Question

y = √1-sin2x/1+sin2x differentiation

Answer 1

Here we have to find the derivative,

$\frac{dy}{dx}$ where $y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$.

Note that

$1-\sin 2x=\sin ^2x+\cos^2 x-2\sin x\cos x=(\sin x-\cos x)^2\\ 1+\sin 2x=\sin ^2x+\cos^2 x+2\sin x\cos x=(\sin x+\cos x)^2$

Thus,

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

Differentiating,

$\frac{dy}{dx}=\pm \frac{d}{dx}(\frac{\sin x-\cos x}{\sin x+\cos x} )\\ \frac{dy}{dx}=\pm \frac{(\sin x+\cos x)^2-(\sin x-\cos x)^2}{(\sin x+\cos x)^2} \\ \frac{dy}{dx}=\pm \frac{4\sin x\cos x}{(\sin x+\cos x)^2} \\ \frac{dy}{dx}=\pm \frac{2\sin 2x}{1+\sin 2x}$