Applications of differentiation

Q: How do you find the equation of a tangent line to a curve \(y = f(x)\) at a given point \((x_1, y_1)\)?

Find the slope \(m_{\text{tangent}} = f'(x_1)\) and use the point-slope form: \(y - y_1 = f'(x_1)(x - x_1)\).

Q: How do you find the gradient and equation of the normal line to a curve at \((x_1, y_1)\)?

The normal slope is \(m_{\text{normal}} = -\frac{1}{f'(x_1)}\) (for \(f'(x_1) \neq 0\)), giving the equation \(y - y_1 = -\frac{1}{f'(x_1)}(x - x_1)\).

Q: How do you determine if a function \(f(x)\) is strictly increasing or strictly decreasing on an interval?

\(f(x)\) is strictly increasing where \(f'(x) > 0\), and strictly decreasing where \(f'(x) < 0\).

Q: How do you classify stationary points using the First Derivative Test?

Examine the sign of \(f'(x)\) across the stationary point: \(+\) to \(-\) indicates a local maximum, \(-\) to \(+\) indicates a local minimum, and no sign change indicates a stationary point of inflexion.

Q: How does the Second Derivative Test classify stationary points?

If \(f'(x_0) = 0\), then \(f''(x_0) > 0\) gives a local minimum, \(f''(x_0) < 0\) gives a local maximum, and \(f''(x_0) = 0\) is inconclusive.

Q: How do you find the global (absolute) maximum and minimum values of a continuous function on a closed interval \([a, b]\)?

Evaluate \(f(x)\) at all stationary points in \((a, b)\) and at the boundary endpoints \(a\) and \(b\); the largest value is the global maximum and the smallest is the global minimum.

Q: How do you find the vertical, horizontal, and oblique asymptotes of a function?

Vertical asymptotes occur where \(\lim_{x \to c} f(x) = \pm\infty\), horizontal asymptotes occur where \(\lim_{x \to \pm\infty} f(x) = L\), and oblique asymptotes occur when \(\lim_{x \to \pm\infty} [f(x) - (mx + c)] = 0\).

Q: What is a point of inflexion and how is it identified?

A point of inflexion is a point on the curve where concavity changes sign, occurring where \(f''(x) = 0\) (or is undefined) and \(f''(x)\) switches sign across the point.

Q: What are the key steps to solve an applied optimization problem in calculus?

Express the objective quantity as a single-variable function using constraints, find stationary points by setting the derivative to zero, and verify the extremum using derivative tests.

Q: How is the Chain Rule applied to solve related rates of change problems with respect to time?

Differentiate the governing geometric or physical formula with respect to time \(t\) using the Chain Rule, such as \(\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}\).

Practice

Mathematics M2 (Algebra and Calculus) · Senior Secondary (HKDSE) · 6 min read · Updated: 24 Aug 2026

M2 Chapter: Applications of Differentiation

Hey everyone! Welcome to one of the most powerful chapters in M2 Calculus: Applications of Differentiation. You might be wondering, "Why did we spend all that time learning how to find derivatives?" Well, this is why!

Think of differentiation as getting a special power: the ability to see the "rate of change" of anything. In this chapter, we'll use that power to:

Find the equations of tangents and normals to a curve at any given point.
Determine intervals of increase and decrease, and find maximum and minimum values to solve optimization problems.
Become masters at sketching complex curves by analyzing asymptotes, concavity, and points of inflexion.
Solve real-world problems involving rates of change over time.

Don't worry if it sounds like a lot. We'll break it down step-by-step. Let's get started!

1. Finding the Equations of Tangents and Normals

Remember that the derivative, \(\frac{dy}{dx}\) or \(f'(x)\), gives the gradient (slope) of the tangent line to a curve \(y = f(x)\) at any point \(x\).

Tangents

A tangent is a straight line touching the curve at a point \((x_1, y_1)\) with gradient \(m_{\text{tangent}} = f'(x_1)\). Its equation in point-slope form is:

\(y - y_1 = m_{\text{tangent}}(x - x_1)\)

Normals

The normal line at a given point is the straight line perpendicular to the tangent line at that point. Since the lines are perpendicular, the product of their gradients is \(-1\):

\(m_{\text{normal}} \times m_{\text{tangent}} = -1 \implies m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}\) (provided \(m_{\text{tangent}} \neq 0\))

The equation of the normal is therefore:

\(y - y_1 = -\frac{1}{f'(x_1)}(x - x_1)\)

Example Walkthrough

Find the equations of the tangent and the normal to the curve \(y = x^3 - 2x + 5\) at \(x = 2\).

