# If $y=\frac{2}{\Large\sqrt {a^2-b^2}}\tan^{-1}\bigg[\sqrt{\frac{a-b}{a+b}}\tan \frac{x}{2} \bigg],$ prove that $\Large\frac{dy}{dx}=\frac{1}{a+b \cos x}, \normalsize a>b>0$

Toolbox:
• $cosx=\large\frac{1-tan^2\large\frac{x}{2}}{1+tan^2\large\frac{x}{2}}$
• Use chain rule and differentiate
$\large\frac{dy}{dx}=\large\frac{2}{\sqrt{a^2-b^2}}$x$\large\frac{1}{1+\large\frac{a-b}{a+b}tan^2\large\frac{x}{2}}$x$\sqrt{\large\frac{a-b}{a+b}}sec^2\large\frac{x}{2}$x$\large\frac{1}{2}$

Simplify which we get
$\large\frac{dy}{dx}=\large\frac{1+tan^2\large\frac{x}{2}}{a(1+tan^2\large\frac{x}{2})+b(1-tan^2\large\frac{x}{2})}$

Devide number and then by $1+tan^2\large\frac{x}{2}$
$\large\frac{dy}{dx}=\large\frac{1}{a+b\large\frac{1-tan^2\large\frac{x}{2}}{1+tan^2\large\frac{x}{2}}}=\large\frac{1}{a+b\: cosx}$

edited Mar 26, 2013