Answer: e 3x (3cos(2x) - 2sin(2x)).
- A e<sup>3x</sup>(3cos(2x) - 2sin(2x))
- B e<sup>3x</sup>(3cos(2x) + 2sin(2x))
- C 3e<sup>3x</sup> cos(2x)
- D e<sup>3x</sup>(2cos(2x) - 3sin(2x))
Correct answer: A. e<sup>3x</sup>(3cos(2x) - 2sin(2x))
Explanation: Product rule: d/dx[e<sup>3x</sup>]cos(2x) + e<sup>3x</sup> d/dx[cos(2x)] = 3e<sup>3x</sup>cos(2x) + e<sup>3x</sup>(-2sin(2x)) = e<sup>3x</sup>(3cos(2x) - 2sin(2x)).
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.