Question
If $$y = {\tan ^{ - 1}}\sqrt {\frac{{x + 1}}{{x - 1}}} $$ then $$\frac{{dy}}{{dx}}$$ is equal to :
A.
$$\frac{{ - 1}}{{2\left| x \right|\sqrt {{x^2} - 1} }}$$
B.
$$\frac{{ - 1}}{{2x\sqrt {{x^2} - 1} }}$$
C.
$$\frac{1}{{2x\sqrt {{x^2} - 1} }}$$
D.
none of these
Answer :
$$\frac{{ - 1}}{{2\left| x \right|\sqrt {{x^2} - 1} }}$$
Solution :
Let $$x = \sec \,\theta $$
$$\eqalign{
& {\text{Then }}y = {\tan ^{ - 1}}\sqrt {\frac{{\sec \,\theta + 1}}{{\sec \,\theta - 1}}} \cr
& = {\tan ^{ - 1}}\sqrt {\frac{{1 + \cos \,\theta }}{{1 - \cos \,\theta }}} \cr
& = {\tan ^{ - 1}}\left( {\cot \frac{\theta }{2}} \right) \cr
& \therefore \,\,y = {\tan ^{ - 1}}\left\{ {\tan \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right\} = \frac{\pi }{2} - \frac{1}{2}{\sec ^{ - 1}}x \cr
& \therefore \,\,\frac{{dy}}{{dx}} = - \frac{1}{2}.\frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }} \cr} $$