Class 12 Continuity And Differentiability · Long Answer3 Marksbranch_sensitive_differentiation

Class 12 Maths · Continuity And Differentiability · Long Answer #84

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

✅ Step-by-Step Solution

Final Answer: du/dv=sgn(x)/[2sgn(1−2x²)].

The arcsec argument is at least 1, and its principal angle is u=arcsin|x|. Thus du/dx=sgn(x)/√(1−x²) for x≠0.
The chain rule with √(1−4x²(1−x²))=|1−2x²| gives dv/dx=2sgn(1−2x²)/√(1−x²).
Take their ratio: du/dv=sgn(x)/[2sgn(1−2x²)]. Both derivatives exist on the stated intervals.
The derivative is given by that sign formula; in particular it equals 1/2 only on the positive inner interval and the negative outer interval.

[0.5 marks] The arcsec argument is at least 1, and its principal angle is u=arcsin|x|. Thus du/dx=sgn(x)/√(1−x²) for x≠0.
[1 mark] The chain rule with √(1−4x²(1−x²))=|1−2x²| gives dv/dx=2sgn(1−2x²)/√(1−x²).
[1 mark] Take their ratio: du/dv=sgn(x)/[2sgn(1−2x²)]. Both derivatives exist on the stated intervals.
[0.5 marks] The derivative is given by that sign formula; in particular it equals 1/2 only on the positive inner interval and the negative outer interval.

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