Source: Advanced Calculus I, University of Central Florida
Tags: implicit differentiation, related rates, optimization, max and min, derivatives, advanced techniques, calculus I, UCF
Difficulty: Intermediate Prerequisites: Chain rule, basic differentiation rules (power, product, quotient), understanding of equations involving two variables.
Once you have the chain rule in hand, three powerful techniques open up: implicit differentiation, related rates, and optimization. These move you from "find the derivative of a formula" to "solve a real problem using derivatives." Implicit differentiation lets you handle equations where y is tangled up with x rather than isolated. Related rates let you connect the speeds at which different quantities change. Optimization uses derivatives to find the best possible value of something. These three topics come up repeatedly in exams and in applied fields like engineering, physics, and economics.
Implicit differentiation finds dy/dx when y is not written as an explicit function of x. Related rates use the chain rule to link how fast different quantities are changing at the same moment. Optimization finds the maximum or minimum value of a function by setting the derivative equal to zero and checking the result.
Implicit differentiation
A technique for differentiating an equation that defines y in terms of x without isolating y first. You differentiate both sides of the equation with respect to x, applying the chain rule to every y term (since y is a function of x), and then solve for dy/dx. In simple terms, you treat y as a "hidden" function of x and tack on a dy/dx every time you differentiate a y.
Related rates
A method for finding the rate of change of one quantity by relating it to the rate of change of another quantity through a shared equation. Both quantities change with respect to a third variable, usually time. Think of it as: "if this thing is changing at some speed, how fast is that other thing changing?"
Optimization (in calculus)
The process of finding the absolute maximum or minimum value of a function on a given interval, typically by finding critical points where f'(x) = 0 or f'(x) is undefined, then comparing values. In simple terms, it is the mathematical way of asking "what is the best I can do?"
Critical point
A value of x where f'(x) = 0 or f'(x) does not exist. These are the candidates for local maxima, local minima, or inflection points.
Used when an equation relates x and y but y is not isolated on one side.
Procedure:
Differentiate every term on both sides with respect to x.
Every time you differentiate a term involving y, multiply by dy/dx (this is the chain rule, since y is a function of x).
Collect all terms containing dy/dx on one side.
Factor out dy/dx and solve.
Worked example: x² + y² = 25
Differentiate both sides: 2x + 2y · (dy/dx) = 0
Solve for dy/dx: dy/dx = -2x / 2y = -x/y
This gives the slope of the tangent line at any point on the circle x² + y² = 25, as long as y ≠ 0.
When to use it:
Circles, ellipses, and other conic sections
Any equation where isolating y would be messy or impossible (e.g., x³ + y³ = 6xy)
Finding tangent lines to curves not given as y = f(x)
Used when two or more quantities change over time and are linked by an equation.
Procedure:
Draw a diagram if possible and label all changing quantities.
Write the equation that relates the quantities.
Differentiate both sides with respect to time t (using the chain rule on every variable).
Substitute the known values and rates.
Solve for the unknown rate.
Worked example: right triangle
A right triangle has legs x and y and hypotenuse z, with x² + y² = z².
Differentiate with respect to t: 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt).
If you know x, y, z, dx/dt, and dy/dt at a specific instant, you can solve for dz/dt.
Common related rates scenarios:
Expanding/shrinking circles or spheres (area and volume changing with radius)
Ladders sliding down walls (Pythagorean theorem)
Filling or draining tanks (volume and height)
Shadow length problems (similar triangles)
Used to find the maximum or minimum value a function can take.
Procedure:
Write the function to be optimised (the "objective function").
If there is a constraint, use it to reduce the objective function to one variable.
Find f'(x) and set it equal to zero to find critical points.
Use the second derivative test or evaluate endpoints to determine whether each critical point is a maximum or minimum.
Answer the question in context.
Why it matters practically:
Maximising profit or revenue given cost constraints
Minimising material usage for a container of fixed volume
Minimising travel time or distance
Implicit differentiation of x² + y² = r² (circle):
2x + 2y(dy/dx) = 0, so dy/dx = -x/y
Related rates, Pythagorean theorem:
x² + y² = z², so 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
First derivative test for optimization:
Find critical points where f'(x) = 0. If f' changes from positive to negative, it is a local maximum. If f' changes from negative to positive, it is a local minimum.
