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Class 12 Continuity And Differentiability · Long Answer3 Marksbranch_sensitive_differentiation

Class 12 Maths · Continuity And Differentiability · Long Answer #67

Differentiate u=arctan[(3x−x³)/(1−3x²)] with respect to v=arctan[x/√(1−x²)], for |x|<1/√3.

✅ Step-by-Step Solution

Final Answer: 3√(1−x²)/(1+x²)
  1. Let θ=arctan x. Then |θ|<π/6 and |3θ|<π/2, so the triple-angle identity gives u=3arctan x.
  2. Thus du/dx=3/(1+x²). Also v=arcsin x because the angle lies in (−π/2,π/2), so dv/dx=1/√(1−x²).
  3. The derivative with respect to v is (du/dx)/(dv/dx)=3√(1−x²)/(1+x²).
  4. Therefore du/dv=3√(1−x²)/(1+x²).

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Let θ=arctan x. Then |θ|<π/6 and |3θ|<π/2, so the triple-angle identity gives u=3arctan x.
  • [1 mark] Thus du/dx=3/(1+x²). Also v=arcsin x because the angle lies in (−π/2,π/2), so dv/dx=1/√(1−x²).
  • [1 mark] The derivative with respect to v is (du/dx)/(dv/dx)=3√(1−x²)/(1+x²).
  • [0.5 marks] Therefore du/dv=3√(1−x²)/(1+x²).

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