Step 1: Find the point.
When \(x = 2\), \(y = (2)^3 - 2(2) + 5 = 8 - 4 + 5 = 9\).
So, the point of contact is \((x_1, y_1) = (2, 9)\).

Step 2: Find the derivative.
\(\frac{dy}{dx} = 3x^2 - 2\)

Step 3: Find the gradients.
At \(x = 2\), \(m_{\text{tangent}} = 3(2)^2 - 2 = 10\).
Consequently, \(m_{\text{normal}} = -\frac{1}{10}\).

Step 4: Find the equations.
Tangent:
\(y - 9 = 10(x - 2) \implies y = 10x - 11\)
Normal:
\(y - 9 = -\frac{1}{10}(x - 2) \implies x + 10y - 92 = 0\)

Key Takeaway

Use \(f'(a)\) to find the tangent slope \(m_{\text{tangent}}\), and take the negative reciprocal \(-\frac{1}{m_{\text{tangent}}}\) for the normal slope \(m_{\text{normal}}\).

2. Increasing/Decreasing Functions and Extrema

Intervals of Increase and Decrease

If \(f'(x) > 0\) for all \(x\) in an interval, \(f(x)\) is strictly increasing on that interval.
If \(f'(x) < 0\) for all \(x\) in an interval, \(f(x)\) is strictly decreasing on that interval.

Stationary Points and the First Derivative Test

A point where \(f'(x) = 0\) is called a stationary point. We can classify stationary points by examining the sign of \(f'(x)\) across the point using a sign table:

Local Maximum: \(f'(x)\) changes from positive (\(+\)) to negative (\(-\)).
Local Minimum: \(f'(x)\) changes from negative (\(-\)) to positive (\(+\)).
Stationary Point of Inflexion: \(f'(x) = 0\), but \(f'(x)\) does not change sign (e.g. \(+ \to 0 \to +\) or \(- \to 0 \to -\)).

The Second Derivative Test

If \(f'(x_0) = 0\):

If \(f''(x_0) > 0\), the curve is concave up (☺), so \(x_0\) is a local minimum.
If \(f''(x_0) < 0\), the curve is concave down (☹), so \(x_0\) is a local maximum.
If \(f''(x_0) = 0\), the test is inconclusive; use the First Derivative Test instead.

Example Walkthrough

Find and classify the stationary points of \(f(x) = 2x^3 - 3x^2 - 12x + 1\).

Step 1: \(f'(x) = 6x^2 - 6x - 12 = 6(x - 2)(x + 1) = 0 \implies x = 2\) or \(x = -1\).

Step 2: \(f''(x) = 12x - 6\).

Step 3:
At \(x = 2\): \(f''(2) = 18 > 0 \implies\) local minimum at \((2, -19)\).
At \(x = -1\): \(f''(-1) = -18 < 0 \implies\) local maximum at \((-1, 8)\).

Global (Absolute) Extrema on a Closed Interval \([a, b]\)

Find all stationary points inside \((a, b)\) and evaluate \(f(x)\) at each.
Evaluate \(f(a)\) and \(f(b)\) at the endpoints.
The greatest value is the global maximum; the least is the global minimum.

3. Curve Sketching

The Complete Sketching Checklist

For a function \(y = f(x)\):

Domain: Note restricted values where denominators equal zero.
Intercepts: \(y\)-intercept at \(x = 0\); \(x\)-intercept(s) where \(f(x) = 0\).
Symmetry: If \(f(-x) = f(x)\), it is even (axisymmetric about \(y\)-axis). If \(f(-x) = -f(x)\), it is odd (rotational symmetry about origin).
Asymptotes:
- Vertical Asymptote (VA): Line \(x = c\) where \(\lim_{x \to c^\pm} f(x) = \pm\infty\).
- Horizontal Asymptote (HA): Line \(y = L\) where \(\lim_{x \to \pm\infty} f(x) = L\). (Note: Curves may cross horizontal or oblique asymptotes at finite values of \(x\)).
- Oblique (Slant) Asymptote (OA): Line \(y = mx + c\) where \(\lim_{x \to \pm\infty} [f(x) - (mx + c)] = 0\). For rational functions \(\frac{P(x)}{Q(x)}\) where \(\deg(P) = \deg(Q) + 1\), find it via polynomial long division.
First Derivative & Monotonicity: Use \(f'(x)\) to find stationary points and intervals where the curve is increasing or decreasing.
Concavity & Points of Inflexion:
- \(f''(x) > 0\): Concave upwards.
- \(f''(x) < 0\): Concave downwards.
- Point of Inflexion: A point on the curve where concavity changes sign (where \(f''(x) = 0\) or \(f''(x)\) is undefined, with a sign change in \(f''(x)\)).
Sketch: Combine asymptotes, intercepts, extrema, and concavity into a clear curve.

