AP EAMCET · Maths · Differentiation
If \(y=\sin ^{-1}\left[x \sqrt{1-x}-\sqrt{x} \cdot \sqrt{1-x^2}\right]\) and \(0 < x < 1\), then \(\frac{d y}{d x}\) is equal to
- A \(\frac{1}{2 \sqrt{1-x^2}}-\frac{1}{x \sqrt{1-x^2}}\)
- B \(\frac{1}{\sqrt{1-x^2}}-\frac{1}{2 \sqrt{x-x^2}}\)
- C \(\frac{1}{2 \sqrt{x-x^2}}+\frac{1}{\sqrt{1-x^2}}\)
- D \(\frac{-1}{\sqrt{1-x^2}}-\frac{1}{x \sqrt{1-x^2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{1-x^2}}-\frac{1}{2 \sqrt{x-x^2}}\)
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