Differentiability is a stronger condition than continuity: a differentiable function is continuous, but a continuous function can have corners, as in Calculus 1. At a point where two formulas meet, the definition splits into one-sided limits.
🔑 Key idea
Differentiable at $a$ implies continuous at $a$. At a break point, $f'(a)$ exists exactly when both one-sided limits of the difference quotient exist and agree. For pieces that meet at $c$, that means: the pieces agree at $c$ (continuity) and their slopes agree at $c$.
💡 Why it works
For $x \ne a$, $$f(x) - f(a) = \frac{f(x) - f(a)}{x - a}\cdot(x - a).$$ As $x \to a$ the first factor tends to $f'(a)$ and the second to $0$, so by the product law $f(x) - f(a) \to f'(a) \cdot 0 = 0$: $f$ is continuous at $a$. $\blacksquare$ The converse fails: $|x|$ is continuous at $0$, but its quotient $\frac{|h|}{h}$ is $1$ for $h > 0$ and $-1$ for $h < 0$, so the one-sided limits disagree.
Gluing $x^{2}$ (for $x \le 1$) to a line: adjust $a$ and $b$ until the pieces meet and have the same slope. The answer is the tangent line $y = 2x - 1$.
A useful way to read differentiability: near $a$, the function is a line plus something much smaller than the distance to $a$. Precisely, $f(a + h) = f(a) + f'(a)h + r(h)$ with $\frac{r(h)}{h} \to 0$. A corner breaks this, because no single line fits the graph from both sides; a jump breaks it more badly, because no line through $(a, f(a))$ even comes close on one side. That is why every differentiable function is continuous, and why the converse must fail.
Worked examples
✏️ Worked example: Making a function differentiable
Find $a$ and $b$ so that $f(x) = x^{2}$ for $x \le 2$ and $f(x) = ax + b$ for $x > 2$ is differentiable at $2$.
The pieces must meet: $4 = 2a + b$.
The slopes must agree: the left derivative is $2x = 4$ at $x = 2$, the right one is $a$. So $a = 4$.
Then $b = 4 - 8 = -4$.
Answer: $a = 4$, $b = -4$: the line is the tangent line $y = 4x - 4$.
✏️ Worked example: A corner, computed
For $f(x) = |x - 1|(x + 2)$, find the one-sided derivatives at $x = 1$.
For $h > 0$ this tends to $3$; for $h < 0$ to $-3$.
Answer: Right: $3$; left: $-3$. Continuous, not differentiable.
⚠️ Common mistake
Matching only one condition. For glued pieces, students often match the slopes and forget that the pieces must meet (then $f$ is not even continuous, so not differentiable), or check that the pieces meet and stop (then there may be a corner). Both equations are needed. And remember the logic: differentiable implies continuous; continuous does not imply differentiable.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.