Question:

For \(1 \leq x<\infty\), let \(f(x) = \sin^{-1}\left(\frac{1}{x}\right) + \cos^{-1}\left(\frac{1}{x}\right)\). Then \(f'(x) =\)

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Remember that the derivative of any constant value is always zero, which simplifies solving problems involving trigonometric identities and their derivatives.
Updated On: Mar 12, 2025
  • \(\frac{2}{x^2\sqrt{1-x^2}}\)
  • \(\frac{-2}{x^2\sqrt{1-x^2}}\)
  • \(\frac{2}{x\sqrt{1-x^2}}\)
  • \(\frac{-2}{x\sqrt{1-x^2}}\)
  • 0
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Solution and Explanation

First, recognize a key identity involving the inverse sine and cosine functions: \[ \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2} \quad {for} \quad -1 \leq y \leq 1 \] Given that \( \frac{1}{x} \) for \( x \geq 1 \) always lies in the range \([0, 1]\), this identity applies, making \( f(x) \) a constant: \[ f(x) = \frac{\pi}{2} \] The derivative of a constant is zero: \[ f'(x) = 0 \]
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