Differentiability and Where Derivatives Fail

You’ll be able to

State the relationship between differentiability and continuity in the correct direction
Identify corners, cusps, vertical tangents, and discontinuities as failures of differentiability
Use one-sided derivatives to test differentiability of a piecewise function

One implication, and it runs one way only

The theorem is: differentiable at a ⟹ continuous at a. The converse is false, and the exam tests the converse being false far more often than it tests the theorem. So the two statements to have ready are: if a function is differentiable at a point it must be continuous there, and continuity at a point guarantees nothing about differentiability. The contrapositive is the version that does useful work on a multiple-choice question — if f is not continuous at a, then f is not differentiable at a — because a discontinuity is usually easier to spot than a failure of the derivative limit.

Four ways differentiability fails

Know all four by sight, because the exam shows graphs. A corner: the one-sided derivatives are finite but unequal, the standard case being |x| at x = 0, where the slope is −1 on the left and +1 on the right. A cusp: the one-sided slopes go to +∞ and −∞, as for x^(2/3) at x = 0. A vertical tangent: the slope grows without bound with the same sign from both sides, as for x^(1/3) at x = 0 — the function is continuous and even has a tangent line, but that line is vertical so the derivative does not exist as a number. A discontinuity of any kind: jump, hole, or asymptote. Note that the first three are all continuous, which is the point — continuity is not enough.

Testing a piecewise function at the seam

For a piecewise function, differentiability at the breakpoint x = a requires two checks in order, and skipping the first is the classic error. First, continuity: the two branches must produce the same value at a, so set the left expression equal to the right expression at a. Second, matching slopes: differentiate each branch and set the two derivatives equal at a. When a problem gives two unknown constants, this is exactly why — one equation from continuity, one from the derivatives, two unknowns. Doing only the derivative check can produce an answer for a function that is not even continuous, which cannot be differentiable no matter what the slopes do.

One-sided derivatives

f′(a) exists ⟺ lim(h→0⁻) [f(a+h) − f(a)]/h = lim(h→0⁺) [f(a+h) − f(a)]/h, both finite

Both one-sided difference quotients must exist, be finite, and agree. A corner fails the agreement; a cusp or vertical tangent fails finiteness.

Worked example

Find a and b so that f is differentiable everywhere, where f(x) = x² for x ≤ 2 and f(x) = ax + b for x > 2.

1.Continuity at x = 2 requires the two branches to agree there: 2² = a(2) + b, so 4 = 2a + b.
2.Differentiate each branch: for x < 2, f′(x) = 2x; for x > 2, f′(x) = a.
3.Matching slopes at x = 2 requires lim from the left of f′ to equal the value from the right: 2(2) = a, so a = 4.
4.Substitute a = 4 into the continuity equation: 4 = 2(4) + b = 8 + b, so b = −4.
5.Check: the left branch gives f(2) = 4 and the right gives 4(2) − 4 = 4, so it is continuous; slopes are 4 and 4, so it is differentiable.

Answer: a = 4 and b = −4. The line y = 4x − 4 is precisely the tangent to y = x² at x = 2, which is what smooth joining means.

Tip

The answer to that example is worth noticing: the linear branch turned out to be the tangent line to the parabola at the seam. That is not a coincidence — joining a curve to a line smoothly is the same problem as finding the tangent. If you get a different line, you have made an arithmetic error.

Checkpoint

Which statement is true?

Checkpoint

The function f(x) = x^(1/3) fails to be differentiable at x = 0 because

Answer the 2 checkpoints as you read.