Calculus I · The Slope of a Single Point

Where this is heading

You now have the engine of all of calculus — but computing each derivative from the definition is laborious, and you may already be tired of expanding $(a+h)^n$ and rationalizing roots. Chapter 4 fixes that: it proves a handful of rules (power, product, quotient, chain) — each derived once from the very definition you just learned — so that you can differentiate almost anything in seconds, never touching a difference quotient again. (In the Diamond problems below you’ll even derive the power rule $\tfrac{d}{dx}x^n = n x^{n-1}$ for yourself, straight from the binomial theorem.) The same definition powers everything after: Chapter 5 computes the derivative of $e^x$ from first principles; Chapter 6’s Fermat’s lemma is built from the one-sided difference quotients you just met at the corner of $|x|$; Chapter 8 finds every peak and valley by setting the derivative to zero, exactly as the thrown ball went still at the top of its arc.

One limit — the slope of a single point — and the rest of the book is its consequences.

Problems

Six tiers. Fluency and Practice are doing* — compute derivatives from the definition until the three-step dance is automatic. Concept & counterexample and ◆ Diamond are thinking — where derivatives fail, and the proofs. Teach it is teaching. Looking ahead opens Chapter 4. Do all of Fluency; the Concept tier is where the continuity/differentiability distinction sets. (Every derivative here is computed from the definition — you have no rules yet. That’s the point.)*

Fluency — do these from the definition, fast

1. Using the difference quotient, find the derivative of: (a) $f(x) = x^2$; (b) $f(x) = 3x^2$; (c) $f(x) = x^2 + 5$.
2. From the definition, find the derivative of: (a) $f(x) = 5$ (a constant); (b) $f(x) = 7x - 2$ (a line); (c) $f(x) = -x$.
3. From the definition, find the derivative of: (a) $f(x) = x^3$; (b) $f(x) = x^4$. (Expand $(a+h)^n$.)
4. From the definition, find the derivative of $f(x) = 3x^2 - 4x + 1$, then evaluate it at $x = 2$.
5. From the definition, find the derivative of: (a) $f(x) = \tfrac1x$; (b) $f(x) = \tfrac{1}{x^2}$. (Combine over a common denominator.)
6. From the definition, find the derivative of $f(x) = \sqrt{x}$, then its value at $x = 9$. (Rationalize.)
7. For $f(x) = x^2$, find the slope at $x = 0$, $x = 1$, and $x = -2$. Where is the slope zero?
8. Find the equation of the tangent line to $y = x^2$ at: (a) $x = 1$; (b) $x = 3$.
9. A ball’s height is $s(t) = -5t^2 + 30t$ meters. From the definition, find its velocity $v(t)$, then $v(0)$ and the time when $v = 0$.

Practice — harder from-scratch, tangents, and motion

1. From the definition, find the derivative of $f(x) = \tfrac{1}{x+1}$.
2. From the definition, find the derivative of $f(x) = \sqrt{x + 3}$.
3. From the definition, find the derivative of $f(x) = x^2 + \tfrac1x$. (Handle the two pieces in one difference quotient.)
4. Find the tangent line to $y = \tfrac1x$ at $x = 2$, and the tangent line to $y = \sqrt{x}$ at $x = 4$.
5. For $f(x) = x^3 - 3x$, find $f'(x)$ from the definition, then find where the tangent line is horizontal (where $f'(x) = 0$). (You’re previewing how Chapter 8 finds peaks and valleys.)
6. A particle’s position is $s(t) = t^2 - 6t + 5$. Find $v(t)$ from the definition, the time the particle is momentarily at rest, and whether it’s moving left or right at $t = 1$.
7. Find the point on $y = x^2$ where the tangent line has slope $10$. (Set the derivative equal to $10$ and solve.)
8. The derivative of $\sqrt{x}$ is $\tfrac{1}{2\sqrt{x}}$. Use the tangent line at $x = 100$ to estimate $\sqrt{101}$ without a calculator. (This is the “best linear approximation” idea at work.)

