Question

d/dx tan⁻¹ √1+x²-√1-x²/√1-x²+√1+x²=......... | x | < 1 ,Select Proper option from the given options.
(a) 1/√1-x⁴
(b) -x/√1-x⁴
(c) 1/2√1-x⁴
(d) x²/1-x⁴

Answer 1

I hope this helps ya

Answer 2

we have to find value of $\frac{d}{dx}tan^{-1}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}$

Let $y=tan^{-1}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}$

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

squaring both sides,
tan²y = $\frac{1-\sqrt{1-x^4}}{1+\sqrt{1-x^4}}$

(1 - tan²y) = √(1 - x⁴) (tan²y + 1)

(1 - tan²y)/(tan²y + 1) = √(1 - x⁴)

(1 - tan²y)/sec²y = √(1 - x⁴)

cos²y - sin²y = √(1 - x⁴)

cos2y = √(1 - x⁴) [ we know, cos2r = cos²r - sin²r]

now , differentiate with respect to x,

$-2sin2y. dy/dx = \frac{1}{\sqrt{1-x^4}}(-2x^3)$

we know, sin2y = 2tany/(1 + tan²y)

= 2{√(1 + x²) - √(1 - x²)}{√(1 + x²) + √(1 - x²)}/[{√(1-x²) + √(1 + x²)}² + {√(1 + x²) - √(1 - x²)}²]

= 4x²/{2 + 2√(1 - x⁴) + 2 - 2√(1 - x⁴)}

= 4x²/4
= x² hence, sin2y = x²

now, -2x². dy/dx = -2x³/√(1 - x⁴)

dy/dx = x/√(1 - x⁴)