A function is differentiable at a point if its derivative exists there — meaning the curve is smooth, no corner or break.
Quick check
Which statement is true?
Where differentiability fails
- Corners — like |x| at 0; the slope jumps.
- Cusps — like x^(2/3) at 0; the slope shoots to ±∞.
- Vertical tangents — like ∛x at 0.
- Discontinuities — any break automatically kills differentiability.
'Smooth' is the intuition: a curve is differentiable at a point if you could lay a single well-defined tangent line there. Corners and breaks ruin that.
Your turn
Is f(x) = |x − 2| differentiable at x = 2?
Differentiable ⇒ continuous, but continuous doesn't imply differentiable. |x| is continuous at 0 but has a corner.