If y={\Sec}^{-1} x, then \frac{d^2 y}{d x^2}=

Correct Answer is : 1−2x2x|x|(x2−1)32

{1-2 x^2}/{x|x| (x^2-1) ^{\frac{3}{2}}}

{1-x^2}/{x^2 (x^2-1) ^{\frac{3}{2}}}

{1-x^2}/{-x^2 (x^2-1) ^{\frac{3}{2}}}

{1+2 x^2}/{x|x| (x^2-1) ^{\frac{3}{2}}}

TG EAMCET 3 May 2025 Shift 2 Paper

Question : 65 of 160

Marks: +1, -0

If y=Sec−1x, then ‌

d2y

dx2

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