Up until now, you have mostly dealt with explicit functions. These are equations where \(y\) is isolated on one side, like \(y = x^2 + 5x\). It’s clear: if you change \(x\), you can directly calculate what happens to \(y\).

But what if the relationship is messy? What if \(x\) and \(y\) are tangled together, like in the equation of a circle: \(x^2 + y^2 = 25\)? This is an implicit function. In this chapter, we will learn how to differentiate these "tangled" equations and apply that knowledge to real-world scenarios where different quantities change at the same time—known as related rates.

Did you know? Implicit differentiation is vital in computer graphics and physics, where paths of objects are often defined by complex curves that aren't easily written as \(y = ...\)


1. Implicit Differentiation

When we cannot (or do not want to) isolate \(y\), we differentiate the entire equation term by term with respect to \(x\). The secret ingredient here is the Chain Rule.

The Golden Rule of Implicit Differentiation

Whenever you differentiate a term containing \(y\), you must multiply by \(\frac{dy}{dx}\). This is because we assume \(y\) is a function of \(x\), even if we can't see the exact formula for it.

For example:
\(\frac{d}{dx}(x^2) = 2x\)
\(\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}\)

Step-by-Step Process

Step 1: Differentiate both sides of the equation with respect to \(x\). Remember to use the Product Rule or Quotient Rule if \(x\) and \(y\) are multiplied or divided!

Step 2: Every time you differentiate a \(y\) term, attach a \(\frac{dy}{dx}\).

Step 3: Group all terms containing \(\frac{dy}{dx}\) on one side of the equation.

Step 4: Factor out \(\frac{dy}{dx}\).

Step 5: Solve for \(\frac{dy}{dx}\) by dividing.

Example: Finding the gradient of a curve

Find \(\frac{dy}{dx}\) for the curve \(x^2 + y^2 - 3xy = 10\).

1. Differentiate \(x^2 \rightarrow 2x\).
2. Differentiate \(y^2 \rightarrow 2y\frac{dy}{dx}\).
3. Differentiate \(-3xy\) using the Product Rule: \(-(3 \cdot y + 3x \cdot \frac{dy}{dx})\).
4. Differentiate \(10 \rightarrow 0\).

The equation becomes: \(2x + 2y\frac{dy}{dx} - 3y - 3x\frac{dy}{dx} = 0\).

Rearrange: \(\frac{dy}{dx}(2y - 3x) = 3y - 2x\)

Final answer: \(\frac{dy}{dx} = \frac{3y - 2x}{2y - 3x}\)

Common Mistake: Forgetting the Product Rule on terms like \(xy\). It is easy to accidentally write the derivative as just \(1 \cdot \frac{dy}{dx}\), but you must treat \(x\) and \(y\) as two separate functions multiplied together!


In Related Rates problems, we look at how two or more variables change with respect to time (\(t\)). Because these variables are related to each other (e.g., the radius and volume of a balloon), their rates of change are also related.

The Mathematical Tool: The Chain Rule

Most problems rely on this version of the chain rule:
\(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\)

This tells us that the rate at which \(y\) changes over time is the rate at which \(y\) changes with \(x\), multiplied by the rate at which \(x\) changes over time.

Problem-Solving Strategy (The "D.R.E.A.M." Method)

D - Diagram: Draw a picture and label the variables (e.g., \(r\) for radius, \(h\) for height).

R - Rates: Write down the rates you know and the rate you need to find using Leibniz notation (e.g., \(\frac{dV}{dt} = 5\), find \(\frac{dr}{dt}\)).

E - Equation: Find an equation that links your variables (e.g., Volume of a sphere \(V = \frac{4}{3}\pi r^3\)).

A - Atmosphere: Identify the "instant" in time mentioned in the question (e.g., "when \(r = 10\)"). Note: Do not plug this number in yet!

M - Math (Differentiate): Differentiate your equation with respect to time (\(t\)) and solve for the unknown rate.

Example: The Inflating Balloon

A spherical balloon is being inflated at a rate of \(10 \text{ cm}^3\text{s}^{-1}\). How fast is the radius increasing when the radius is \(5 \text{ cm}\)?

1. We know \(\frac{dV}{dt} = 10\). We want to find \(\frac{dr}{dt}\) when \(r = 5\).
2. Relationship: \(V = \frac{4}{3}\pi r^3\).
3. Differentiate with respect to \(t\): \(\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}\).
4. Plug in what we know: \(10 = 4\pi (5)^2 \cdot \frac{dr}{dt}\).
5. Solve: \(\frac{dr}{dt} = \frac{10}{100\pi} = \frac{1}{10\pi} \text{ cm s}^{-1}\).

Quick Review:
If a quantity is increasing, its rate is positive.
If a quantity is decreasing (like water leaking out of a tank), its rate is negative.


3. Optimization Problems (HL Extension)

Optimization involves finding the maximum or minimum value of a function. In the HL syllabus, you may be asked to optimize a situation where the relationship between variables is implicit or involves related rates.

Key Concepts

1. A maximum or minimum occurs when the derivative is zero (\(\frac{dy}{dx} = 0\)).
2. In implicit equations, you find \(\frac{dy}{dx}\) as shown in Section 1, then set the numerator to zero.
3. Always check the constraints of the problem (e.g., lengths cannot be negative).

Analogy: Finding the "Peak"
Imagine walking along a curvy mountain path defined by an implicit equation. To find the highest point, you aren't looking for where \(x\) or \(y\) is zero, but where your "vertical change" stops for a split second before you start heading down again. That "stop" is where the gradient \(\frac{dy}{dx} = 0\).


Summary and Key Takeaways

Implicit Differentiation: Used when \(y\) is not isolated. Differentiate everything, and remember that \(\frac{d}{dx}(y) = \frac{dy}{dx}\). Use the product rule for terms like \(xy\).

Related Rates: Used when variables change over time. Use the Chain Rule to link \(\frac{dy}{dt}\), \(\frac{dx}{dt}\), and \(\frac{dy}{dx}\). Always identify your "knowns" and "unknowns" first.

Optimization: Set the derivative to zero to find the best (maximum) or worst (minimum) outcome. In implicit differentiation, this usually means setting the numerator of your \(\frac{dy}{dx}\) expression to zero.

Don't worry if this seems tricky at first! The hardest part is often just setting up the initial equation. Once you have the equation, the differentiation steps follow the same pattern every time.