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📐 Mathematics  ·  Continuity and Differentiability  ·  JEE

Differentiate y = sin(x<sup>2</sup> + 1) using the chain rule.

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

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

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

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