One implication, and its false converse AB & BC

If $f'(a)$ exists, then $f$ is continuous at $a$. The argument is two lines and worth following, because it shows exactly what the derivative buys you:

$$\lim_{x \to a}\left[f(x) - f(a)\right] = \lim_{x \to a}\left[\frac{f(x) - f(a)}{x - a} \cdot (x - a)\right] = f'(a) \cdot 0 = 0.$$

The multiplication by $(x - a)$ is legal because $x \neq a$ inside a limit. If the difference of the outputs tends to zero, then $\lim_{x \to a} f(x) = f(a)$, which is continuity at $a$.

The useful form of this in practice is the contrapositive: if $f$ is not continuous at $a$, then $f'(a)$ does not exist. That single sentence disposes of every discontinuity without any further work.

Continuity does not give you differentiability. Take $f(x) = |x|$ at $a = 0$. It is continuous there: the left and right limits are both 0, and $f(0) = 0$. But the difference quotient has two different one-sided limits.

$$\frac{|0 + h| - |0|}{h} = \frac{|h|}{h} = \begin{cases} 1 & h > 0 \\ -1 & h < 0\end{cases}$$

A two-sided limit requires the two sides to agree, and $1 \neq -1$, so $f'(0)$ does not exist. The graph has no break at the origin; what it has is a sudden change of direction. Topic 1.11 flagged this in passing, and this is where it is cashed in.

So differentiability is the stronger condition. Every differentiable function is continuous, and continuous functions can still fail to be differentiable at isolated points, or, in exotic cases outside this course, at every point at once.

On the AP exam the failures come in exactly four flavours, and naming the right one is part of the answer.

1. Corner. Both one-sided slopes are finite but unequal. $|x|$ at $0$, where they are $-1$ and $1$.
2. Cusp. The one-sided slopes run off to infinity in opposite directions. $x^{2/3}$ at $0$, where the left slopes go to $-\infty$ and the right slopes to $+\infty$; the two branches meet in a point.
3. Vertical tangent. Both one-sided slopes run off to infinity in the same direction. $x^{1/3}$ at $0$. The tangent line exists and is vertical; a vertical line has no slope, so $f'(0)$ does not exist.
4. Discontinuity. Any break at all, whether a hole, a jump, or an asymptote. Ruled out by the contrapositive above.

The distinction that gets missed most often is cusp against vertical tangent. Both send the slopes to infinity; only the cusp sends them opposite ways. Sketching the two one-sided behaviours separately settles it every time.

In all four cases the correct wording is that the derivative does not exist. Writing $f'(0) = \infty$ claims a value where there is none.

A piecewise definition that changes rule at $x = c$ is differentiable there only if it clears two hurdles, and they must be checked in this order:

1. Continuity. The two pieces must agree in value at $c$. Without this, nothing else matters.
2. Matching slopes. The derivatives of the two pieces must also agree at $c$.

Worked. Let $f(x) = x^2$ for $x \leq 1$ and $f(x) = mx + b$ for $x > 1$. Continuity at 1 requires $m + b = 1$. Matching slopes requires $m = 2$, since the left piece has slope $2x = 2$ there. Together, $m = 2$ and $b = -1$.

Checking only the slopes would accept $m = 2$ with any $b$, producing a function with the right steepness on both sides of a jump. Checking only the values would accept a corner. Both hurdles are needed, and continuity comes first because differentiability implies it.