AP EAMCET · Maths · Inverse Trigonometric Functions
If \(f(x)=\operatorname{Sec}^{-1}\left(\frac{1}{2 x^2-1}\right)\) and \(g(x)=\operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\), then the derivative of \(f(x)\) with respect to \(g(x)\) is
- A \(\frac{1+x^2}{4 \sqrt{1-x^2}}\)
- B \(\frac{\left(1-x^2\right)}{4 \sqrt{1+x^2}}\)
- C \(-\frac{4\left(1-x^2\right)}{\sqrt{1+x^2}}\)
- D \(-\frac{4\left(1+x^2\right)}{\sqrt{1-x^2}}\)
Answer & Solution
Correct Answer
(D) \(-\frac{4\left(1+x^2\right)}{\sqrt{1-x^2}}\)
Step-by-step Solution
Detailed explanation
\(f(x) = \operatorname{Sec}^{-1}\left(\frac{1}{2 x^2-1}\right) = \operatorname{Cos}^{-1}(2x^2-1)\) \(f'(x) = -\frac{1}{\sqrt{1-(2x^2-1)^2}} \cdot (4x) = -\frac{4x}{\sqrt{4x^2-4x^4}} = -\frac{4x}{2x\sqrt{1-x^2}} = -\frac{2}{\sqrt{1-x^2}}\)…
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