Differentiation: Composite, Implicit, and Inverse Functions
~3 weeks · September into early October · 5–10% · the chain rule is a third of the courseChain rule — the most important rule in the course
(f ∘ g)(x) means f(g(x)): do g first, then feed the result into f. g is the inner function, f the outer.
Examples, and the distinction matters:
- sin(x²) — inner x², outer sine
- (sin x)² — inner sine, outer squaring
- e3x — inner 3x, outer e□
- √(x²+1) — inner x²+1, outer √
Test for spotting the inner function: what's the thing you'd compute first if you were plugging in a number?
Use the linear-approximation definition. Near a point, g magnifies a small input change by the factor g′. The result then enters f, which magnifies its input change by f′ — evaluated at g's output, which is g(a), not a. Two magnifications in sequence multiply.
In Leibniz notation the du's appear to cancel: (dy/du)(du/dx) = dy/dx. That's a genuinely good intuition and it's why Leibniz's notation won. The naive proof — multiply and divide by Δu — breaks if Δu happens to be 0, and patching that is why textbooks get evasive here. The patch is routine; the intuition is correct.
The chain rule is roughly a third of the course wearing different hats. Implicit differentiation is the chain rule with y treated as a function of x. Related rates is the chain rule with t as the hidden variable. u-substitution in Unit 6 is the chain rule run backwards. Each is taught as a separate topic; they're one rule.
Mechanically
Differentiate the outer function, leaving the inner one untouched inside it, then multiply by the derivative of the inner. Repeat per layer.
Differentiate sin³(2x). Read it as: cube ∘ sine ∘ doubling.
| outermost is cubing → 3·sin²(2x) | power rule on the outside; inside untouched |
| × derivative of sin(2x)'s outer layer → cos(2x) | next layer in |
| × derivative of 2x → 2 | innermost |
| = 6 sin²(2x) cos(2x) | multiply the three factors |
Differentiate √(x³ + 5x).
| rewrite as (x³ + 5x)1/2 | always convert roots to exponents first |
| outer: ½(x³+5x)−1/2 | power rule, inside untouched |
| × inner: (3x² + 5) | |
| = (3x² + 5) / (2√(x³ + 5x)) | tidy the negative exponent into a denominator |
Two shortcuts worth memorizing in chain-rule form, because they occur constantly:
Assembling a messy derivative
Name the outermost structure before writing anything. That decides which rule opens the problem; the others get nested inside.
Differentiate f(x) = x²·sin(3x) / (ex + 1).
| outermost structure: a quotient | so the quotient rule frames everything |
| top = x²·sin(3x) → needs the product rule | and sin(3x) inside needs the chain rule |
| d/dx[top] = 2x·sin(3x) + x²·3cos(3x) | product rule; the 3 comes from the chain |
| d/dx[bottom] = ex | the 1 differentiates away |
| f′ = { [2x sin3x + 3x² cos3x](ex+1) − x² sin3x · ex } / (ex+1)² | assemble: (top′·bottom − top·bottom′)/bottom² |
Stop there. Don't simplify unless the problem asks. Simplification is where errors enter and it almost never earns points.
The trigonometry, rebuilt from scratch
Trig is where rust shows worst, and it's load-bearing for the rest of the year — every integration technique in Unit 6, all of Unit 9 (polar), and the Taylor series in Unit 10 rest on it. This section is the whole of what BC actually uses.
Put a point on the circle of radius 1, at angle θ counterclockwise from the positive x-axis. Then cos θ is its x-coordinate and sin θ is its y-coordinate. That's the definition; everything else is a consequence.
- sin²θ + cos²θ = 1 — the Pythagorean theorem applied to the radius. The most-used identity in calculus.
- Both live in [−1, 1]; both repeat every 2π.
- cos is even: cos(−θ) = cos θ. sin is odd: sin(−θ) = −sin θ. (Reflecting across the x-axis keeps the x-coordinate, flips the y.)
