Answer: 1/sqrt(1-x 2 ).
- A 1/sqrt(1-x<sup>2</sup>)
- B -1/sqrt(1-x<sup>2</sup>)
- C 1/(1+x<sup>2</sup>)
- D 1/sqrt(x<sup>2</sup>-1)
Correct answer: A. 1/sqrt(1-x<sup>2</sup>)
Explanation: The derivative of sin inverse x is 1/sqrt(1-x<sup>2</sup>), valid for -1 < x < 1.
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.