Question:medium

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

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When differentiating expressions involving $|x|$, it is often safest to consider the cases $x>0$ and $x<0$ separately. After finding the derivative for one case (e.g., $x>0$), you can often generalize the result by replacing terms like $x^2$ with $x|x|$ to make it valid for both cases.

Updated On: Mar 30, 2026
  • $\frac{1-2x^2}{x|x|(x^2-1)^{3/2}}$
  • $\frac{1-x^2}{x^2(x^2-1)^{3/2}}$
  • $-\frac{1-x^2}{x^2(x^2-1)^{3/2}}$
  • $\frac{1+2x^2}{x|x|(x^2-1)^{3/2}}$
Show Solution

The Correct Option is A

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