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📐 Mathematics  ·  Continuity and Differentiability  ·  JEE

If y = log(x + sqrt(x<sup>2</sup>+1)), find dy/dx.

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

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

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

Explanation: Let u = x + sqrt(x<sup>2</sup>+1). du/dx = 1 + x/sqrt(x<sup>2</sup>+1) = (sqrt(x<sup>2</sup>+1)+x)/sqrt(x<sup>2</sup>+1) = u/sqrt(x<sup>2</sup>+1). dy/dx = (1/u)(du/dx) = 1/sqrt(x<sup>2</sup>+1).

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 →