Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist AB & BC
If $f'(a)$ exists then $f$ is continuous at $a$, because $f(x) - f(a)$ can be written as the difference quotient times $(x - a)$, and that product tends to $f'(a) \cdot 0 = 0$. The contrapositive is the working tool: a discontinuity at $a$ rules out $f'(a)$ immediately. The converse fails, and $|x|$ at $0$ is the standard counterexample: continuous there, but with one-sided difference quotients of $-1$ and $1$.
Two errors dominate. The first is running the implication backwards, so an unbroken graph is assumed to be differentiable everywhere. The second is misnaming the failure: a cusp called a vertical tangent or the reverse, a jump overlooked entirely, or an infinite slope reported as a value of the derivative rather than as the derivative failing to exist. For a piecewise rule both hurdles have to be cleared: the pieces must agree in value at the junction, and then in slope.
The work
3 ways in · any order
Lesson
Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist
›
Proves that differentiability forces continuity, kills the converse with the absolute value function, drills telling a cusp from a vertical tangent, and works the two conditions a piecewise rule must clear at its junction.
Diagnostic
10-item topic check
›
Ten items spanning the two failure modes of this topic: assuming an unbroken graph must be differentiable, and misnaming or missing the features, corners, cusps, vertical tangents and discontinuities, where a derivative does not exist.
Targeted Practice
Drill a single misconception
›
Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears the misconception and moves you to the next.