Quick Example: Sketching \(y = \frac{x^2}{x-2}\)

Domain: \(x \neq 2\).
Intercepts: \((0, 0)\).
Asymptotes: VA is \(x = 2\). Long division gives \(\frac{x^2}{x-2} = x + 2 + \frac{4}{x-2}\), so OA is \(y = x + 2\).
Extrema: \(f'(x) = \frac{x(x-4)}{(x-2)^2} = 0 \implies x = 0, x = 4\). Local maximum at \((0, 0)\); local minimum at \((4, 8)\).
Concavity: \(f''(x) = \frac{8}{(x-2)^3}\). Concave down for \(x < 2\), concave up for \(x > 2\). No points of inflexion since \(x = 2\) is not in the domain.

4. Real-World Applications

Optimization Problems

Define variables, draw a diagram if applicable, and write down the objective function.
Use constraint equations to express the objective function in terms of a single variable.
Find stationary points by setting the first derivative to zero.
Justify the maximum or minimum using the second derivative test (or first derivative test).
Answer with appropriate units.

Example: Maximizing Garden Area

You have \(40\text{ m}\) of fencing to enclose a rectangular garden. Find the maximum possible area.

1. Let length be \(L\) and width be \(W\). Perimeter: \(2L + 2W = 40 \implies L = 20 - W\).
2. Area: \(A(W) = W(20 - W) = 20W - W^2\).
3. Differentiate: \(\frac{dA}{dW} = 20 - 2W = 0 \implies W = 10\text{ m}\).
4. Second derivative: \(\frac{d^2A}{dW^2} = -2 < 0\), confirming a maximum.
5. The maximum area is \(A = 10(20 - 10) = 100\text{ m}^2\).

Rates of Change & Chain Rule

For related rates changing with respect to time \(t\), use the Chain Rule:

\(\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}\)

Key Takeaway

Formulate a single-variable function to optimize, verify nature using derivatives, and use the Chain Rule to link rates of change over time.

Quick check

Can you answer these now?

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

How do you find the equation of a tangent line to a curve \(y = f(x)\) at a given point \((x_1, y_1)\)?

Find the slope \(m_{\text{tangent}} = f'(x_1)\) and use the point-slope form: \(y - y_1 = f'(x_1)(x - x_1)\).

How do you find the gradient and equation of the normal line to a curve at \((x_1, y_1)\)?

The normal slope is \(m_{\text{normal}} = -\frac{1}{f'(x_1)}\) (for \(f'(x_1) \neq 0\)), giving the equation \(y - y_1 = -\frac{1}{f'(x_1)}(x - x_1)\).

How do you determine if a function \(f(x)\) is strictly increasing or strictly decreasing on an interval?

\(f(x)\) is strictly increasing where \(f'(x) > 0\), and strictly decreasing where \(f'(x) < 0\).

How do you classify stationary points using the First Derivative Test?

Examine the sign of \(f'(x)\) across the stationary point: \(+\) to \(-\) indicates a local maximum, \(-\) to \(+\) indicates a local minimum, and no sign change indicates a stationary point of inflexion.

How does the Second Derivative Test classify stationary points?

If \(f'(x_0) = 0\), then \(f''(x_0) > 0\) gives a local minimum, \(f''(x_0) < 0\) gives a local maximum, and \(f''(x_0) = 0\) is inconclusive.

How do you find the global (absolute) maximum and minimum values of a continuous function on a closed interval \([a, b]\)?

Evaluate \(f(x)\) at all stationary points in \((a, b)\) and at the boundary endpoints \(a\) and \(b\); the largest value is the global maximum and the smallest is the global minimum.

How do you find the vertical, horizontal, and oblique asymptotes of a function?

Vertical asymptotes occur where \(\lim_{x \to c} f(x) = \pm\infty\), horizontal asymptotes occur where \(\lim_{x \to \pm\infty} f(x) = L\), and oblique asymptotes occur when \(\lim_{x \to \pm\infty} [f(x) - (mx + c)] = 0\).

What is a point of inflexion and how is it identified?

A point of inflexion is a point on the curve where concavity changes sign, occurring where \(f''(x) = 0\) (or is undefined) and \(f''(x)\) switches sign across the point.

What are the key steps to solve an applied optimization problem in calculus?

Express the objective quantity as a single-variable function using constraints, find stationary points by setting the derivative to zero, and verify the extremum using derivative tests.

How is the Chain Rule applied to solve related rates of change problems with respect to time?

Differentiate the governing geometric or physical formula with respect to time \(t\) using the Chain Rule, such as \(\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}\).

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