Question

If $y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}then{\left(\frac{dy}{dx}\right)}_{x=0}=$

• A. $\frac{1}{2}$
• B. $1$
• C. $-2$
• D. $2$

Answer 1

Answer: (C)

$-2$

Given,

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

$\frac{dy}{dx}=\frac{d}{dx}\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$

$f=\sqrt{u},u=\frac{1-\sin \left(2x\right)}{1+\sin \left(2x\right)}$

$=\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(\frac{1-\sin \left(2x\right)}{1+\sin \left(2x\right)}\right)$

$=\frac{1}{2\sqrt{u}}(-\frac{4\cos \left(2x\right)}{{(1+\sin \left(2x\right))}^2})$

$=\frac{1}{2\sqrt{\frac{1-\sin \left(2x\right)}{1+\sin \left(2x\right)}}}(-\frac{4\cos \left(2x\right)}{{(1+\sin \left(2x\right))}^2})$

$=-\frac{2\cos \left(2x\right)}{\sqrt{-\sin \left(2x\right)+1}{(\sin \left(2x\right)+1)}^{\frac{3}{2}}}$

${\frac{dy}{dx}}_{(x=0)}=-\frac{2\cos \left(2\right(0\left)\right)}{\sqrt{-\sin \left(2\right(0\left)\right)+1}{(\sin \left(2\right(0\left)\right)+1)}^{\frac{3}{2}}}=-2$