Calculus I · A Toolkit for Slopes

Where this is heading

You now hold the complete differentiation toolkit, and the rest of the book simply uses it. Chapter 5 turns the inverse-function rule on $\ln x$ and the exponential on $e^x$ to build the whole theory of growth and decay. Chapter 6 (the Mean Value Theorem) and everything after assume you can differentiate anything — and now you can. Chapter 7’s Taylor series demanded the trig and exponential derivatives you just proved. Chapter 8 finds maxima and minima by setting derivatives to zero, which means differentiating first.

And one quiet thing worth noticing: every rule in this chapter takes a function and returns its derivative as a new function — which you can feed right back in. Differentiate $x^4$ to get $4x^3$, differentiate again to get $12x^2$: that’s the second derivative, the curvature that runs Chapters 7 and 8. The toolkit doesn’t just compute slopes; it lets you climb the whole tower of derivatives, as high as you need to go.

Problems

Six tiers. Fluency and Practice are doing* — apply the rules until choosing the right one is automatic. Concept & counterexample and ◆ Diamond are thinking — the rule-traps and the proofs (each rule derived from the definition). Teach it is teaching. Looking ahead opens Chapter 5. Do all of Fluency — and there’s a lot of it, more than any other chapter, on purpose: differentiation is a mechanical skill, and the only way to own it is volume. These rules should become as automatic as multiplication.*

Fluency — differentiate, fast; this tier is the whole point

1. Power rule, whole-number powers: (a) $x^9$; (b) $4x^6$; (c) $x^{100}$; (d) $7$; (e) $x$.
2. Power rule, negative and fractional powers: (a) $x^{-3}$; (b) $\tfrac{1}{x^5}$; (c) $x^{1/3}$; (d) $\sqrt[4]{x}$; (e) $x^{-1/2}$; (f) $\tfrac{1}{\sqrt{x}}$.
3. Rewrite first, then power rule: (a) $\tfrac{1}{x^2} + \sqrt{x}$; (b) $\tfrac{x^3 - 1}{x}$ (split the fraction); (c) $x^2\sqrt{x}$ (combine into one power first).
4. Polynomials, term by term: (a) $5x^3 - 2x^2 + 7x - 4$; (b) $\tfrac13 x^6 + 4x$; (c) $9x^5 - x^4 + 2x - 11$; (d) $(2x-1)(x+3)$ (expand first).
5. Product rule: (a) $x^2 e^x$; (b) $x\sin x$; (c) $(x^2+1)(x^3-2)$; (d) $x^3\cos x$; (e) $e^x\cos x$.
6. Product rule, then evaluate at $x = 0$: (a) $(x+1)e^x$; (b) $x^2\sin x$.
7. Quotient rule: (a) $\tfrac{x}{x^2+1}$; (b) $\tfrac{x^2}{x-1}$; (c) $\tfrac{\sin x}{x}$; (d) $\tfrac{e^x}{x}$; (e) $\tfrac{x-1}{x+1}$.
8. Quotient rule with trig: (a) $\tan x = \tfrac{\sin x}{\cos x}$; (b) $\tfrac{\cos x}{x}$; (c) $\tfrac{1}{\sin x}$.
9. Chain rule, powers: (a) $(x^2+1)^5$; (b) $(1-x^2)^{10}$; (c) $(3x+2)^7$; (d) $(x^3 - x)^4$.
10. Chain rule, roots: (a) $\sqrt{3x+2}$; (b) $\sqrt{x^2+1}$; (c) $\sqrt{1-x^2}$; (d) $\tfrac{1}{\sqrt{x^2+4}}$.
11. Chain rule, exponentials: (a) $e^{4x}$; (b) $e^{-x}$; (c) $e^{x^2}$; (d) $e^{-x^2/2}$; (e) $e^{3x+1}$.
12. Chain rule, trig: (a) $\sin(3x)$; (b) $\cos(x^2)$; (c) $\sin(x^2+1)$; (d) $\cos(5x)$; (e) $\tan(2x)$.
13. Trig, basic: (a) $\cos x$; (b) $\tan x$; (c) $3\sin x - 2\cos x$; (d) $x + \cos x$; (e) $x^2 + \sin x$.
14. Powers of trig (chain): (a) $\sin^2 x$; (b) $\cos^3 x$; (c) $\tan^2 x$.
15. Mixed — pick the right rule first: (a) $x^2 e^x$; (b) $\tfrac{x}{\cos x}$; (c) $\sin(x^3)$; (d) $(x^2+x)^6$; (e) $e^x + e^{-x}$.
16. Evaluate the derivative at the given point: (a) $f(x)=x^3-x$ at $x=2$; (b) $f(x)=e^x\sin x$ at $x=0$; (c) $f(x)=\tfrac1x$ at $x=\tfrac12$; (d) $f(x)=\sqrt{x}$ at $x=4$.
17. Find the equation of the tangent line: (a) $y=(x^2-1)^3$ at $x=1$; (b) $y=e^x$ at $x=0$; (c) $y=\sin x$ at $x=0$; (d) $y=\tfrac1x$ at $x=1$.
18. Differentiate, then simplify: (a) $\tfrac{e^x}{x}$; (b) $x^2\cos x$; (c) $\sqrt{x}\,e^x$; (d) $\tfrac{\sin x}{1+\cos x}$.
19. Second derivatives — find $f''$: (a) $x^4$; (b) $\sin x$; (c) $e^{2x}$; (d) $x^3 - 2x$.
20. Higher-order: (a) the third derivative of $x^5$; (b) the fourth derivative of $\sin x$; (c) the second derivative of $\tfrac1x$.