The values worth knowing cold, at θ = 0, π/6, π/4, π/3, π/2:
cos: 1, √3/2, √2/2, 1/2, 0
Cosine is sine read backwards. And if you write the sine row as √0/2, √1/2, √2/2, √3/2, √4/2, the whole table is a single pattern.
sec θ = 1/cos θ · csc θ = 1/sin θ
The reciprocal pairing is deliberately confusing: secant pairs with cosine; cosecant pairs with sine. Crossed over. Read it as "the one with the co- goes with the one without."
The names are literal geometry, shown below: tan θ is the length of a segment on the tangent line, and sec θ is the length of a segment on a secant line. They were measured lengths in a diagram for centuries before anyone thought of them as functions.
Start from sin²θ + cos²θ = 1 and divide the whole equation by cos²θ:
Divide the original by sin²θ instead:
Don't store these separately — regenerate them in five seconds from the one you do remember. tan² + 1 = sec² is the one that runs all through Unit 6 integration.
cos(A + B) = cos A cos B − sin A sin B
Sanity check with B = 0: sin(A) = sin A·1 + cos A·0 ✓.
Where they come from: rotating by A+B is the same as rotating by A then by B, and working out the resulting coordinates gives exactly these. The minus sign in the cosine formula is the one people drop.
Setting B = A gives the double angle forms: sin 2θ = 2 sin θ cos θ, and cos 2θ = cos²θ − sin²θ. Rearranged with the Pythagorean identity: sin²θ = (1 − cos 2θ)/2 and cos²θ = (1 + cos 2θ)/2 — the only way to integrate sin² and cos² in Unit 6.
The sine derivative, derived
| [ sin(x+h) − sin x ] / h | the definition |
| = [ sin x cos h + cos x sin h − sin x ] / h | angle addition |
| = sin x·(cos h − 1)/h + cos x·(sin h)/h | group the sin x terms; split into two known pieces |
| (cos h − 1)/h → 0 and (sin h)/h → 1 | the two special limits from Unit 1, which existed for exactly this moment |
| = sin x·0 + cos x·1 = cos x |
This is the payoff for the sector-squeeze diagram. Those two limits weren't arbitrary exercises — they're precisely the two pieces the sine derivative decomposes into.
Cosine goes the same way, using cos(x+h) = cos x cos h − sin x sin h, and lands on −sin x. The minus comes from the minus in the angle-addition formula.
A point moving around the unit circle at unit speed is at (cos t, sin t). Its velocity vector is tangent to the circle, same length, rotated 90° counterclockwise from the position vector. Rotating (cos t, sin t) by 90° gives (−sin t, cos t).
Read off the components: the x-component says d/dt[cos t] = −sin t, and the y-component says d/dt[sin t] = cos t. Both derivatives at once, no limits, no identities — geometry did the work.
This is also the seed of eit = cos t + i sin t, which she'll derive herself in Unit 10.
The rest of the trig derivatives, each in two lines
| tan x = sin x / cos x | quotient rule |
| = [cos x·cos x − sin x·(−sin x)] / cos²x | note the double negative |
| = (cos²x + sin²x)/cos²x = 1/cos²x | Pythagorean identity on top |
| = sec²x |
| sec x = (cos x)−1 | write it as a power so the chain rule applies |
| = −1·(cos x)−2·(−sin x) | power rule outside, derivative of cosine inside |
| = sin x / cos²x | the two minus signs cancel |
| = (1/cos x)·(sin x/cos x) = sec x tan x | split deliberately into the standard form |
Cotangent and cosecant follow identically and pick up minus signs.
sec → sec·tan · csc → −csc·cot
Memory aid: every function whose name begins with "co" has a minus sign in its derivative. Cosine, cotangent, cosecant. That halves the table.
