Question:medium
If \( x^2 + y^2 = 1 \), then
If \( x^2 + y^2 = 1 \), then
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For implicit equations, always differentiate carefully using product rule when variables are mixed.
Updated On: May 8, 2026
- \( yy'' + (y')^2 + 1 = 0 \)
- \( yy'' + 2(y')^2 + 1 = 0 \)
- \( yy'' - 2(y')^2 + 1 = 0 \)
- \( yy'' + (y')^2 - 1 = 0 \)
- \( yy'' - (2y')^2 - 1 = 0 \)
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The Correct Option is A
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