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(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 ddxxn=nxn−1\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 exe^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||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)=x2f(x) = x^2; (b) f(x)=3x2f(x) = 3x^2; (c) f(x)=x2+5f(x) = x^2 + 5.
  2. From the definition, find the derivative of: (a) f(x)=5f(x) = 5 (a constant); (b) f(x)=7x−2f(x) = 7x - 2 (a line); (c) f(x)=−xf(x) = -x.
  3. From the definition, find the derivative of: (a) f(x)=x3f(x) = x^3; (b) f(x)=x4f(x) = x^4. (Expand (a+h)n(a+h)^n.)
  4. From the definition, find the derivative of f(x)=3x2−4x+1f(x) = 3x^2 - 4x + 1, then evaluate it at x=2x = 2.
  5. From the definition, find the derivative of: (a) f(x)=1xf(x) = \tfrac1x; (b) f(x)=1x2f(x) = \tfrac{1}{x^2}. (Combine over a common denominator.)
  6. From the definition, find the derivative of f(x)=xf(x) = \sqrt{x}, then its value at x=9x = 9. (Rationalize.)
  7. For f(x)=x2f(x) = x^2, find the slope at x=0x = 0, x=1x = 1, and x=−2x = -2. Where is the slope zero?
  8. Find the equation of the tangent line to y=x2y = x^2 at: (a) x=1x = 1; (b) x=3x = 3.
  9. A ball’s height is s(t)=−5t2+30ts(t) = -5t^2 + 30t meters. From the definition, find its velocity v(t)v(t), then v(0)v(0) and the time when v=0v = 0.

Practice — harder from-scratch, tangents, and motion

  1. From the definition, find the derivative of f(x)=1x+1f(x) = \tfrac{1}{x+1}.
  2. From the definition, find the derivative of f(x)=x+3f(x) = \sqrt{x + 3}.
  3. From the definition, find the derivative of f(x)=x2+1xf(x) = x^2 + \tfrac1x. (Handle the two pieces in one difference quotient.)
  4. Find the tangent line to y=1xy = \tfrac1x at x=2x = 2, and the tangent line to y=xy = \sqrt{x} at x=4x = 4.
  5. For f(x)=x3−3xf(x) = x^3 - 3x, find f′(x)f'(x) from the definition, then find where the tangent line is horizontal (where f′(x)=0f'(x) = 0). (You’re previewing how Chapter 8 finds peaks and valleys.)
  6. A particle’s position is s(t)=t2−6t+5s(t) = t^2 - 6t + 5. Find v(t)v(t) from the definition, the time the particle is momentarily at rest, and whether it’s moving left or right at t=1t = 1.
  7. Find the point on y=x2y = x^2 where the tangent line has slope 1010. (Set the derivative equal to 1010 and solve.)
  8. The derivative of x\sqrt{x} is 12x\tfrac{1}{2\sqrt{x}}. Use the tangent line at x=100x = 100 to estimate 101\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. 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)=0f'(a) = 0, the graph has a horizontal tangent at aa. (d) A function can be differentiable at a point where it is not continuous.
  1. Find the error. A student computes the derivative of x2x^2 as: “(a+h)2−a2h=2ah+h2h\tfrac{(a+h)^2 - a^2}{h} = \tfrac{2ah + h^2}{h}, and setting h=0h = 0 gives 00\tfrac{0}{0}, so the derivative doesn’t exist.” What did they do wrong?
  2. Construct a function that is continuous everywhere but not differentiable at x=0x = 0. Explain, using one-sided difference quotients, why the derivative fails to exist there.
  3. The function |x||x| is continuous at 00 but not differentiable there. Is the reverse possible — a function differentiable at 00 but not continuous there? Justify with the theorem from the chapter.
  4. Is it possible? A function whose derivative is 00 at every point. What must such a function look like, and why? (You’ll prove this rigorously in Chapter 6 — guess the answer now.)
  5. Sketch (or describe) the graph of f′f' for a function ff that increases, levels off at a peak, then decreases. What is the sign of f′f' in each region, and what is f′f' at the peak?

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

  1. ◆ Prove that differentiable implies continuous: if f′(a)f'(a) exists, then ff is continuous at aa. (Write f(a+h)−f(a)f(a+h) - f(a) as the difference quotient times hh, and take h→0h \to 0.)
  2. ◆ Prove that |x||x| is not differentiable at 00 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=xn+nxn−1h+(n2)xn−2h2+⋯(x+h)^n = x^n + n x^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \cdots, show from the definition that ddxxn=nxn−1\tfrac{d}{dx}x^n = n x^{n-1} for every positive integer nn. (Every term after the first carries an hh and dies in the limit — the same “everything but one term survives” pattern as Chapter 1’s n1/nn^{1/n} and Chapter 7’s Taylor coefficients.)
  4. ◆◆ A function with a vertical tangent. Show that f(x)=x1/3f(x) = x^{1/3} has difference quotient at 00 equal to h1/3h=h−2/3\tfrac{h^{1/3}}{h} = h^{-2/3}, and that this →+∞\to +\infty as h→0h \to 0. Conclude the derivative doesn’t exist at 00 — but for a different reason than |x||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→0+f(a+h)−f(a)h\lim_{h\to0^+}\tfrac{f(a+h)-f(a)}{h} and the left-hand as the same with h→0−h\to0^-. Show ff is differentiable at aa exactly when both exist and are equal, and use this to re-explain the |x||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 hh turns an average rate into an instantaneous one.
  2. Debug your classmate. They write: “The derivative of x2x^2 is (x+h)2−x2h\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 1x\tfrac1x from the definition, narrating why you combine fractions before letting h→0h \to 0, so a peer could redo it for 1x+2\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||x| at 00 as the example that separates them.

Looking ahead — Chapter 4 in your hands early

  1. You’ve now computed ddxx2=2x\tfrac{d}{dx}x^2 = 2x, ddxx3=3x2\tfrac{d}{dx}x^3 = 3x^2, ddxx4=4x3\tfrac{d}{dx}x^4 = 4x^3. Predict ddxx5\tfrac{d}{dx}x^5 and ddxx10\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 ddx(3x2−4x+1)=6x−4\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)f(x) + g(x) relate to f′(x)f'(x) and g′(x)g'(x)?
  3. Chapter 5 will compute ddxex\tfrac{d}{dx}e^x from this same definition — and the answer will be exe^x itself. Using the definition, write down the limit you’d have to evaluate, and identify the one piece of it (eh−1h\tfrac{e^h - 1}{h}) whose value will decide the whole thing.

Questions to talk about

No answer key. These are for arguing over, with someone who read the same chapter.

  1. A rate seems to need two moments, yet a speedometer shows a speed at one instant. State the paradox in your own words before reading how the limit resolves it.
  2. Every derivative here is earned from the definition rather than handed over. The chapter admits this is a terrible way to use derivatives and the right way to learn them. When is the slow way worth it?
  3. The derivative turns out to be a function itself — a machine returning the slope at every point at once. Why is that a bigger idea than finding the slope at one point?

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