If $y = an^{- 1}{\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}},\ \ \ then\frac{dy}{dx}\ }is\ equal\ to$

Question

If $y = 	an^{- 1}{\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}},\ \ \ then\frac{dy}{dx}\ }is\ equal\ to$

Dr. Abubacker · Accepted Answer

(d)
Put $x^{2} = \cos{2	heta\ }$in the given equation, we get
$y = 	an^{- 1}\frac{\sqrt{1 + \cos{2	heta}} - \sqrt{1 - \cos{2	heta}}}{\sqrt{1 + \cos{2	heta}} + \sqrt{1 - \cos{2	heta}}}$
$= 	an^{- 1}\frac{\cos{	heta - sin	heta}}{\cos{	heta + \sin	heta}}$
$= 	an^{- 1}{	an\left( \frac{\pi}{4} - 	heta ight)}$
$\Rightarrow \ y = \frac{\pi}{4} - 	heta = \frac{\pi}{4} - \frac{1}{2}\cos^{- 1}x^{2}$
$\Rightarrow \ \frac{dy}{dx} = 0 - \frac{1}{2}\left( - \frac{(2x)}{\sqrt{1 - x^{4}}} ight) = \frac{x}{\sqrt{1 - x^{4}}}$

Q. If $y = \tan^{- 1}{\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}},\ \ \ then\frac{dy}{dx}\ }is\ equal\ to$

Explanation

Reviewed by Dr. Abubacker

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