Pure Mathematics P2: Chapter on Differentiation
Welcome to the world of Differentiation! This chapter is one of the foundational pillars of Pure Mathematics. It might seem abstract at first, but it is fundamentally about understanding change—specifically, how fast things change and where those changes lead.
In simple terms, differentiation allows us to find the gradient (or steepness) of a curve at any specific point. If you understand the gradient, you can solve real-world problems involving speed, growth rates, and optimization (finding the biggest or smallest possible result). Ready to dive in? Let's go!
1. The Fundamental Idea of Differentiation
1.1 What is the Derivative?
The derivative is a fancy name for the instantaneous rate of change of a function. Think of driving a car: your average speed might be 50 km/h for the entire trip, but your instantaneous speed (what your speedometer shows) changes every second. Differentiation finds that instantaneous rate of change.
Geometrically, the derivative is the gradient of the tangent line to the curve \(y = f(x)\) at a given point.
Key Notations:
- The derivative of \(y\) with respect to \(x\) is written as \(\frac{dy}{dx}\).
- If you are working with a function \(f(x)\), the derivative is written as \(f'(x)\).
- The second derivative (the derivative of the derivative) is written as \(\frac{d^2y}{dx^2}\) or \(f''(x)\).
(Accessibility Note: Don't worry, you are not expected to differentiate from first principles—just learn and apply the rules!)
Key Takeaway:
The derivative \(\frac{dy}{dx}\) is the formula for calculating the gradient of the curve \(y\) at any point \(x\). It tells us the rate of change.
2. Rules for Polynomials and Powers (P1.3)
The most basic rule you must master is the Power Rule. It applies to any term in the form \(ax^n\), where \(n\) is a rational number (it can be positive, negative, or a fraction).
2.1 The Power Rule
If \(y = ax^n\), then the derivative is given by:
$$\frac{dy}{dx} = anx^{n-1}$$
Step-by-Step Trick:
- Multiply: Bring the existing power (\(n\)) down and multiply it by the coefficient (\(a\)).
- Reduce: Decrease the power by 1 (\(n-1\)).
Example: If \(y = 5x^4\).
1. Multiply: \(5 \times 4 = 20\).
2. Reduce: \(4 - 1 = 3\).
So, \(\frac{dy}{dx} = 20x^3\).
2.2 Handling Different Forms
Before you use the Power Rule, you must ensure every term is written in the form \(ax^n\). This is a common place for students to make mistakes!
- Fractions: Use negative indices.
e.g., If \(y = \frac{3}{x^2}\), rewrite it as \(y = 3x^{-2}\). Then \(\frac{dy}{dx} = (3)(-2)x^{-3} = -6x^{-3}\) or \(-\frac{6}{x^3}\). - Roots: Use fractional indices.
e.g., If \(y = \sqrt{x}\), rewrite it as \(y = x^{1/2}\). Then \(\frac{dy}{dx} = \frac{1}{2}x^{-1/2}\) or \(\frac{1}{2\sqrt{x}}\). - Constants: If \(y = c\) (where \(c\) is a constant, like \(y=7\)), the gradient is zero.
\(\frac{d}{dx}(c) = 0\). (A horizontal line has zero slope!)
2.3 Sums and Differences
If a function is a sum or difference of several terms (a polynomial), you differentiate each term separately:
If \(f(x) = g(x) + h(x)\), then \(f'(x) = g'(x) + h'(x)\).
Example: If \(y = 2x^3 - 4x + 9\), then \(\frac{dy}{dx} = 6x^2 - 4 + 0 = 6x^2 - 4\).
Key Takeaway:
Always rewrite the function using indices (\(x^n\)) before applying the Power Rule. Differentiate term by term.
3. Standard Function Derivatives (P2.6)
When you move into A-Level mathematics, you need to know how to differentiate trigonometric, exponential, and logarithmic functions.
Important Note on Linear Combinations: If the input of the function is a linear term, \(kx\), you differentiate as normal, and then multiply the result by the constant \(k\).
Standard Derivatives Table:
Function \(f(x)\) | Derivative \(f'(x)\)
\(e^x\) | \(e^x\)
\(e^{kx}\) | \(ke^{kx}\)
\(\ln x\) | \(\frac{1}{x}\)
\(\sin x\) | \(\cos x\)
\(\sin kx\) | \(k\cos kx\)
\(\cos x\) | \(-\sin x\)
\(\cos kx\) | \(-k\sin kx\)
\(\tan x\) | \(\sec^2 x\)
Did you know? The function \(y = e^x\) is the only function (up to a constant multiple) that is its own derivative. It perfectly models exponential growth!
Key Takeaway:
Memorise the standard derivatives. Remember to multiply by the constant \(k\) if the function is \(f(kx)\).
4. Essential Differentiation Rules (P2.6)
Sometimes functions are combined in tricky ways—multiplied together, divided, or nested inside one another. For these, we use three essential rules.
4.1 The Chain Rule (Function of a Function)
Use this rule when you have a function nested inside another function, like a Russian doll. This is often written as \(y = f(g(x))\).
If \(y\) is a function of \(u\), and \(u\) is a function of \(x\), then:
$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$
Step-by-Step Process:
- Identify the "outside" function (the main operation, e.g., raising to a power).
- Identify the "inside" function (\(u\)).
- Differentiate the outside, leaving the inside alone (\(\frac{dy}{du}\)).
- Multiply by the derivative of the inside (\(\frac{du}{dx}\)).
Example: Differentiate \(y = (x^2 + 5)^3\).
1. Outside: \(u^3\). Inside: \(u = x^2 + 5\).
2. \(\frac{dy}{du} = 3u^2\).
3. \(\frac{du}{dx} = 2x\).
