Applications of differentiation

Q: What is a stationary point and how do you find its coordinates?

A stationary point is where the gradient is zero (\(\frac{dy}{dx} = 0\)). Solve this for \(x\), then find \(y\) using the original equation.

Q: How do you use the second derivative test to determine the nature of a stationary point?

If \(\frac{d^2y}{dx^2} 0\), it is a Minimum point.

Additional Mathematics (4049) · GCE O-Level · 5 min read · Updated: 7 Sept 2026

Welcome to Applications of Differentiation!

In the previous chapter, you learned how to differentiate functions to find the derivative, \( \frac{dy}{dx} \). Now, we are going to use that skill to solve real-world problems! Think of differentiation as a superpower that lets you see how things are changing at any exact moment—whether it’s the speed of a car, the growth of a plant, or the slope of a mountain. Don't worry if it seems complex; we will take it step-by-step.

1. Tangents and Normals

The most fundamental use of the derivative \( \frac{dy}{dx} \) is finding the gradient (slope) of a curve at a specific point.

The Tangent

A tangent is a straight line that just touches a curve at a single point. The gradient of the tangent at point \( (x_1, y_1) \) is exactly the value of \( \frac{dy}{dx} \) at that point.

Step-by-step to find the equation of a tangent:
1. Find the derivative \( \frac{dy}{dx} \).
2. Substitute the \( x \)-coordinate of the point into \( \frac{dy}{dx} \) to find the gradient, \( m \).
3. Use the straight-line formula: \( y - y_1 = m(x - x_1) \).

The Normal

A normal is a straight line that is perpendicular (at 90 degrees) to the tangent at that same point. From your coordinate geometry knowledge, you know that if two lines are perpendicular, the product of their gradients is \( -1 \).

The Rule: Gradient of Normal \( = -\frac{1}{\text{Gradient of Tangent}} \)

Quick Tip: If the tangent's gradient is \( \frac{2}{3} \), the normal's gradient is \( -\frac{3}{2} \). Just flip it and change the sign!

Key Takeaway: Differentiation gives you the slope of the curve. Use \( y - y_1 = m(x - x_1) \) to find the actual equation of the line.

2. Increasing and Decreasing Functions

Sometimes we just want to know if a graph is going "up" or "down" as we move from left to right.

Increasing Function: The gradient is positive. Formally: \( \frac{dy}{dx} > 0 \).
Decreasing Function: The gradient is negative. Formally: \( \frac{dy}{dx} < 0 \).

Analogy: Imagine walking on a hill. If you are walking uphill, your "rate of change" of height is positive (increasing). If you are walking downhill, it's negative (decreasing).

3. Stationary Points

A stationary point is a point on the graph where the gradient is zero. This means the curve is perfectly flat for a split second. At these points: \( \frac{dy}{dx} = 0 \).

Types of Stationary Points:

1. Maximum Turning Point: The peak of a hill (the graph stops going up and starts going down).
2. Minimum Turning Point: The bottom of a valley (the graph stops going down and starts going up).
3. Stationary Point of Inflexion: The graph flattens out but then continues in the same direction.

The Second Derivative Test

How do we know if a stationary point is a Max or a Min without drawing it? We use the second derivative, \( \frac{d^2y}{dx^2} \).

If \( \frac{d^2y}{dx^2} < 0 \) (negative) at the point, it is a Maximum point.
If \( \frac{d^2y}{dx^2} > 0 \) (positive) at the point, it is a Minimum point.
If \( \frac{d^2y}{dx^2} = 0 \), the test fails! You must check the signs of the gradient slightly to the left and right of the point to determine its nature.

Did you know? A memory trick for the second derivative: A negative second derivative looks like a sad face (the shape of a Max point). A positive second derivative looks like a happy face (the shape of a Min point).

4. Connected Rates of Change

In the real world, things change together. For example, as you blow air into a balloon, the radius increases, which causes the volume to increase over time.

We use the Chain Rule to link these changes:

\( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \)

Example Scenario:
If you know how fast the radius of a circle is growing (\( \frac{dr}{dt} \)), and you want to find how fast the area is growing (\( \frac{dA}{dt} \)), you use:
\( \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} \)

Common Mistake to Avoid: Always check if the rate is "increasing" (positive value) or "decreasing/leaking/shrinking" (negative value). If water is leaking at \( 5 \text{ cm}^3/\text{s} \), then \( \frac{dV}{dt} = -5 \).

5. Maxima and Minima Problems (Optimization)

This is where calculus gets very practical. Businesses use this to maximize profit, and engineers use it to minimize material waste.

Step-by-Step Approach:
1. Write an equation: Find an expression for the quantity you want to maximize or minimize (e.g., Volume \( V \)).
2. One variable only: If your equation has two variables (like \( r \) and \( h \)), use other information in the question to substitute one out so the equation is in terms of one variable only.
3. Differentiate: Find the derivative (e.g., \( \frac{dV}{dr} \)).
4. Set to Zero: Let the derivative equal zero and solve for the variable.
5. Check Nature: Use the second derivative test to prove it is indeed a maximum or minimum as requested.

Key Takeaway: "Maximum" or "Minimum" is a code word for "Set the derivative to zero!"

Summary Checklist

1. Tangent Gradient: \( m = \frac{dy}{dx} \)
2. Normal Gradient: \( m_{normal} = -\frac{1}{m_{tangent}} \)
3. Increasing: \( \frac{dy}{dx} > 0 \)
4. Decreasing: \( \frac{dy}{dx} < 0 \)
5. Stationary Point: \( \frac{dy}{dx} = 0 \)
6. Max Point: \( \frac{d^2y}{dx^2} < 0 \)
7. Min Point: \( \frac{d^2y}{dx^2} > 0 \)
8. Chain Rule: \( \frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} \)

Note: For applications involving displacement, velocity, and acceleration, please refer to the Kinematics chapter. While those are applications of differentiation, they are handled as a specific topic in the syllabus!

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

How do you find the gradient of a tangent to a curve at a specific point?

The gradient of the tangent at point \((x_1, y_1)\) is the value of the derivative \(\frac{dy}{dx}\) calculated at that point.

What are the steps to find the equation of a tangent line?

1. Find \(\frac{dy}{dx}\). 2. Calculate gradient \(m\) by substituting \(x_1\). 3. Use \(y - y_1 = m(x - x_1)\).

What is the relationship between the gradient of a tangent and the gradient of a normal?

The normal is perpendicular to the tangent, so the product of their gradients is \(-1\). Gradient of Normal \( = -\frac{1}{\text{Gradient of Tangent}} \).

How is an increasing function defined in terms of its derivative?

A function is increasing when its gradient is positive, expressed as \(\frac{dy}{dx} > 0\).

How is a decreasing function defined in terms of its derivative?

A function is decreasing when its gradient is negative, expressed as \(\frac{dy}{dx} < 0\).

What is a stationary point and how do you find its coordinates?

A stationary point is where the gradient is zero (\(\frac{dy}{dx} = 0\)). Solve this for \(x\), then find \(y\) using the original equation.

How do you use the second derivative test to determine the nature of a stationary point?

If \(\frac{d^2y}{dx^2} < 0\), it is a Maximum point. If \(\frac{d^2y}{dx^2} > 0\), it is a Minimum point.

How do you connect the rates of change of two related variables using the Chain Rule?

Use the formula \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).

What is the general strategy for solving optimization (maxima/minima) problems?

Express the quantity in one variable, differentiate, set the derivative to zero, and verify the nature using the second derivative test.

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