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📐 Mathematics  ·  Continuity and Differentiability  ·  JEE

If y = sin-1(2x sqrt(1-x<sup>2</sup>)) for -1/sqrt(2) <= x <= 1/sqrt(2), find dy/dx.

Answer: 2/sqrt(1-x 2 ).

  • A 2/sqrt(1-x<sup>2</sup>)
  • B -2/sqrt(1-x<sup>2</sup>)
  • C 1/sqrt(1-x<sup>2</sup>)
  • D 2/(1-x<sup>2</sup>)

Correct answer: A. 2/sqrt(1-x<sup>2</sup>)

Explanation: Substituting x = sin(theta), 2x sqrt(1-x<sup>2</sup>) = sin(2theta), so y = sin-1[sin(2theta)] = 2theta = 2sin-1(x) in this range, giving dy/dx = 2/sqrt(1-x<sup>2</sup>).

f(x) = |x|: Continuous but NOT Differentiable at 0x=0 (sharp "kink")LHD = -1RHD = +1LHD ≠ RHD at the kink → not differentiable, even though the curve has no gap (continuous)

f(x)=|x| is perfectly continuous at x=0 (no break, no jump), but the curve has a sharp corner there: approaching from the left gives slope -1 and from the right gives slope +1 - since these one-sided derivatives disagree, the function is not differentiable at that point.

Concept context

When a function has no breaks (continuity) and when it has a well-defined slope (differentiability), plus rules for differentiating implicit, inverse trig, exponential, log, and parametric functions.

Read the full Continuity and Differentiability notes →