If f ( x ) = Tan − 1 [ 1 + x 2 − 1 − x 2 1 + x 2 + 1 − x 2 ] for 0 ∣ x ∣ 1 then f ′ ( x ) =

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If $f(x)=\operatorname{Tan}^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]$ for $0 < |x| < 1$ then $f^{\prime}(x)=$

MathematicsDifferentiationAP EAMCETAP EAMCET 2017 (25 Apr Shift 1)

Options:

A $\frac{x}{\sqrt{1-x^2}}$
B $\frac{-x}{\sqrt{1-x^2}}$
C $\frac{x}{\sqrt{1-x^4}}$
D $\frac{-x}{\sqrt{1-x^4}}$

Solution:

2994 Upvotes Verified Answer

The correct answer is: $\frac{-x}{\sqrt{1-x^4}}$

No solution. Refer to answer key.

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