Class 12 Continuity And Differentiability · Long Answer3 Marksbranch_sensitive_differentiation

Class 12 Maths · Continuity And Differentiability · Long Answer #75

Differentiate u=arctan[√(1−x²)/x] with respect to v=arccos[2x√(1−x²)] on intervals where 0<|x|<1 and |x|≠1/√2.

✅ Step-by-Step Solution

Final Answer: du/dv=1/2 if |x|<1/√2; −1/2 if |x|>1/√2.

On either side of 0, direct chain differentiation gives du/dx=−1/√(1−x²). For negative x the principal representation of u differs by π from the source substitution, but this derivative remains the same.
For w=2x√(1−x²), w′=2(1−2x²)/√(1−x²), and √(1−w²)=|1−2x²|. Therefore dv/dx=−2sgn(1−2x²)/√(1−x²).
Dividing gives du/dv=1/[2sgn(1−2x²)], which is +1/2 inside the thresholds and −1/2 outside them.
Thus the derivative is 1/2 for 0<|x|<1/√2 and −1/2 for 1/√2<|x|<1.

[0.5 marks] On either side of 0, direct chain differentiation gives du/dx=−1/√(1−x²). For negative x the principal representation of u differs by π from the source substitution, but this derivative remains the same.
[1 mark] For w=2x√(1−x²), w′=2(1−2x²)/√(1−x²), and √(1−w²)=|1−2x²|. Therefore dv/dx=−2sgn(1−2x²)/√(1−x²).
[1 mark] Dividing gives du/dv=1/[2sgn(1−2x²)], which is +1/2 inside the thresholds and −1/2 outside them.
[0.5 marks] Thus the derivative is 1/2 for 0<|x|<1/√2 and −1/2 for 1/√2<|x|<1.

Practise interactive questions, flashcards, visual mind maps, and timed board tests for Continuity And Differentiability.