Difficulty: Intermediate | Prerequisites: Part 1 of these notes (derivative definitions and core rules), basic trigonometry, Pythagorean theorem.
Once you know how to take a derivative, the next question is: what can you do with one? This set of notes covers four major applications from Unit 2. Implicit differentiation lets you handle equations where y is tangled up with x instead of isolated. Tangent line approximation uses the derivative to estimate function values without a calculator. Related rates connect multiple changing quantities through a shared equation. L'Hôpital's Rule uses derivatives to resolve indeterminate limits. Each of these is a staple of the AP exam free-response section.
Implicit differentiation handles equations not solved for y. Tangent line approximation uses a derivative to estimate nearby function values. Related rates link two or more changing quantities through differentiation with respect to time. L'Hôpital's Rule resolves 0/0 or ∞/∞ limits by differentiating the numerator and denominator separately.
Explicit function
A function where y is isolated on one side: y = f(x). You can read off the output directly.
Think of it as: y is already solved for you.
Implicit function
An equation relating x and y where y is not isolated, such as x³ + y³ = xy. You work with both variables at once.
In simple terms, y is buried inside the equation and you differentiate both sides as they are.
Implicit differentiation
The technique of differentiating both sides of an implicit equation with respect to x, applying the chain rule to every y term (attaching dy/dx), and then solving for dy/dx algebraically.
Think of it as: treat y as a function of x that you cannot see, so every time you differentiate a y term, tag on dy/dx.
Tangent line approximation (linearisation)
Using the tangent line at a known point to estimate the function's value at a nearby point. The formula is: y ≈ f(c) + f'(c)(x − c), or equivalently, f(x) ≈ f(c) + f'(c)·Δx.
In simple terms, zoom in close enough and the curve looks like its tangent line, so the tangent line's y-value is a good stand-in.
Related rates
A class of problem where two or more quantities are changing with respect to time (t), linked by a geometric or algebraic equation. You differentiate that equation with respect to t and solve for the unknown rate.
Think of it as: everything is moving, and the equation connecting the quantities tells you how their speeds relate to each other.
L'Hôpital's Rule
If lim(x→a) f(x)/g(x) produces the indeterminate form 0/0 or ∞/∞, then the limit equals lim(x→a) f'(x)/g'(x), provided that new limit exists. You may apply the rule repeatedly if the result is still indeterminate.
In simple terms, when a limit gives you 0/0 or ∞/∞, differentiate top and bottom separately and try again.
Explicit: y = x (y is isolated).
Implicit: xy = 4 (y is not isolated, or "not given outright").
When a function is implicit, you cannot simply apply the power rule to y, because y is itself a function of x.
Differentiate every term on both sides with respect to x.
Whenever you differentiate a term containing y, attach dy/dx (chain rule, since y depends on x).
Gather all dy/dx terms on one side, factor out dy/dx, and solve.
All standard derivative rules (power, product, quotient, chain) still apply.
Example: x³ + y³ = xy.
3x² + 3y²·(dy/dx) = x·(dy/dx) + y·1
3y²·(dy/dx) − x·(dy/dx) = y − 3x²
dy/dx·(3y² − x) = y − 3x²
dy/dx = (y − 3x²) / (3y² − x)
Formula: y ≈ f(c) + f'(c)·(x − c), or equivalently f(c) + f'(c)·Δx.
You need three ingredients: the function value at a known point, the derivative at that known point, and the small step Δx to the target.
Example: Estimate √16.5.
Use f(x) = √x, c = 16, Δx = 0.5.
f(16) = 4, f'(x) = 1/(2√x), so f'(16) = 1/8 = 0.125.
Approximation: 4 + 0.125·(0.5) = 4.0625.
The closer x is to c, the better the approximation.
Setting up and solving a related rates problem follows a consistent sequence:
Identify all quantities that change with time and assign them variables.
Any fixed lengths, volumes, or areas are constants, not variables.
Write an equation connecting the variables. Simplify to use the fewest variables possible.
Differentiate the entire equation with respect to time t (this is where the chain rule does heavy lifting, since every variable is a function of t).
Plug in the known values and the known rates only after differentiating.
Solve for the unknown rate.
Worked example: rocket and observer.
An observer stands 3000 ft from a launch pad. A rocket travels vertically at 880 ft/sec. Find the rate of change of the angle of elevation θ when the rocket is 4000 ft high.
The relationship: tan θ = y / 3000, where y is the rocket's height.
Differentiate with respect to t: sec²θ · (dθ/dt) = (1/3000)·(dy/dt).
When y = 4000: the hypotenuse is 5000 (by Pythagoras, 3000-4000-5000 triangle), so sec θ = 5/3.
Substitute: (5/3)² · (dθ/dt) = (1/3000)·880.
(25/9)·(dθ/dt) = 880/3000.
dθ/dt = (9/25)·(880/3000) = 0.1056 rad/sec (approximately 0.11 rad/sec).
Applies only to indeterminate forms 0/0 or ∞/∞. Do not use it on limits that are not indeterminate.
The rule: lim(x→a) f(x)/g(x) = lim(x→a) f'(x)/g'(x), provided the right-hand limit exists.
This is not the quotient rule. You differentiate the numerator and denominator independently, not as a single fraction.
You can apply the rule more than once if the result is still indeterminate.
Example 1: lim(x→2) (x − 2) / (3x³ − 6x² + x − 2).
