Difficulty: Intermediate to Advanced | Prerequisites: Chain rule, product rule, quotient rule, trig derivatives, curve sketching notes
This final set of exam-review topics covers four techniques that apply differentiation in different directions. Linear approximation uses the tangent line to estimate function values. Newton's method uses the tangent line iteratively to find roots. Implicit differentiation handles equations where y is not isolated. Related rates connect the rates of change of two or more quantities through a shared equation. These are all derivative applications, but each has its own setup and its own pitfalls.
Linear approximation replaces a function with its tangent line near a known point to estimate nearby values. Newton's method repeats that idea to zero in on roots. Implicit differentiation finds dy/dx when y is tangled up with x in an equation you cannot solve for y. Related rates use the chain rule to connect how fast two quantities change when they are linked by a geometric or physical equation.
Linear approximation (linearisation)
The approximation L(x) = f(a) + f'(a)(x – a), which uses the tangent line at x = a to estimate f(x) for values of x near a.
In simple terms, zoom in close enough to any smooth curve and it looks like a straight line. The tangent line is that straight line, and you use it as a stand-in for the curve.
Newton's method
An iterative root-finding algorithm. Starting from an initial guess x₁, the next approximation is x₂ = x₁ – f(x₁)/f'(x₁). Each step uses the tangent line to jump closer to the root.
Think of it as repeatedly sliding down the tangent line to the x-axis and using the landing spot as your new guess.
Implicit differentiation
A technique for finding dy/dx when the relationship between x and y is given as an equation (like x²y³ + x³y⁴ = 11) rather than as y = f(x). You differentiate both sides with respect to x, treating y as a function of x and applying the chain rule wherever y appears, then solve for dy/dx.
In simple terms, every time you differentiate a y-term, you tack on a dy/dx because y depends on x.
Related rates
Problems where two or more quantities change with time, linked by an equation. You differentiate the linking equation with respect to time t, substitute the known rates and values, and solve for the unknown rate.
Think of it as: the equation connects the quantities, and differentiating with respect to time connects their speeds.
The linearisation formula is L(x) = f(a) + f'(a)(x – a).
f(x) = x cos x. Evaluate at a = 0: f(0) = 0.
f'(x) = –x sin x + cos x (product rule). Evaluate at a = 0: f'(0) = 0 + 1 = 1.
L(x) = 0 + 1 · (x – 0) = x.
So near x = 0, x cos x is approximately equal to x. The approximation is best close to a = 0 and deteriorates as you move away.
The iteration formula is x_{n+1} = x_n – f(x_n)/f'(x_n).
f(x) = x⁴ – 4x + 1, so f'(x) = 4x³ – 4.
x₁ = 0. f(0) = 1, f'(0) = –4.
x₂ = 0 – (1)/(–4) = 0 + 1/4 = 1/4.
Each iteration gives a better approximation of the root. On an exam, you are typically asked to find x₂ (one iteration) or x₃ (two iterations).
Differentiate both sides with respect to x, using the product rule on each term and the chain rule on every y factor.
d/dx[x²y³]: x² · 3y²(dy/dx) + y³ · 2x.
d/dx[x³y⁴]: x³ · 4y³(dy/dx) + y⁴ · 3x².
Set the sum equal to 0 (since d/dx[11] = 0).
Collect all terms with dy/dx on one side: (3x²y² + 4x³y³)(dy/dx) = –2xy³ – 3x²y⁴.
Solve: dy/dx = –(2xy³ + 3x²y⁴)/(3x²y² + 4x³y³).
Factor: dy/dx = –(2y + 3xy²)/(3x + 4x²y).
The answer will usually contain both x and y. That is normal for implicit differentiation.
A balloon rises so that the angle of elevation θ from an observer 100 m away changes at dθ/dt = 0.1 rad/min when θ = π/6. Find dh/dt.
The linking equation: tan θ = h/100, so h = 100 tan θ.
Differentiate with respect to t: dh/dt = 100 sec²θ · dθ/dt.
Substitute θ = π/6 and dθ/dt = 0.1: dh/dt = 100 · sec²(π/6) · 0.1.
sec(π/6) = 2/√3, so sec²(π/6) = 4/3.
dh/dt = 100 · (4/3) · 0.1 = 40/3 m/min.
The key steps in any related rates problem: draw a picture, write the equation linking the quantities, differentiate with respect to time, substitute known values, solve.
Linear approximation
L(x) = f(a) + f'(a)(x – a)
Newton's method iteration
x_{n+1} = x_n – f(x_n) / f'(x_n)
Implicit differentiation pattern
Whenever you differentiate a term containing y with respect to x, apply the chain rule: d/dx[g(y)] = g'(y) · dy/dx. Then collect all dy/dx terms on one side and solve.
