Calculus I · The Slope of a Single Point

The tangent line: a function’s best straight-line stand-in

The derivative hands you the slope of the tangent at $a$, and with a slope and a point you can write the tangent line:

$$\boxed{\;y = f(a) + f'(a)(x - a)\;}$$

Example. The tangent to $y = x^2$ at $x = 1$: here $f(1) = 1$ and $f'(1) = 2(1) = 2$, so the line is $y = 1 + 2(x - 1) = 2x - 1$. Near $x = 1$, the curve $x^2$ and the line $2x - 1$ are almost indistinguishable.

That “almost indistinguishable near $a$” is a profound idea in disguise, and we’ll name it fully in Chapter 7: the tangent line is the best straight-line approximation of a curve near a point. Zoom in on any smooth curve far enough and it looks straight — and the straight line it looks like is the tangent, whose slope is the derivative. To know a function’s derivative at $a$ is to know the simplest possible model of it right there.

◆ Layer 2 — When a function has no slope, and what that has to do with Chapter 2

The derivative is a limit, and limits don’t always exist — so neither do derivatives. The cleanest example is one you already know the shape of: $f(x) = |x|$ at $x = 0$. It’s a perfectly continuous function (you can draw it without lifting your pen), but it has a sharp corner at the origin, and a corner has no single slope. Watch the difference quotient confirm it. At $a = 0$, the quotient is $\dfrac{|0 + h| - 0}{h} = \dfrac{|h|}{h}$, which is the sign function from Chapter 2: - from the right ($h > 0$): $\tfrac{|h|}{h} = +1$; - from the left ($h < 0$): $\tfrac{|h|}{h} = -1$.

The two one-sided limits disagree, so the two-sided limit — the derivative — does not exist. This is exactly the $\tfrac{|x|}{x}$ jump from Chapter 2, reappearing as the slope of $|x|$. The corner of one graph is the jump of its derivative.

So differentiability is genuinely stronger than continuity — every differentiable function is continuous (proved below), but not every continuous function is differentiable. Corners (like $|x|$), cusps, and vertical tangents are all continuous-but-not-differentiable. This is the third floor of the “smoothness has consequences” story from Chapters 2 and 6, seen from below: differentiability asks for more than continuity, and a function can pass the first test and fail the second.

◆ Layer 2 — Proof that differentiable $\Rightarrow$ continuous

Suppose $f'(a)$ exists. We must show $f$ is continuous at $a$, i.e. $f(a+h) \to f(a)$ as $h \to 0$. The trick is to write the change in $f$ as the difference quotient times $h$: $$f(a+h) - f(a) = \underbrace{\frac{f(a+h) - f(a)}{h}}_{\to\, f'(a)} \cdot \; h.$$ As $h \to 0$, the first factor approaches the finite number $f'(a)$ and the second approaches $0$, so the product approaches $f'(a)\cdot 0 = 0$. Therefore $f(a+h) \to f(a)$: $f$ is continuous at $a$. $\blacksquare$ (The converse fails — $|x|$ is the counterexample — so “differentiable” sits strictly above “continuous.”)

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