Question

Differentiate the function w.r.t.x:
$\rm \sqrt{\frac{1+\sin 2x}{1-\sin 2x}}$

Answer 1

√1+sin2x/1-sin2x

√(sinx + cosx)^2/(sinx - cosx)^2

sinx + cosx / sinx - cosx

Answer 2

Given that :

$\sqrt{ \frac{1 + \sin2x }{1 - \sin2x} } \\ \\ = > \sqrt{ \frac{ { \sin}^{2}x + { \cos}^{2}x + 2 \sin(x) \cos(x) }{ { \sin}^{2} x + { \cos }^{2} x - 2 \sin(x) \cos( x) } } \\ \\ = > \sqrt{ \frac{ {[ \sin(x) + \cos(x) ]}^{2} }{ {[ \sin(x) - \cos(x) ] }^{2} } } \\ \\ = > \frac{ \sin(x) + \cos(x) }{ \sin(x ) - \cos(x) }$

Now, on differentiate with respect to x :

$\frac{[ \sin(x) - \cos(x)][( \cos(x) - \sin(x) ] - [ \sin(x) + \cos(x) ][ \cos(x) + \sin(x)] }{ {[ \sin(x) - \cos(x)] }^{2} } \\ \\ = > \frac{ - 2 { \sin}^{2}x - 2 { \cos}^{2} x }{ {[ \sin(x) - \cos(x)] }^{2} } \\ \\ = > \frac{ - 2( { \sin}^{2} x + { \cos}^{2} x)}{ {[ \sin(x) - \cos(x) ]}^{2} } \\ \\ = > \frac{ - 2}{ {[ \sin(x) - \cos(x) ]}^{2} }$