Class 12 Continuity And Differentiability · Long Answer3 Marksbranch_sensitive_differentiation

Class 12 Maths · Continuity And Differentiability · Long Answer #142

Differentiate u=arctan[x/√(1−x²)] with respect to v=arcsin[2x√(1−x²)], on |x|<1 excluding |x|=1/√2.

✅ Step-by-Step Solution

Final Answer: +1/2 for |x|<1/√2; −1/2 for 1/√2<|x|<1.

Let θ=arcsin x∈(−π/2,π/2). Then u=arctan(tanθ)=θ and u′=1/√(1−x²).
For v, the argument is sin2θ. Chain differentiation yields v′=2sgn(1−2x²)/√(1−x²), because the square root in the inverse derivative is |1−2x²|.
Dividing gives du/dv=1/[2sgn(1−2x²)]. The sign changes at the excluded corners ±1/√2.
Thus the derivative is +1/2 on the inner interval and −1/2 on the two outer intervals.

[0.5 marks] Let θ=arcsin x∈(−π/2,π/2). Then u=arctan(tanθ)=θ and u′=1/√(1−x²).
[1 mark] For v, the argument is sin2θ. Chain differentiation yields v′=2sgn(1−2x²)/√(1−x²), because the square root in the inverse derivative is |1−2x²|.
[1 mark] Dividing gives du/dv=1/[2sgn(1−2x²)]. The sign changes at the excluded corners ±1/√2.
[0.5 marks] Thus the derivative is +1/2 on the inner interval and −1/2 on the two outer intervals.

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