Question

Find d/dx tan⁻¹ √1+x²+√1-x²/√1+x²+√1-x² | x | < 1

Answer 1

Answer:

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Answer 2

Correct answer is $\frac{\mathrm{d} }{\mathrm{d} x}f_{x} = 1$

Explanation :

Let, $\frac{\mathrm{d}}{\mathrm{d}x}f_{x} = \frac{\mathrm{d} }{\mathrm{d} x}tan^{-1}\frac{\sqrt(1 + x^{2}) + \sqrt(1 - x^{2})}{\sqrt(1 + x^{2})- \sqrt(1 - x^{2})}$  

$f_{x} = tan^{-1}\frac{\sqrt(1 + x^{2}) + \sqrt(1 - x^{2})}{\sqrt(1 + x^{2})- \sqrt(1 - x^{2})}$

Let, $x^{2} = \sin2x$

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

$f_{x} = tan^{-1}(\frac{2\sin x}{2\cos x}) = tan^{-1}(\tan x) = x$

Correct answer is $\frac{\mathrm{d} }{\mathrm{d} x}f_{x} = 1$ .