Second derivative test:
At a critical point c: if f''(c) > 0, the point is a local minimum. If f''(c) < 0, the point is a local maximum. If f''(c) = 0, the test is inconclusive.
Implicit differentiation is used in thermodynamics when state equations (like the ideal gas law PV = nRT) link several variables and you need to know how one responds to changes in another. Optimization is everywhere in engineering and business: designing a can that uses the least aluminium for a given volume, choosing a price point that maximises revenue, or finding the angle that maximises the range of a projectile.
Forgetting dy/dx when differentiating y terms in implicit differentiation. If you differentiate y² and write 2y instead of 2y(dy/dx), you have missed the entire point of the technique. The chain rule is doing the work here.
Confusing "at a specific instant" with "for all time" in related rates. You plug in known values only after differentiating, never before. Substituting before differentiating treats changing quantities as constants and produces wrong answers.
Thinking critical points are always maxima or minima. A critical point could be an inflection point (like x = 0 for f(x) = x³). Always verify with the first or second derivative test.
Forgetting to check endpoints in optimization on a closed interval. The absolute maximum or minimum of a continuous function on [a, b] can occur at an endpoint, not just at a critical point.
⚠️ Implicit differentiation is commonly tested with circles and ellipses. Be ready to find tangent lines to curves given implicitly.
⚠️ Related rates problems almost always require a clearly labelled diagram. Marks are often given for setting up the equation, not just the final answer.
⚠️ Optimization problems on exams typically combine a constraint equation with an objective function. Reducing to one variable is the step most students struggle with.
⚠️ The distinction between local and absolute extrema matters. Read the problem carefully: "find the maximum" on a closed interval requires checking endpoints.
True or false: When differentiating implicitly, you differentiate y terms as if y were a constant.
A: False. You treat y as a function of x and apply the chain rule, attaching dy/dx to every y derivative.
Fill in the blank: In a related rates problem, you should substitute known numerical values ______ differentiating.
A: After.
True or false: If f'(c) = 0, then f has a local maximum at c.
A: False. f'(c) = 0 means c is a critical point, but it could be a minimum or neither. You must test further.
Q: Use implicit differentiation to find dy/dx for the equation x³ + y³ = 6xy.
A: Differentiate both sides: 3x² + 3y²(dy/dx) = 6y + 6x(dy/dx). Rearrange: 3y²(dy/dx) – 6x(dy/dx) = 6y – 3x². Factor: (3y² – 6x)(dy/dx) = 6y – 3x². So dy/dx = (6y – 3x²) / (3y² – 6x) = (2y – x²) / (y² – 2x).
Q: A circular oil slick has a radius increasing at 2 m/min. How fast is the area increasing when the radius is 10 m?
A: A = πr², so dA/dt = 2πr(dr/dt) = 2π(10)(2) = 40π m²/min.
Q: A farmer has 200 m of fencing and wants to enclose a rectangular field along a river (no fence needed on the river side). What dimensions maximise the area?
A: Let x = the side perpendicular to the river (two of these) and y = the side parallel. Constraint: 2x + y = 200, so y = 200 – 2x. Area = x(200 – 2x) = 200x – 2x². A'(x) = 200 – 4x = 0, so x = 50. Then y = 100. The maximum area is 5 000 m².
Q: Why must you differentiate before substituting in a related rates problem?
A: Because the variables are changing with time. Substituting first replaces them with constants, and the derivative of a constant is zero. You lose the rate-of-change information you need.
Implicit differentiation connects directly to the chain rule, since every y-term differentiation is a chain rule application with y as the inner function. Related rates extend to multivariable calculus, where partial derivatives formalise the idea of rates of change in multiple directions simultaneously. Optimization is the foundation for the entire field of mathematical optimisation, including linear programming, constrained optimisation with Lagrange multipliers, and machine learning loss minimisation.
implicit differentiation, implicit derivative, dy/dx, related rates, rate of change, optimization, maxima, minima, critical points, first derivative test, second derivative test, endpoints, closed interval, applied derivatives, calculus I advanced topics, UCF Advanced Calculus