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📐 Mathematics  ·  Continuity and Differentiability  ·  JEE

If y = tan-1[(sqrt(1+x<sup>2</sup>) - 1)/x], find dy/dx.

Answer: 1/(2(1+x 2 )).

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

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

Explanation: Substituting x = tan(theta), the expression simplifies to tan-1[tan(theta/2)] = theta/2 = (1/2)tan-1(x), so dy/dx = (1/2) times 1/(1+x<sup>2</sup>) = 1/(2(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 →