TS EAMCET · Maths · Inverse Trigonometric Functions
If \(y=\operatorname{Sec}^{-1} x\), then \(\frac{d^2 y}{d x^2}=\)
- A \(\frac{1-2 x^2}{x|x|\left(x^2-1\right)^{\frac{3}{2}}}\)
- B \(\frac{1-x^2}{x^2\left(x^2-1\right)^{\frac{3}{2}}}\)
- C \(\frac{1-x^2}{-x^2\left(x^2-1\right)^{\frac{3}{2}}}\)
- D \(\frac{1+2 x^2}{x|x|\left(x^2-1\right)^{\frac{3}{2}}}\)
Answer & Solution
Correct Answer
(A) \(\frac{1-2 x^2}{x|x|\left(x^2-1\right)^{\frac{3}{2}}}\)
Step-by-step Solution
Detailed explanation
\( \frac{dy}{dx} = \frac{1}{|x|\sqrt{x^2-1}} = (|x|(x^2-1)^{1/2})^{-1} \) \( \frac{d^2y}{dx^2} = -1 \cdot (|x|(x^2-1)^{1/2})^{-2} \cdot \left( \frac{d}{dx}(|x|)\sqrt{x^2-1} + |x|\frac{d}{dx}((x^2-1)^{1/2}) \right) \)…
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