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📐 Mathematics  ·  Continuity and Differentiability  ·  JEE

Find dy/dx if y = (sin x)<sup>x.</sup>

Answer: y[x cot x + ln(sin x)].

  • A y[x cot x + ln(sin x)]
  • B y[x cot x]
  • C y[ln(sin x)]
  • D x(sin x)<sup>x-1</sup> cos x

Correct answer: A. y[x cot x + ln(sin x)]

Explanation: Take ln: ln y = x ln(sin x). Differentiating: (1/y)(dy/dx) = ln(sin x) + x cot x. So dy/dx = y[x cot x + ln(sin x)].

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 →