Exponentials and logarithms
| [ ex+h − ex ] / h | the definition |
| = [ ex·eh − ex ] / h | exponent law: ea+b = eaeb |
| = ex · (eh − 1)/h | factor out ex — it doesn't involve h at all |
| = ex · 1 = ex | the Unit 1 special limit (eh−1)/h → 1 |
The factoring step is the real content: an exponential's rate of change is proportional to its current value, because ex comes out front no matter what. The constant of proportionality is that limit — and e is defined as the base making it exactly 1.
| ax = ex ln a | since eln a = a, raise both sides to the x |
| d/dx = ex ln a · ln a | chain rule; the inner function is x·ln a, whose derivative is the constant ln a |
| = ax · ln a | convert back |
Sanity check against the Unit 1 diagram: the slope of 2x at x=0 should be ln 2 ≈ 0.693, and 3x should be ln 3 ≈ 1.099. Those are exactly the numbers on that picture.
| y = ln x, so ey = x | rewrite using the inverse relationship |
| ey·y′ = 1 | differentiate both sides in x; the left needs the chain rule since y depends on x |
| y′ = 1/ey | solve |
| = 1/x | because ey is x |
This is implicit differentiation, used before it's formally introduced. It's the standard way to differentiate any inverse function: write the inverse relationship, differentiate, solve.
Look at what just happened. The derivatives of powers are powers: x³ → 3x². x² → 2x. x¹ → 1. x⁰ → 0. x−1 → −x−2.
Nothing in that list produces x−1. Run it backwards: what has derivative 1/x? No power of x does — the power rule always lowers the exponent by one, and to land on −1 you'd need to start at 0, but x⁰ is constant with derivative 0.
There's a hole in the power rule, and ln x is what fills it. That's not a coincidence or a curiosity; it's why logarithms show up in integration constantly. Unit 6 will define ln x as an area under the hyperbola 1/x, which explains the coincidence properly.
Logarithmic differentiation
Use it when the variable is in both the base and the exponent, or when facing a monstrous product or quotient.
d/dx[xx] is not x·xx−1 — the power rule needs a constant exponent. It's also not xx·ln x — that rule needs a constant base. Both requirements fail, so you need a different technique entirely.
Differentiate y = xsin x.
| ln y = sin x · ln x | take ln of both sides; the log law drags the exponent down |
| y′/y = cos x·ln x + sin x·(1/x) | left side by chain rule; right side by product rule |
| y′ = y·( cos x ln x + sin x/x ) | multiply through by y |
| = xsin x( cos x ln x + sin x/x ) | substitute back what y was |
Differentiate y = (x²+1)³ (x−4)⁵ / √(3x+2). Doing this with product and quotient rules is a page of algebra.
| ln y = 3ln(x²+1) + 5ln(x−4) − ½ln(3x+2) | logs turn products into sums, quotients into differences, powers into coefficients — all three laws at once |
| y′/y = 3·(2x)/(x²+1) + 5/(x−4) − ½·3/(3x+2) | each term is now a simple d/dx[ln u] = u′/u |
| y′ = y·[ 6x/(x²+1) + 5/(x−4) − 3/(2(3x+2)) ] | multiply back by the original y |
This is the technique's real value: logarithms convert multiplicative structure into additive structure, and additive structure is what differentiation handles easily. Same reason slide rules worked.
Derivative of an inverse function
f−1 undoes f: if f(3) = 7 then f−1(7) = 3. Graphically the inverse is the reflection across the line y = x — every point (a,b) becomes (b,a).
That reflection is the formula. Reflecting a line across y = x swaps rise and run, turning slope m into slope 1/m. So the inverse's derivative is the reciprocal of the original's — evaluated at the corresponding point.
Note f−1 means the inverse function, not the reciprocal 1/f. Unfortunate notation, universally used.
f is differentiable and increasing, with f(2) = 5 and f′(2) = 3. Find (f−1)′(5).
| need the a with f(a) = 5 | the formula evaluates f′ at f−1(5), so find that first |
| f(2) = 5, so f−1(5) = 2 | read the given fact backwards |
| f′(2) = 3 | given |
| (f−1)′(5) = 1/3 | reciprocal |
The arithmetic is trivial. The whole difficulty is keeping straight that you evaluate f′ at 2, not at 5. Write down f−1(5) = 2 explicitly as a separate step and the confusion disappears.