Direct substitution gives 0/0.
Differentiate top and bottom: 1 / (9x² − 12x + 1).
Evaluate at x = 2: 1 / (36 − 24 + 1) = 1/13.
Example 2: lim(x→0) (cos x − 1) / x².
Direct substitution gives 0/0.
First application: −sin x / 2x. At x = 0 this is 0/0 again.
Second application: −cos x / 2. At x = 0 this is −1/2.
Example 3: lim(x→−2) (x + 2) / (x² + 2x − 3).
Direct substitution gives 0/0.
Differentiate: 1 / (2x + 2). At x = −2: 1 / (−4 + 2) = 1 / (−2) = −1/2.
Note: the lecture notes record this answer as 1. The factoring approach gives (x+2)/[(x+3)(x−1)], and at x = −2 that is 0/(−1)(−3) = 0, which suggests the original expression is not actually 0/0 and L'Hôpital's may not have been needed. Double-check your own substitution on problems like this.
Technique | Key Formula |
|---|---|
Tangent line approximation | y ≈ f(c) + f'(c)·(x − c) |
Implicit differentiation | Differentiate both sides w.r.t. x, attach dy/dx to every y-term, solve for dy/dx |
Related rates | Differentiate the connecting equation w.r.t. t, substitute known values after differentiating |
L'Hôpital's Rule | lim f(x)/g(x) = lim f'(x)/g'(x) when the form is 0/0 or ∞/∞ |
Related rates problems model real scenarios: how fast a shadow lengthens as you walk away from a lamp, how quickly water level rises in a conical tank, how rapidly an angle of elevation changes as a plane flies overhead. Engineers and physicists set up these equations routinely. Tangent line approximation is how calculators and computers estimated roots and trig values before modern algorithms, and it remains the conceptual foundation of numerical methods.
Students often try to plug in known values before differentiating in related rates. This collapses a variable into a constant and kills the term you need. Always differentiate first, then substitute.
L'Hôpital's Rule is frequently confused with the quotient rule. The quotient rule differentiates a single fraction as one object. L'Hôpital's differentiates the numerator and denominator as separate functions.
In implicit differentiation, students forget to apply the product rule when x and y are multiplied together (e.g., the xy term in x³ + y³ = xy).
Tangent line approximation gives an estimate, not an exact value. The further the target point is from the known point, the worse the estimate becomes.
⚠️ Related rates problems appear almost every year on the AP Calculus AB free-response section. The setup phase (choosing the right equation, identifying constants vs variables) is where most marks are lost.
⚠️ Implicit differentiation is tested both on its own and inside related rates problems. Being comfortable solving for dy/dx algebraically is essential.
⚠️ L'Hôpital's Rule is a quick-marks topic: if you spot the 0/0 or ∞/∞ form and apply the rule correctly, the problem is usually short. Missing the applicability condition (not indeterminate) is the most common error.
⚠️ Tangent line approximation questions often appear as part of a multi-part free-response question. They are typically worth a few points and are straightforward if you remember the formula.
True or False: In implicit differentiation, you only attach dy/dx to terms that contain y alone (not mixed xy terms).
Fill in the blank: In a related rates problem, you should differentiate ______ plugging in known numerical values.
True or False: L'Hôpital's Rule can be applied to any limit involving a fraction.
Fill in the blank: The tangent line approximation formula is y ≈ f(c) + ______.
True or False: L'Hôpital's Rule and the quotient rule are the same technique.
Answers: 1. False (you attach dy/dx to every y-term, including mixed terms, via the product and chain rules). 2. Before. 3. False (only for 0/0 or ∞/∞ indeterminate forms). 4. f'(c)·(x − c). 5. False.
Q: Use implicit differentiation to find dy/dx for x³ + y³ = xy.
A: Differentiate both sides: 3x² + 3y²·(dy/dx) = x·(dy/dx) + y. Rearrange: dy/dx·(3y² − x) = y − 3x². So dy/dx = (y − 3x²) / (3y² − x).
Q: Approximate √16.5 using a tangent line approximation.
A: Use f(x) = √x at c = 16. f(16) = 4, f'(16) = 1/8. Approximation: 4 + (1/8)(0.5) = 4.0625.
Q: A rocket rises at 880 ft/sec. An observer is 3000 ft away horizontally. How fast is the angle of elevation changing when the rocket is 4000 ft high?
A: tan θ = y/3000. Differentiate: sec²θ·(dθ/dt) = (1/3000)·(dy/dt). With y = 4000, hypotenuse = 5000, sec θ = 5/3. So (25/9)·(dθ/dt) = 880/3000, giving dθ/dt ≈ 0.11 rad/sec.
Q: Evaluate lim(x→0) (cos x − 1)/x² using L'Hôpital's Rule.
A: First application gives (−sin x)/(2x), still 0/0. Second application gives (−cos x)/2. At x = 0: −1/2.
Q: When should you not use L'Hôpital's Rule?
A: When the limit is not in an indeterminate form (0/0 or ∞/∞). If direct substitution produces a determinate value, just use that value.
Implicit differentiation reappears in multivariable calculus when dealing with surfaces and partial derivatives. Related rates problems draw on geometry (Pythagorean theorem, volume and area formulas) and trigonometry, connecting this unit to prerequisite material. L'Hôpital's Rule ties back to limits (Unit 1) and forward to series convergence tests, where indeterminate forms arise again.
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