Concept & counterexample — think; true/false with justification, or build the example

1. 1. The derivative at a point is a number, but the derivative is also a function. (b) Every continuous function is differentiable. (c) If $f'(a) = 0$, the graph has a horizontal tangent at $a$. (d) A function can be differentiable at a point where it is not continuous.
2. Find the error. A student computes the derivative of $x^2$ as: “$\tfrac{(a+h)^2 - a^2}{h} = \tfrac{2ah + h^2}{h}$, and setting $h = 0$ gives $\tfrac{0}{0}$, so the derivative doesn’t exist.” What did they do wrong?
3. Construct a function that is continuous everywhere but not differentiable at $x = 0$. Explain, using one-sided difference quotients, why the derivative fails to exist there.
4. The function $|x|$ is continuous at $0$ but not differentiable there. Is the reverse possible — a function differentiable at $0$ but not continuous there? Justify with the theorem from the chapter.
5. Is it possible? A function whose derivative is $0$ at every point. What must such a function look like, and why? (You’ll prove this rigorously in Chapter 6 — guess the answer now.)
6. Sketch (or describe) the graph of $f'$ for a function $f$ that increases, levels off at a peak, then decreases. What is the sign of $f'$ in each region, and what is $f'$ at the peak?

◆ Diamond — the proofs, and the rule you’ll meet next chapter

1. ◆ Prove that differentiable implies continuous: if $f'(a)$ exists, then $f$ is continuous at $a$. (Write $f(a+h) - f(a)$ as the difference quotient times $h$, and take $h \to 0$.)
2. ◆ Prove that $|x|$ is not differentiable at $0$ by computing the left- and right-hand limits of the difference quotient and showing they disagree.
3. ◆◆ Derive the power rule. Using the binomial theorem $(x+h)^n = x^n + n x^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \cdots$, show from the definition that $\tfrac{d}{dx}x^n = n x^{n-1}$ for every positive integer $n$. (Every term after the first carries an $h$ and dies in the limit — the same “everything but one term survives” pattern as Chapter 1’s $n^{1/n}$ and Chapter 7’s Taylor coefficients.)
4. ◆◆ A function with a vertical tangent. Show that $f(x) = x^{1/3}$ has difference quotient at $0$ equal to $\tfrac{h^{1/3}}{h} = h^{-2/3}$, and that this $\to +\infty$ as $h \to 0$. Conclude the derivative doesn’t exist at $0$ — but for a different reason than $|x|$. (One has a corner; this has a vertical tangent. Describe the difference geometrically.)
5. ◆◆ One-sided derivatives. Define the right-hand derivative as $\lim_{h\to0^+}\tfrac{f(a+h)-f(a)}{h}$ and the left-hand as the same with $h\to0^-$. Show $f$ is differentiable at $a$ exactly when both exist and are equal, and use this to re-explain the $|x|$ corner. (These one-sided difference quotients are precisely the tool behind Chapter 6’s Fermat’s lemma.)

Teach it — explain it, write it up, or fix it

1. In three or four sentences with the secant-to-tangent picture, teach a friend what a derivative is and how shrinking $h$ turns an average rate into an instantaneous one.
2. Debug your classmate. They write: “The derivative of $x^2$ is $\tfrac{(x+h)^2 - x^2}{h}$.” Explain what they’ve left out, and write the correct two final steps.
3. Write a complete worked solution finding the derivative of $\tfrac1x$ from the definition, narrating why you combine fractions before letting $h \to 0$, so a peer could redo it for $\tfrac{1}{x+2}$.
4. Design a problem. Invent a function whose derivative-from-the-definition gives a clean answer, and write the full solution key showing all three steps of the dance.
5. In a few sentences, teach the difference between continuous and differentiable, using $|x|$ at $0$ as the example that separates them.

Looking ahead — Chapter 4 in your hands early

1. You’ve now computed $\tfrac{d}{dx}x^2 = 2x$, $\tfrac{d}{dx}x^3 = 3x^2$, $\tfrac{d}{dx}x^4 = 4x^3$. Predict $\tfrac{d}{dx}x^5$ and $\tfrac{d}{dx}x^{10}$ from the pattern — then notice that doing each from the definition would be miserable, which is exactly why Chapter 4 proves the rule once and for all.
2. You found $\tfrac{d}{dx}(3x^2 - 4x + 1) = 6x - 4$ — the derivative of a sum, term by term. Predict the rule: how does the derivative of $f(x) + g(x)$ relate to $f'(x)$ and $g'(x)$?
3. Chapter 5 will compute $\tfrac{d}{dx}e^x$ from this same definition — and the answer will be $e^x$ itself. Using the definition, write down the limit you’d have to evaluate, and identify the one piece of it ($\tfrac{e^h - 1}{h}$) whose value will decide the whole thing.

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