Class 12 Continuity And Differentiability · Long Answer3 Marksbranch_sensitive_differentiation
Class 12 Maths · Continuity And Differentiability · Long Answer #82
For |x|>1, prove d(arcsec x)/dx=1/[|x|√(x²−1)].
✅ Step-by-Step Solution
Final Answer: 1/[|x|√(x²−1)]
- Let y=arcsec x on the principal range [0,π]∖{π/2}, so sec y=x.
- Differentiate: sec y tan y·y′=1. For x>1, both sec y,tan y are positive; for x<−1, both are negative.
- Thus sec y tan y=|sec y|·|tan y|=|x|√(x²−1), positive on both domain components.
- Therefore y′=1/[|x|√(x²−1)].
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Let y=arcsec x on the principal range [0,π]∖{π/2}, so sec y=x.
- [1 mark] Differentiate: sec y tan y·y′=1. For x>1, both sec y,tan y are positive; for x<−1, both are negative.
- [1 mark] Thus sec y tan y=|sec y|·|tan y|=|x|√(x²−1), positive on both domain components.
- [0.5 marks] Therefore y′=1/[|x|√(x²−1)].
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