Inverse trig derivatives, derived
| y = arcsin x means sin y = x | the inverse relationship |
| cos y · y′ = 1 | differentiate both sides in x, chain rule on the left |
| y′ = 1/cos y | but the answer must be in terms of x, not y |
| cos y = √(1 − sin²y) = √(1 − x²) | Pythagorean identity; positive root because arcsin outputs angles in [−π/2, π/2] where cosine is ≥ 0 |
| y′ = 1/√(1 − x²) |
| y = arctan x means tan y = x | |
| sec²y · y′ = 1 | differentiate; derivative of tan is sec² |
| y′ = 1/sec²y = 1/(1 + tan²y) | the identity tan² + 1 = sec², rederived from sin²+cos²=1 |
| = 1/(1 + x²) | since tan y = x |
Notice what happened: differentiating a transcendental function (arctan) produced a purely algebraic one (1/(1+x²)).
Run that backwards and you get the fact that ∫ dx/(1+x²) = arctan x — an innocent-looking rational function whose antiderivative is a trig function. There is no way to guess that from the integrand. It's a large part of why integration in Unit 6 is a bag of tricks rather than an algorithm, and it's why these two derivatives must be memorized in both directions.
Higher derivatives
f″ is the derivative of f′; f‴ the derivative of f″; beyond three, write f(4). In Leibniz notation d²y/dx².
Physically: position → velocity → acceleration → jerk. Each differentiation asks "how fast is the previous thing changing."
Find the first four derivatives of f(x) = sin x.
| f′ = cos x | |
| f″ = −sin x | |
| f‴ = −cos x | |
| f(4) = sin x — back to the start | the derivatives of sine cycle with period 4 |
That four-cycle is what produces the alternating signs in sine's Taylor series, and it's why eiθ = cos θ + i sin θ works — the powers of i cycle with period 4 too. File it away for April.
Piecewise differentiability — the two-unknown problem
Promised in Unit 1, and now the tools exist.
f(x) = x² + 1 for x < 2, and ax + b for x ≥ 2. Find a and b making f differentiable at 2.
| differentiable ⟹ continuous, so first match values | you get continuity for free as a requirement |
| 2² + 1 = 5 and 2a + b | evaluate both pieces at the seam |
| equation 1: 2a + b = 5 | |
| now match slopes: left piece has derivative 2x | at x = 2 that's 4 |
| right piece has derivative a | constant |
| equation 2: a = 4 | |
| a = 4, b = 5 − 8 = −3 | substitute into equation 1 |
Order matters for sanity: solve the slope equation first when it's simpler, then back-substitute. And always state that differentiability requires continuity — on the AP that reasoning is worth a point on its own.
Implicit differentiation
Some curves aren't functions. x² + y² = 25 fails the vertical line test — for most x there are two y's. You still want dy/dx.
The move is to treat y as an unknown function of x and differentiate. But if the curve isn't a function, what justifies pretending it is?
The answer is the Implicit Function Theorem: near almost any point on such a curve, a small enough piece of it is the graph of a function, even though the whole curve isn't. A circle isn't a function; the top half near (3,4) is. You're differentiating that local piece.
The exceptions are exactly the points where the curve is locally vertical. And this is the elegant part — the formula announces its own failure at precisely those points, because that's where its denominator hits zero. For the circle, dy/dx = −x/y blows up at y = 0, which is (±5, 0), the two points where the circle is vertical. The algebra knows where the theorem stops working.
BC never mentions any of this. It's worth having, because "why is this legal" is the obvious question and the textbook answer is silence.
| x² + y² = 25 | |
| 2x + 2y·y′ = 0 | d/dx[y²] = 2y·y′ — chain rule, because y is a function of x |
| y′ = −2x/(2y) | |
| = −x/y |
Check it geometrically: the tangent to a circle is perpendicular to the radius. The radius to (x,y) has slope y/x, so the tangent should have slope −x/y. ✓ Agreement, and a good demonstration that implicit differentiation isn't a trick.
