Class 12 Continuity And Differentiability · Long Answer3 Marksbranch_sensitive_differentiation

Class 12 Maths · Continuity And Differentiability · Long Answer #137

For y=arcsin[(6x−4√(1−4x²))/5], find y′ on the interior real domain.

✅ Step-by-Step Solution

Final Answer: −2/√(1−4x²) for −1/2<x<−3/10; +2/√(1−4x²) for −3/10<x<1/2; not differentiable at −3/10.

Put θ=arcsin(2x)∈(−π/2,π/2), and φ=arctan(4/3), so cosφ=3/5,sinφ=4/5. The argument is sin(θ−φ).
The angle θ−φ crosses −π/2 when θ=φ−π/2, giving 2x=−3/5 and x=−3/10. Below this point the principal arcsine is −π−(θ−φ); above it the principal arcsine is θ−φ.
Since θ′=2/√(1−4x²), differentiation gives the negative sign below −3/10 and positive sign above. At −3/10 the opposite nonzero one-sided derivatives give a corner.
The derivative is the stated piecewise formula; the domain endpoints require separate one-sided treatment.

[0.5 marks] Put θ=arcsin(2x)∈(−π/2,π/2), and φ=arctan(4/3), so cosφ=3/5,sinφ=4/5. The argument is sin(θ−φ).
[1 mark] The angle θ−φ crosses −π/2 when θ=φ−π/2, giving 2x=−3/5 and x=−3/10. Below this point the principal arcsine is −π−(θ−φ); above it the principal arcsine is θ−φ.
[1 mark] Since θ′=2/√(1−4x²), differentiation gives the negative sign below −3/10 and positive sign above. At −3/10 the opposite nonzero one-sided derivatives give a corner.
[0.5 marks] The derivative is the stated piecewise formula; the domain endpoints require separate one-sided treatment.

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