify and get

$ \large\frac{dy}{dx}=\large\frac{sec^2\large\frac{x}{2}}{(b+a)-(b-a)tan^2\large\frac{x}{2}}$