Differentiability and Where Derivatives Fail

You’ll be able to

State the four ways a function can fail to be differentiable at a point
Explain why differentiability implies continuity and why the converse is false
Use one-sided derivatives to decide whether a piecewise function is differentiable at its seam

The derivative is a limit, so it can fail to exist

f′(a) is defined as lim(h→0) [f(a + h) − f(a)]/h. Like any limit, it exists only when the one-sided versions agree and are finite. That single sentence generates the whole list of failures: if the two one-sided difference quotients approach different numbers, the derivative does not exist; if they both blow up, it does not exist either. Nothing about "the graph looks smooth" is part of the definition — the one-sided limits are.

The four failure modes

A corner has one-sided derivatives that are finite but unequal (|x| at 0: −1 on the left, +1 on the right). A cusp has one-sided derivatives running to +∞ and −∞ (x^(2/3) at 0). A vertical tangent has both one-sided derivatives running to the same infinity (x^(1/3) at 0) — the tangent line exists, but its slope is not a number. A discontinuity of any kind kills the derivative outright, because the numerator f(a + h) − f(a) does not shrink to zero.

Differentiable ⇒ continuous, and never the reverse

If f′(a) exists then f is continuous at a: the difference quotient can only settle on a finite number if the numerator goes to 0 as the denominator does. Reversing this is the single most common false statement in the course. |x| is continuous everywhere and differentiable nowhere at 0. Continuity is a necessary condition for differentiability, not a sufficient one — so on an exam, "f is continuous at x = 2" never justifies "f′(2) exists," while "f′(2) exists" always justifies "f is continuous at 2."

Differentiability at a seam

f differentiable at c ⟺ f continuous at c AND lim(x→c⁻) f′(x) = lim(x→c⁺) f′(x)

Both conditions, in that order. Matching slopes alone is not enough — a function can have equal one-sided slopes across a jump.

Worked example

Let f(x) = x² for x ≤ 2 and f(x) = mx + b for x > 2. Find m and b so that f is differentiable everywhere.

1.Differentiability at x = 2 requires continuity there first: the branches must agree, so 2² = 2m + b, i.e. 4 = 2m + b.
2.Now match the derivatives. On the left f′(x) = 2x, so the left-hand slope at x = 2 is 4. On the right f′(x) = m.
3.Setting the one-sided derivatives equal gives m = 4.
4.Substitute back into the continuity equation: 4 = 2(4) + b, so b = −4.
5.Check: the right branch is 4x − 4, which at x = 2 gives 4 — matching the left branch, with slope 4 on both sides.

Answer: m = 4 and b = −4. Both conditions were needed; solving only the slope equation would leave b undetermined.

Watch out

Doing the slope condition first and the continuity condition never is the classic error on this problem type. Two branches can have identical slopes at the seam and still be separated by a jump — parallel, not joined. Continuity is what joins them.

Checkpoint

The graph of f has a sharp corner at x = 1 but no break there. Which statement is true?

Checkpoint

At x = 0, the function g(x) = x^(1/3) has lim(x→0) g′(x) = +∞ from both sides. What does the graph do there?

On the exam

When a free-response question hands you a piecewise function with unknown constants and says "differentiable," it is asking for two equations, not one. Write the continuity equation and the slope equation, then solve the system.

Answer the 2 checkpoints as you read.