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Code Scholars Tutoring · AP Calculus BC

Where derivatives fail to exist

Unit 2 · Topics 2.4 · Detailed notes, a worked example, and a self-check with reasoning.

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Understand the idea

Differentiability at a point implies continuity there; the converse fails. Corners have different one-sided slopes, cusps may have unbounded one-sided slopes, and vertical tangents have no finite derivative. Discontinuities automatically prevent differentiability. For a piecewise function first enforce continuity, then compare the limiting derivative from each side.

f differentiable at a ⟹ f continuous at af\text{ differentiable at }a\ \Longrightarrow\ f\text{ continuous at }a

Worked example

Explain why f(x)=∣x∣\displaystyle f(x)=|x| is continuous but not differentiable at zero.

Both one-sided function limits equal f(0)=0\displaystyle f(0)=0, so it is continuous. The left-hand slope is −1\displaystyle -1 and right-hand slope is 1\displaystyle 1. Because they differ, the derivative limit does not exist.

Try it yourself

Is f(x)=x1/3\displaystyle f(x)=x^{1/3} differentiable at 0\displaystyle 0?

Show solution and reasoning

No finite derivative exists. Its difference quotient is h−2/3\displaystyle h^{-2/3} for h≠0\displaystyle h\ne0, which tends to +∞\displaystyle +\infty. The graph has a vertical tangent.

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