Question

Differentiate the following function with respect to $x$.
$y=\sqrt{\frac{1+\sin 2x}{1-\sin 2x}}$

Answer 1

$y=\sqrt{\frac{1+\sin 2x}{1-\sin 2x}}=\sqrt{\frac{1+2\sin x\cos x}{1-2\sin x\cos x}}[\because \sin 2x=2\sin x\cos x{\sin }^{2}x1{\cos }^{2}x=1]$

$=\sqrt{\frac{{\cos }^{2}x-1{\sin }^{2}x+2\sin x\cos x}{{\cos }^{2}x+{\sin }^{2}x+2\sin x\cos x}}$

$=\frac{\cos x+\sin x}{\cos x-\sin x}$

$\frac{dy}{dx}=\frac{(\cos x-\sin x)(-\sin x+\cos x)-(\cos x+\sin x)(-\sin x-\cos x)}{(\cos x-\sin x{)}^2}$ by quotient mule

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

$=\frac{1-2\sin x+\cos x+1+2\sin x2\cos x}{(\cos x-\sin x{)}^2}$

$\frac{dy}{x}=\frac{2}{(\cos x-\sin x{)}^2}$