Practice — combine the rules

1. A chain inside a product: $f(x) = x^2(x^3+1)^4$.
2. A chain inside a chain: $f(x) = \sqrt{1 + \sqrt{x}}$. (Outer root, then inner root.)
3. Order of composition matters: (a) $e^{\sin x}$; (b) $\sin(e^x)$; (c) $(\sin x)^2$. (Note how (a) and (b) differ.)
4. A product of three: $f(x) = x\,e^x\sin x$. (Apply the product rule twice, or group as $x\cdot(e^x\sin x)$.)
5. A quotient with a chain on top: $f(x) = \dfrac{(x^2+1)^3}{x}$.
6. Differentiate $f(x) = \dfrac{e^x}{x^2+1}$ and find where its tangent is horizontal.
7. For $f(x) = x^3 - 6x^2 + 9x$, find $f'(x)$, then all points where the tangent is horizontal (preview of Chapter 8’s critical points).
8. Find where $f(x) = x\,e^{-x}$ has a horizontal tangent, and compute $f''$ there.
9. Inverse-function rule: find the derivative of $g(x)=x^{1/3}$ as the inverse of $f(x)=x^3$, and check it against the power rule. Then do the same for $g(x)=\sqrt{x}$ as the inverse of $x^2$.
10. The hyperbolic functions $\cosh x = \tfrac{e^x + e^{-x}}{2}$ and $\sinh x = \tfrac{e^x - e^{-x}}{2}$. Differentiate each from the rules, and show $\cosh' x = \sinh x$ and $\sinh' x = \cosh x$ — note the absence of a minus sign, unlike $\cos$ and $\sin$. (A Chapter 5 preview, built entirely from $e^x$.)

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

1. 1. $(fg)' = f'g'$. (b) $\left(\tfrac{f}{g}\right)' = \tfrac{f'}{g'}$. (c) $(f\circ g)' = f'\cdot g'$ (composition). (d) The derivative of $e^{kx}$ is $e^{kx}$.
2. Find the error. A student writes “$\tfrac{d}{dx}(x^2\sin x) = 2x\cos x$.” Identify the rule they misused and give the correct derivative.
3. Find the error. Another writes “$\tfrac{d}{dx}(x^2+1)^5 = 5(x^2+1)^4$.” What did they forget, and what’s the correct answer?
4. Construct two functions $f$ and $g$ for which $(fg)'$ happens to equal $f'g'$ at some point, illustrating that the naive product rule is occasionally accidentally right (and explain why you can’t rely on it).
5. Explain, using the chain rule, why $\tfrac{d}{dx}e^{\sin x}$ and $\tfrac{d}{dx}\sin(e^x)$ are different — what does swapping the inner and outer function do?
6. Is it possible? A nonzero function that equals its own derivative. (You differentiated one in this chapter — which, and is it the only such function up to a constant multiple?)
7. Without computing it fully, use the chain-rule factor structure to predict the sign of $\tfrac{d}{dx}e^{-x^2}$ for $x > 0$ and for $x < 0$ (look at the $-2x$ that the chain rule produces). What does this tell you about the shape of the bell curve — where it rises, where it falls, and where it peaks?

