CUET · MATHS · PYQ PAPER 2025
If \(y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^2}, x \in(-1,0)\), then \(\frac{d y}{d x}\) is equal to
- A 0
- B \(\frac{1}{\sqrt{1-x^2}}\)
- C \(\frac{2}{\sqrt{1-x^2}}\)
- D \(\frac{-2}{\sqrt{1-x^2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{2}{\sqrt{1-x^2}}\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x} = \frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-(\sqrt{1-x^2})^2}} \cdot \frac{-x}{\sqrt{1-x^2}}\) \(= \frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{x^2}} \cdot \frac{-x}{\sqrt{1-x^2}}\) \(= \frac{1}{\sqrt{1-x^2}} + \frac{1}{-x} \cdot \frac{-x}{\sqrt{1-x^2}}\) (since…
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