4. \(\frac{dy}{dx} = (3u^2) \times (2x) = 3(x^2 + 5)^2 (2x) = 6x(x^2 + 5)^2\).
4.2 The Product Rule
Use this when differentiating the product of two functions, \(y = uv\).
$$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$
Mnemonic: Differentiate the first part, leave the second part alone. THEN, leave the first part alone, and differentiate the second part. Add them together.
Example: Differentiate \(y = x^2 \sin x\).
Let \(u = x^2\), so \(\frac{du}{dx} = 2x\).
Let \(v = \sin x\), so \(\frac{dv}{dx} = \cos x\).
$$\frac{dy}{dx} = (x^2)(\cos x) + (\sin x)(2x)$$
4.3 The Quotient Rule
Use this when differentiating a fraction where both the numerator and the denominator are functions of \(x\), \(y = \frac{u}{v}\).
$$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$
Mnemonic: "Low D High, minus High D Low, over the square of the Low." (Where Low = \(v\) and High = \(u\)).
Common Mistake to Avoid: The order matters in the Quotient Rule because there is a minus sign. Always start with \(v\) (the denominator).
Key Takeaway:
The Chain, Product, and Quotient Rules are your tools for handling combined functions. Identify which rule you need first!
5. Advanced Techniques (P2.6)
5.1 Implicit Differentiation
Sometimes we have equations where \(y\) cannot easily be written as a function of \(x\) (e.g., \(x^2 + y^2 = 25\)). When we differentiate these, we use implicit differentiation.
The Rule: When differentiating any term involving \(y\), differentiate the term normally with respect to \(y\), but then multiply the result by \(\frac{dy}{dx}\).
Example: Differentiate \(y^3 + 2xy = 5\).
1. \(\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}\)
2. \(\frac{d}{dx}(2xy)\) requires the Product Rule!
\(u=2x\), \(\frac{du}{dx}=2\). \(v=y\), \(\frac{dv}{dx}=1 \cdot \frac{dy}{dx}\).
Product Rule result: \((2x)(\frac{dy}{dx}) + (y)(2) = 2x\frac{dy}{dx} + 2y\).
3. \(\frac{d}{dx}(5) = 0\).
4. Putting it together and solving for \(\frac{dy}{dx}\):
$$3y^2 \frac{dy}{dx} + 2x\frac{dy}{dx} + 2y = 0$$
$$\frac{dy}{dx}(3y^2 + 2x) = -2y$$
$$\frac{dy}{dx} = \frac{-2y}{3y^2 + 2x}$$
(Syllabus Alert: You will only be asked for the first derivative of implicitly defined curves.)
5.2 Parametric Differentiation
If \(x\) and \(y\) are both expressed in terms of a third parameter (usually \(t\) or \(\theta\)), we use the following relationship to find \(\frac{dy}{dx}\):
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} \text{ or } \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$$
Process:
- Find \(\frac{dx}{dt}\).
- Find \(\frac{dy}{dt}\).
- Divide the two results.
Example: If \(x = t^2\) and \(y = 2t + 1\).
1. \(\frac{dx}{dt} = 2t\).
2. \(\frac{dy}{dt} = 2\).
3. \(\frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t}\).
Key Takeaway:
Implicit differentiation requires multiplying by \(\frac{dy}{dx}\) every time you differentiate a \(y\) term. Parametric differentiation uses the division formula to link \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\).
6. Applications of Differentiation (P1.3 & P2.6)
The derivative is not just a theoretical quantity; it has huge practical applications, particularly in finding equations of lines and locating turning points.
6.1 Tangents and Normals
We use the derivative \(\frac{dy}{dx}\) to find the gradient of the tangent at a specific point \((x_1, y_1)\). The equation of the line is then found using \(y - y_1 = m(x - x_1)\).
- Gradient of Tangent (\(m_T\)): Find \(\frac{dy}{dx}\) and substitute the \(x\)-coordinate.
- Gradient of Normal (\(m_N\)): The normal is perpendicular to the tangent, so its gradient is the negative reciprocal: \(m_N = -\frac{1}{m_T}\).
Quick Review: Remember from Coordinate Geometry that if two lines are perpendicular, the product of their gradients is \(-1\).
6.2 Stationary Points (Maxima and Minima)
A stationary point (also called a critical point or turning point) occurs where the gradient is zero—the curve momentarily flattens out.
To find stationary points:
Step 1: Find \(\frac{dy}{dx}\).
Step 2: Set \(\frac{dy}{dx} = 0\) and solve for \(x\). These are the \(x\)-coordinates of the stationary points.
Step 3: Substitute these \(x\)-values back into the original equation \(y = f(x)\) to find the corresponding \(y\)-coordinates.
6.3 Classifying Stationary Points (Second Order Derivative)
Once you have found the stationary points, you need to know if they are maxima (peaks) or minima (troughs). We use the second derivative, \(\frac{d^2y}{dx^2}\).
Test for Maxima/Minima:
- If \(\frac{d^2y}{dx^2} < 0\), the point is a Maximum (Think of a concave/sad shape).
- If \(\frac{d^2y}{dx^2} > 0\), the point is a Minimum (Think of a convex/happy shape).
- If \(\frac{d^2y}{dx^2} = 0\), the test is inconclusive (it might be a point of inflection, but you won't be tested on this concept).
6.4 Increasing and Decreasing Functions
Differentiation also tells us where a function is rising or falling.
- The function is Increasing if the gradient is positive: \(\frac{dy}{dx} > 0\).
- The function is Decreasing if the gradient is negative: \(\frac{dy}{dx} < 0\).
Key Takeaway:
Differentiation is used to solve practical problems (optimization) by finding where the gradient is zero (\(\frac{dy}{dx}=0\)) and classifying those points using the second derivative (\(\frac{d^2y}{dx^2}\)).