◆ Diamond — derive the rules you’ve been using

1. ◆ Prove the product rule from the definition, using the add-and-subtract trick. State clearly where you use the continuity of $f$ (and why Chapter 3 guarantees it).
2. ◆ Prove the quotient rule by writing $\tfrac{f}{g} = f\cdot g^{-1}$ and applying the product and chain rules. (You’ll need $\tfrac{d}{dx}g^{-1} = -g^{-2}g'$ first.)
3. ◆ Derive $\tfrac{d}{dx}\sin x = \cos x$ from the definition, using the angle-addition formula and the two Chapter 2 limits $\tfrac{\sin h}{h}\to1$, $\tfrac{\cos h - 1}{h}\to 0$. Then do the same for $\cos x$.
4. ◆ Prove the inverse-function rule $g'(x) = \tfrac{1}{f'(g(x))}$ by differentiating $f(g(x)) = x$ with the chain rule. Then use it to derive $\tfrac{d}{dx}\ln x = \tfrac1x$ (taking $\ln$ as the inverse of $e^x$, and $\tfrac{d}{dx}e^x = e^x$ as given) — the result Chapter 5 will lean on.
5. ◆◆ Prove the power rule for negative integers: using the quotient rule on $\tfrac{1}{x^n}$, show $\tfrac{d}{dx}x^{-n} = -n\,x^{-n-1}$, confirming the power rule extends below zero.
6. ◆◆ Prove the power rule for rational exponents, closing the loan from this chapter. For $y = x^{p/q}$, raise both sides to the $q$ to get $y^q = x^p$, differentiate both sides (chain rule on the left, power rule on the right), and solve for $y'$ — you should land on exactly $\tfrac{p}{q}\,x^{p/q - 1}$. (This trick — differentiating an equation neither side of which is solved for $y$ — is implicit differentiation, and you’ve just invented it.)
7. ◆◆ The chain rule, carefully. The heuristic $\tfrac{\Delta y}{\Delta x} = \tfrac{\Delta y}{\Delta u}\cdot\tfrac{\Delta u}{\Delta x}$ breaks when $\Delta u = 0$ for some small $\Delta x$ (you’d be dividing by zero). Explain why this case can arise, and sketch how a rigorous proof gets around it. (You don’t need the full proof — just identify the gap the heuristic hides.)

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

1. In three or four sentences with the gear-ratio picture, teach a friend the chain rule — why rates of change multiply through a composition.
2. Debug your classmate. They insist “$(fg)' = f'g'$ — derivatives just distribute over multiplication.” Give them a single concrete counterexample (try $f = g = x$) that settles it in one line, then state the correct rule.
3. Write a complete worked solution differentiating $\sqrt{x^2+1}$, naming the outer and inner functions and narrating each chain-rule step, so a peer could redo it for $\sqrt{x^3+1}$.
4. Design a problem. Invent a function that requires two rules together (say, a chain inside a product), and write the full solution key.
5. In a few sentences, teach why the trigonometric derivatives depend on radians — trace it back through the $\tfrac{\sin h}{h}\to1$ limit to the reason mathematicians measure angles in radians at all.

Looking ahead — Chapter 5 in your hands early

1. You proved (Diamond 41) that $\tfrac{d}{dx}\ln x = \tfrac1x$ using the inverse rule. Predict how you’d differentiate $a^x$ for a general base $a$, given the hint that $a^x = e^{(\ln a)x}$. (Chapter 5 confirms it.)
2. The number $e$ is the one base for which $\tfrac{e^h - 1}{h}\to 1$, making $e^x$ its own derivative. Predict what the derivative of $2^x$ would be — specifically, what constant multiple of $2^x$ — and why that constant is not $1$. (Chapter 5 will name the constant.)
3. You differentiated $\sin x$ twice in problem 19 and found you return to $\pm\sin x$ after four derivatives. Predict the $100$th derivative of $\sin x$. (Hint: the cycle has length four — what is $100$ divided by $4$?)
4. You found (problem 30) that $\cosh' = \sinh$ and $\sinh' = \cosh$. Predict $\cosh''$ and $\sinh''$, and notice each hyperbolic function is its own second derivative: $y'' = y$. Which functions from this chapter instead satisfy $y'' = -y$? (That contrast — $y''=y$ for the hyperbolics, $y''=-y$ for sine and cosine — is the seed of the exponential–trig duality behind Euler’s formula in Chapter 7.)

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