Find dy/dx for x²y + y³ = 6.
| d/dx[x²y] = 2xy + x²y′ | product rule — x² and y are both functions of x |
| d/dx[y³] = 3y²·y′ | chain rule |
| 2xy + x²y′ + 3y²y′ = 0 | the right side, being constant, differentiates to 0 |
| y′(x² + 3y²) = −2xy | collect every y′ term on one side and factor |
| y′ = −2xy / (x² + 3y²) |
Reading the answer: horizontal tangents where the numerator is zero (and the denominator isn't); vertical tangents where the denominator is zero (and the numerator isn't).
To get y″, differentiate y′ again (usually a quotient rule) — and then substitute the expression for y′ back in. Leaving a bare y′ in the final answer is the standard lost point. Many problems then simplify further using the original equation, which is worth trying if the result looks ugly.
Hyperbolic functions. Standard in every college Calc I, absent from AP, and they cost about twenty minutes.
Every trig fact has a hyperbolic twin, with one sign flipped:
| Circular | Hyperbolic |
|---|---|
| sin²+cos² = 1 | cosh² − sinh² = 1 |
| (sin)′ = cos | (sinh)′ = cosh |
| (cos)′ = −sin | (cosh)′ = +sinh (no minus) |
| (tan)′ = sec² | (tanh)′ = sech² |
Why "hyperbolic": (cos t, sin t) traces the circle x² + y² = 1; (cosh t, sinh t) traces the hyperbola x² − y² = 1. Same construction, different conic. Verify the derivative claim directly — differentiate (ex − e−x)/2 and you get (ex + e−x)/2, which is cosh. No identity needed.
Where they matter: the shape of a hanging chain or power line is a catenary, y = a·cosh(x/a) — not a parabola, though it looks like one. Hyperbolic substitutions also make several integrals in Unit 6 far cleaner than trigonometric ones.
Derive the power rule for rational exponents, rather than being told it extends. Implicit differentiation makes it four lines, and it closes a gap the textbook waves at.
| let y = xp/q, so yq = xp | clear the fractional exponent |
| q·yq−1·y′ = p·xp−1 | differentiate implicitly — both sides now have integer powers, where we already proved the rule |
| y′ = (p/q)·xp−1/yq−1 | |
| yq−1 = xp(q−1)/q | substitute back |
| y′ = (p/q)·x(p/q) − 1 | the power rule, now proved for every rational exponent |
A harder one in the same spirit: differentiate y = xxx. Take logs twice, or take logs once and treat xx as a known derivative from the logarithmic-differentiation section. The answer is xxx·xx·[ln²x + ln x + 1/x]. It's not on any exam; it's a good hour if she likes this sort of thing.
Unit 3 — Composite, Implicit, and Inverse Functions
Chain rule
- (f∘g)′ = f′(g(x))·g′(x) or dy/dx = (dy/du)(du/dx)
- (ln u)′ = u′/u · (√u)′ = u′/(2√u)
The table
- sin → cos · cos → −sin · tan → sec² · cot → −csc²
- sec → sec·tan · csc → −csc·cot — every “co-” function carries a minus
- ex → ex · ax → axln a · ln x → 1/x · logax → 1/(x ln a)
- arcsin → 1/√(1−x²) · arctan → 1/(1+x²) · arcsec → 1/(|x|√(x²−1))
Trig facts the derivatives rest on
- sin²θ + cos²θ = 1 → ÷cos²: tan²+1 = sec² → ÷sin²: 1+cot² = csc²
- sin(A+B) = sinA cosB + cosA sinB · cos(A+B) = cosA cosB − sinA sinB
- sin²θ = (1 − cos2θ)/2 · cos²θ = (1 + cos2θ)/2 — needed in Units 6 and 9
Techniques
- (f−1)′(b) = 1/f′(f−1(b)) — find the a with f(a)=b first
- Logarithmic differentiation: variable in base and exponent, or a big product. Take ln, differentiate, multiply back by y.
- Implicit: every y yields a y′; collect and solve. Horizontal tangent where the numerator vanishes, vertical where the denominator does. For y″, substitute y′ back in.
- Piecewise: differentiability requires matching values and matching slopes at the seam.
Practise Unit 3 → Fresh problems from this unit, with worked solutions.