Differentiation

Pure Mathematics 1: Chapter 1.7 Differentiation Study Notes

Hello! Welcome to the exciting world of Differentiation. This chapter is fundamental—it's essentially the mathematical tool we use to study how quickly things change. If you can master differentiation, you unlock the ability to solve problems about speed, growth, movement, and optimisation.

Ready to jump in? Let’s break down the core concepts for Paper 1!

1. The Concept: Finding the Instantaneous Rate of Change

What is Differentiation? (The Informal Idea)

Imagine driving a car. Your average speed is the total distance divided by the total time. But what if you want your speed at that exact moment? That’s what differentiation tells us: the instantaneous rate of change.

Geometrically, the rate of change of a curve \(y = f(x)\) at a specific point is the gradient of the tangent line at that point.

The syllabus requires an informal understanding that the gradient of the curve is the limit of the gradients of chords as the two endpoints of the chord get infinitely close together. Don't worry about the formal definition using first principles; just understand the idea!

Key Notations:

• The first derivative (the gradient function) is denoted by: \(\frac{dy}{dx}\) or \(f'(x)\).
• The second derivative is denoted by: \(\frac{d^2y}{dx^2}\) or \(f''(x)\). (We use this later to determine the nature of stationary points.)

Did you know? Differentiation forms the backbone of Calculus, developed independently by Isaac Newton and Gottfried Leibniz in the 17th century. It changed mathematics forever!

2. The Rules of Differentiation (Pure Mathematics 1)

In P1, we focus primarily on differentiating polynomials and simple functions involving powers of \(x\).

2.1 The Power Rule

This is the most important rule for P1. If you have a function \(y = x^n\), you find the derivative using the formula:

\[ \frac{d}{dx} (x^n) = nx^{n-1} \]

Step-by-Step Trick:

1. Multiply the term by the power \(n\).
2. Decrease the power by 1.

Examples:

• If \(y = x^5\), then \(\frac{dy}{dx} = 5x^{5-1} = 5x^4\).
• If \(y = 3x^2\), then \(\frac{dy}{dx} = 3(2x^{2-1}) = 6x\).

2.2 Rules for Sums, Differences, and Constants

Differentiation is linear. This means you can differentiate term by term.

• Constant Rule: If \(y = c\) (where \(c\) is a constant number, like 5 or -10), then \(\frac{dy}{dx} = 0\). (A horizontal line has zero gradient.)
• Sum/Difference Rule: If \(y = u \pm v\), then \(\frac{dy}{dx} = \frac{du}{dx} \pm \frac{dv}{dx}\).

Example: If \(y = 4x^3 - 7x + 10\), then:

\[ \frac{dy}{dx} = \frac{d}{dx}(4x^3) - \frac{d}{dx}(7x) + \frac{d}{dx}(10) \]

\[ \frac{dy}{dx} = 12x^2 - 7 + 0 = 12x^2 - 7 \]

2.3 Dealing with Roots and Fractions (Non-Integer Powers)

The power rule works for any rational power \(n\) (positive, negative, or fractional). You must first rewrite the function using index notation.

Essential Rewriting Steps:

• \(\sqrt{x} = x^{1/2}\)
• \(\frac{1}{x^n} = x^{-n}\)
• \(\frac{1}{\sqrt[3]{x}} = x^{-1/3}\)

Example: Differentiate \(y = \frac{2}{x^3} + 5\sqrt{x}\)

Step 1: Rewrite: \(y = 2x^{-3} + 5x^{1/2}\)

Step 2: Differentiate (apply the power rule to each term):

\[ \frac{dy}{dx} = 2(-3)x^{-3-1} + 5(\frac{1}{2})x^{1/2 - 1} \]

\[ \frac{dy}{dx} = -6x^{-4} + \frac{5}{2}x^{-1/2} \]

Step 3: (Optional but neat) Rewrite with positive indices: \(\frac{dy}{dx} = -\frac{6}{x^4} + \frac{5}{2\sqrt{x}}\)

2.4 The Chain Rule (For Composite Functions in P1)

When you have a function inside another function (like \((3x+1)^4\)), you use the Chain Rule. In P1, this typically applies to composite functions of the form \((ax+b)^n\).

The Chain Rule states: If \(y\) is a function of \(u\), and \(u\) is a function of \(x\), then:

\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]

The P1 Simplified Method (Outer and Inner):

If \(y = (ax+b)^n\):

1. Differentiate the outer bracket using the power rule.
2. Multiply the result by the derivative of the inner bracket (which is just \(a\)).

