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Class 12 Continuity And Differentiability · Long Answer3 Marksbranch_sensitive_differentiation

Class 12 Maths · Continuity And Differentiability · Long Answer #140

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

✅ Step-by-Step Solution

Final Answer: −1/2
  1. Let θ=arcsin(x²)∈(0,π/2). Then √(1+x²)=cos(θ/2)+sin(θ/2) and √(1−x²)=cos(θ/2)−sin(θ/2).
  2. The quotient becomes tan(θ/2), so the principal arctangent gives u=θ/2=½arcsin(x²).
  3. Also v=arccos(x²)=π/2−arcsin(x²), hence u=π/4−v/2. This directly expresses u as a differentiable function of v.
  4. Therefore du/dv=−1/2.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Let θ=arcsin(x²)∈(0,π/2). Then √(1+x²)=cos(θ/2)+sin(θ/2) and √(1−x²)=cos(θ/2)−sin(θ/2).
  • [1 mark] The quotient becomes tan(θ/2), so the principal arctangent gives u=θ/2=½arcsin(x²).
  • [1 mark] Also v=arccos(x²)=π/2−arcsin(x²), hence u=π/4−v/2. This directly expresses u as a differentiable function of v.
  • [0.5 marks] Therefore du/dv=−1/2.

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