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Class 12 Continuity And Differentiability · Long Answer3 Marksbranch_sensitive_differentiation

Class 12 Maths · Continuity And Differentiability · Long Answer #144

Differentiate u=arctan[(√(1+x²)−1)/x] with respect to v=arcsin[2x/(1+x²)], for −1<x<1 and x≠0.

✅ Step-by-Step Solution

Final Answer: 1/4
  1. The half-angle identity gives u=½arctan x on the original punctured domain.
  2. For |x|<1, 2arctan x is in (−π/2,π/2), so v=2arctan x.
  3. Therefore u=v/4. Equivalently the derivative ratio is [1/(2(1+x²))]/[2/(1+x²)]=1/4.
  4. Thus du/dv=1/4.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] The half-angle identity gives u=½arctan x on the original punctured domain.
  • [1 mark] For |x|<1, 2arctan x is in (−π/2,π/2), so v=2arctan x.
  • [1 mark] Therefore u=v/4. Equivalently the derivative ratio is [1/(2(1+x²))]/[2/(1+x²)]=1/4.
  • [0.5 marks] Thus du/dv=1/4.

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