Related rates pattern
Given an equation relating quantities that change with time, differentiate every term with respect to t. Each variable gets its own rate: d/dt[x²] = 2x · dx/dt, d/dt[tan θ] = sec²θ · dθ/dt, and so on.
Students often forget to apply the chain rule during implicit differentiation. Every time you differentiate a y-term, you must multiply by dy/dx. Missing even one instance will give the wrong answer.
In Newton's method, students sometimes plug in f(x) instead of f(x)/f'(x), or forget the subtraction. The formula is x_{n+1} = x_n MINUS the fraction, not plus.
In related rates, students sometimes differentiate with respect to x instead of with respect to t. Every variable is a function of time, so d/dt is the correct operator, and every variable picks up its own rate (dx/dt, dy/dt, dθ/dt).
Confusing the point of evaluation "a" in linear approximation with the point you are estimating. a is the known point where you can compute f and f' exactly. The estimate is for a nearby x-value.
⚠️ Newton's method on the exam is typically a "find x₂" problem. You will be given f(x), f'(x), and x₁. One clean iteration is all they ask for. Get the formula right and the arithmetic follows.
⚠️ Implicit differentiation problems almost always ask you to simplify dy/dx by factoring. Leave your answer in terms of both x and y.
⚠️ Related rates: expect a trig setup (right triangle, angle of elevation) or a Pythagorean setup (ladder sliding, expanding circle). Know your trig derivatives cold, especially d/dt[tan θ] = sec²θ · dθ/dt.
⚠️ Linear approximation is the quickest topic on the exam. Memorise L(x) = f(a) + f'(a)(x – a) and practise identifying a (the convenient known point) quickly.
Fill in the blank: the linear approximation of f(x) near x = a is L(x) = f(a) + ______ · (x – a). Answer: f'(a).
True or false: Newton's method always converges to a root. Answer: False. Convergence depends on the initial guess and the behaviour of f near the root. A bad starting point can cause the method to diverge or cycle.
In implicit differentiation, when you differentiate y³ with respect to x, the result is ______. Answer: 3y² · dy/dx.
True or false: in a related rates problem, you should substitute known values before differentiating. Answer: False. Differentiate first, then substitute. Substituting first turns variables into constants and kills the rates you need.
Fill in the blank: sec(π/6) = ______. Answer: 2/√3 (or equivalently 2√3/3).
Q: Find the linear approximation of f(x) = x cos x at a = 0.
A: f(0) = 0. f'(x) = –x sin x + cos x, so f'(0) = 1. L(x) = 0 + 1(x – 0) = x.
Q: Use Newton's method to find x₂ for x⁴ – 4x + 1 = 0, starting with x₁ = 0.
A: f(x) = x⁴ – 4x + 1, f'(x) = 4x³ – 4. x₂ = 0 – (0 – 0 + 1)/(0 – 4) = 0 – (1)/(–4) = 1/4.
Q: Differentiate implicitly: x²y³ + x³y⁴ = 11. Find dy/dx.
A: Differentiating: (3x²y² dy/dx + 2xy³) + (4x³y³ dy/dx + 3x²y⁴) = 0. Collecting dy/dx terms: (3x²y² + 4x³y³) dy/dx = –2xy³ – 3x²y⁴. dy/dx = –(2y + 3xy²)/(3x + 4x²y).
Q: A balloon rises above a point on the ground 100 m from an observer. The angle of elevation θ changes at 0.1 rad/min when θ = π/6. How fast is the balloon rising?
A: h = 100 tan θ. dh/dt = 100 sec²θ · dθ/dt = 100 · (4/3) · 0.1 = 40/3 m/min.
Linear approximation is the foundation of differentials (dx, dy), which appear in integration. Newton's method reappears in numerical analysis and in any applied course that needs root-finding. Implicit differentiation extends to multivariable calculus, where partial derivatives replace dy/dx. Related rates are the single-variable version of problems you will see again in vector calculus and physics.
All four topics rely on the chain rule. If you are shaky on the chain rule, fix that first, because every technique here is the chain rule applied in a different context.
linear approximation, linearisation, linearization, tangent line approximation, L(x) formula, Newton's method, Newton's iteration, root finding calculus, implicit differentiation, implicit diff, dy/dx implicit, related rates, related rates trig, related rates Pythagorean, angle of elevation problem, balloon problem calculus, chain rule applications, MATH 231, calculus 1, exam 2 review