Example: Find \(\frac{dy}{dx}\) for \(y = (5x - 2)^3\)

Outer: Differentiate the power/bracket: \(3(5x - 2)^2\)

Inner: Differentiate the inside (\(5x - 2\)): \(5\)

Result: \(\frac{dy}{dx} = 3(5x - 2)^2 \times 5 = 15(5x - 2)^2\)

3. Applications of Differentiation

3.1 Gradients, Tangents, and Normals

The value of the first derivative \(\frac{dy}{dx}\) at a specific \(x\)-value gives the gradient of the curve at that point.

Steps to find the Equation of a Tangent:

1. Find the coordinates \((x_1, y_1)\) where the tangent meets the curve (if only \(x\) is given, substitute into \(y=f(x)\)).
2. Calculate the gradient \(m_T\) by finding \(\frac{dy}{dx}\) and substituting \(x_1\).
3. Use the straight-line equation form: \(y - y_1 = m_T(x - x_1)\).

Finding the Normal:

The Normal is the line perpendicular (at 90°) to the tangent at that point.

If the gradient of the tangent is \(m_T\), the gradient of the normal, \(m_N\), is:

\[ m_N = -\frac{1}{m_T} \]

Example: If the tangent gradient is 2, the normal gradient is \(-1/2\).

3.2 Increasing and Decreasing Functions

We use the sign of the derivative to describe how the function behaves:

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

4. Stationary Points (Maxima and Minima)

A stationary point (or turning point) is a point on a curve where the gradient is exactly zero. This is where the curve momentarily stops increasing or decreasing.

\[ \text{At a stationary point: } \frac{dy}{dx} = 0 \]

Steps to Locate and Determine the Nature of Stationary Points

Step 1: Find the first derivative, \(\frac{dy}{dx}\).

Step 2: Set \(\frac{dy}{dx} = 0\) and solve for \(x\). These \(x\)-values are the positions of the stationary points.

Step 3: Find the corresponding \(y\)-coordinates by substituting these \(x\)-values back into the original function \(y = f(x)\).

Step 4: Determine the Nature (Is it a maximum, minimum, or neither?):

Method A: Using the Second Derivative Test (\(\frac{d^2y}{dx^2}\))

1. Find the second derivative, \(\frac{d^2y}{dx^2}\).
2. Substitute the \(x\)-value of the stationary point into \(\frac{d^2y}{dx^2}\).

• If \(\frac{d^2y}{dx^2} > 0\), it’s a Minimum Point (U-shape). (Positive means Happy face/Valley.)
• If \(\frac{d^2y}{dx^2} < 0\), it’s a Maximum Point (n-shape). (Negative means Sad face/Hill.)
• If \(\frac{d^2y}{dx^2} = 0\), the test fails, and you must use Method B.

Method B: Using the Sign Test (Required if the second derivative test fails)

We check the sign of the gradient just before and just after the stationary point \(x=a\).

• Maximum: Gradient changes from Positive (+) to Zero (0) to Negative (-). (Slope up, then down.)
• Minimum: Gradient changes from Negative (-) to Zero (0) to Positive (+). (Slope down, then up.)

Note: The P1 syllabus does not require knowledge of Points of Inflexion (where the gradient is zero but the nature test fails, e.g., \(y=x^3\) at \(x=0\)).

Key Takeaway: Stationary points help us find the highest and lowest points of a function, which is crucial for optimisation problems (e.g., maximizing profit or minimizing material needed).

5. Connected Rates of Change

In real-world problems, quantities often change with respect to time (\(t\)). This section links two different rates of change using the Chain Rule structure.

If Quantity A and Quantity B are both changing with respect to time (\(t\)), and A and B are related by a formula, we can find a relationship between their rates:

\[ \frac{dA}{dt} = \frac{dA}{dB} \times \frac{dB}{dt} \]

The term \(\frac{dA}{dB}\) is found by differentiating the formula connecting A and B.

Analogy: Imagine blowing up a balloon. The rate at which you pump air in is \(\frac{dV}{dt}\) (volume change). We want to find how quickly the radius is growing, \(\frac{dr}{dt}\). The connection is the volume formula, \(V = \frac{4}{3}\pi r^3\).

Example Steps (Area of a circle changing):

If the radius of a circle, \(r\), is increasing at a rate of \(0.5 \text{ cm/s}\), find the rate at which the area, \(A\), is increasing when \(r=4\text{ cm}\).

Given rates: \(\frac{dr}{dt} = 0.5\)

Target rate: \(\frac{dA}{dt}\)

Connecting formula: \(A = \pi r^2\)

Step 1: Find the linking derivative: \(\frac{dA}{dr} = 2\pi r\)

Step 2: Use the Chain Rule formula: \(\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}\)

Step 3: Substitute values:

When \(r=4\), \(\frac{dA}{dr} = 2\pi (4) = 8\pi\)

\[ \frac{dA}{dt} = (8\pi) \times (0.5) = 4\pi \text{ cm}^2/\text{s} \]

This shows that differentiation helps us manage multiple related variables changing simultaneously!