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Class 12 Continuity And Differentiability · Long Answer3 Marksbranch_sensitive_differentiation

Class 12 Maths · Continuity And Differentiability · Long Answer #143

For 0<|x|<1, differentiate y=arctan[(√(1+x²)+√(1−x²))/(√(1+x²)−√(1−x²))].

✅ Step-by-Step Solution

Final Answer: −x/√(1−x⁴)
  1. Let θ=arcsin(x²)∈(0,π/2). The sum-over-difference half-angle quotient is cot(θ/2).
  2. Since θ/2∈(0,π/4), the principal arctangent of cot(θ/2) is π/2−θ/2. Thus y=π/2−½arcsin(x²).
  3. Differentiate: y′=−½·2x/√(1−x⁴). The original denominator is zero at x=0, which is excluded.
  4. Therefore y′=−x/√(1−x⁴).

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Let θ=arcsin(x²)∈(0,π/2). The sum-over-difference half-angle quotient is cot(θ/2).
  • [1 mark] Since θ/2∈(0,π/4), the principal arctangent of cot(θ/2) is π/2−θ/2. Thus y=π/2−½arcsin(x²).
  • [1 mark] Differentiate: y′=−½·2x/√(1−x⁴). The original denominator is zero at x=0, which is excluded.
  • [0.5 marks] Therefore y′=−x/√(1−x⁴).

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