Mastering Differentiation (Pure Mathematics 2)

Welcome to the Differentiation chapter of Pure Mathematics 2! If you've already tackled differentiation in P1, you know it's all about finding the rate of change, or the gradient, of a curve.
In P2, we supercharge these skills by introducing new, exciting functions like \(e^x\), \(\ln x\), and advanced trigonometric functions. We will also learn powerful new rules—the Product, Quotient, Implicit, and Parametric rules—which let us differentiate virtually any combination of functions.
Mastering these techniques is crucial, as they form the foundation for Paper 3 Calculus later on!

1. Fundamental P2 Derivatives: Your New Toolkit

In P1, you mostly used the power rule for \(x^n\). For P2, you need to commit these five essential derivatives to memory. They are often tested in combination with the chain rule.

Quick Review: Core Derivatives (\(f(x) \rightarrow f'(x)\))
  • The Exponential Function:
    If \(y = e^x\), then \(\frac{dy}{dx} = e^x\).
    (Analogy: \(e^x\) is the "King of Calculus"—it remains unchanged when differentiated. It is the only function that is its own derivative!)
  • The Natural Logarithm Function:
    If \(y = \ln x\), then \(\frac{dy}{dx} = \frac{1}{x}\).
  • Sine:
    If \(y = \sin x\), then \(\frac{dy}{dx} = \cos x\).
  • Cosine:
    If \(y = \cos x\), then \(\frac{dy}{dx} = -\sin x\).
    (Memory Aid: Any trigonometric function starting with "C" (\(\cos x\), \(\cot x\), \(\csc x\)) has a negative derivative!)
  • Tangent:
    If \(y = \tan x\), then \(\frac{dy}{dx} = \sec^2 x\).
The Composite Function Rule (Chain Rule Revisited)

The Chain Rule is essential for differentiating a function within a function (a composite function). It often appears with the new P2 functions.
If \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \times g'(x)\).

Step-by-Step for Composites:

  1. Differentiate the Outer Function (keeping the inside function exactly the same).
  2. Multiply the result by the derivative of the Inner Function.

Example: Differentiate \(y = e^{4x^2 - 1}\).
1. Outer function is \(e^u\). Derivative is \(e^{4x^2 - 1}\).
2. Inner function is \(4x^2 - 1\). Derivative is \(8x\).
Therefore, \(\frac{dy}{dx} = 8x \cdot e^{4x^2 - 1}\).

Key Takeaway: Master the derivatives of \(e^x\), \(\ln x\), \(\sin x\), \(\cos x\), and \(\tan x\). Always check if you need to apply the Chain Rule to a composite function.

2. Differentiating Products and Quotients

When two functions of \(x\) are multiplied together or divided, you cannot simply differentiate them separately. You must use the Product Rule or the Quotient Rule.

2.1. The Product Rule

Used for differentiating functions in the form \(y = u \cdot v\), where both \(u\) and \(v\) are functions of \(x\).
Formula: \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\) (or written as \(u v' + v u'\))

Process:

  1. Define \(u\) and \(v\), and find their derivatives, \(\frac{du}{dx}\) and \(\frac{dv}{dx}\).
  2. Substitute into the formula.

Example: Find \(\frac{dy}{dx}\) for \(y = x^2 \ln x\).
Let \(u = x^2\), so \(\frac{du}{dx} = 2x\).
Let \(v = \ln x\), so \(\frac{dv}{dx} = \frac{1}{x}\).
\(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = x^2 \left( \frac{1}{x} \right) + (\ln x) (2x)\)
\(\frac{dy}{dx} = x + 2x \ln x\).

2.2. The Quotient Rule

Used for differentiating functions in the form \(y = \frac{u}{v}\), where both \(u\) and \(v\) are functions of \(x\).
Formula: \(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\) (or written as \(\frac{v u' - u v'}{v^2}\))

Memory Aid: Many students remember this as "Low d high minus high d low over low squared." (differentiating the function in the denominator (low) first).

Example: Find \(\frac{dy}{dx}\) for \(y = \frac{2x - 4}{3x + 2}\). (Syllabus Example)
Let \(u = 2x - 4\), so \(\frac{du}{dx} = 2\).
Let \(v = 3x + 2\), so \(\frac{dv}{dx} = 3\).
\(\frac{dy}{dx} = \frac{(3x+2)(2) - (2x-4)(3)}{(3x+2)^2}\)
\(\frac{dy}{dx} = \frac{6x + 4 - (6x - 12)}{(3x+2)^2}\)
\(\frac{dy}{dx} = \frac{16}{(3x+2)^2}\).

Common Mistake to Avoid: In the Quotient Rule, the order matters! Always start the top part of the numerator with the denominator (\(v\)) multiplied by the derivative of the numerator (\(u'\)).

Key Takeaway: Product Rule (\(u v' + v u'\)) for multiplication. Quotient Rule (\(\frac{v u' - u v'}{v^2}\)) for division.

3. Implicit Differentiation

Sometimes, equations defining a curve mix \(x\) and \(y\) together, and it is difficult or impossible to rearrange them into the form \(y = f(x)\). This is an implicit function (e.g., \(x^2 + y^2 = xy + 7\)).
We use Implicit Differentiation to find \(\frac{dy}{dx}\) by differentiating all terms with respect to \(x\).

The Crucial Step for Differentiating \(y\) Terms

Whenever you differentiate a term containing \(y\), you must treat \(y\) as a function of \(x\) and apply the Chain Rule.

If \(f(y)\) is a function of \(y\), then \(\frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx}\).

  • Differentiating \(y^2\) becomes \(2y \cdot \frac{dy}{dx}\).
  • Differentiating \(\sin(y)\) becomes \(\cos(y) \cdot \frac{dy}{dx}\).
Step-by-Step Implicit Differentiation

Example: Find \(\frac{dy}{dx}\) for \(x^2 + y^2 = xy + 7\).

  1. Differentiate term by term with respect to \(x\):
    \(\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(xy) + \frac{d}{dx}(7)\)
    (Note: We use the Product Rule for the \(xy\) term where \(u=x, v=y\))
  2. Apply rules and Chain Rule for \(y\):
    \(2x + 2y \frac{dy}{dx} = \left[ x \cdot \frac{dy}{dx} + y \cdot 1 \right] + 0\)
  3. Collect all \(\frac{dy}{dx}\) terms on one side:
    \(2y \frac{dy}{dx} - x \frac{dy}{dx} = y - 2x\)
  4. Factor out \(\frac{dy}{dx}\) and solve:
    \(\frac{dy}{dx} (2y - x) = y - 2x\)
    \(\frac{dy}{dx} = \frac{y - 2x}{2y - x}\)

Key Takeaway: Implicit differentiation is just the Chain Rule applied to \(y\). Every time you differentiate a \(y\) term, remember to multiply by \(\frac{dy}{dx}\).

4. Parametric Differentiation

In P2, a curve might be defined using a third variable, often \(t\), called the parameter. Instead of \(y = f(x)\), we have \(x\) defined by \(t\) and \(y\) defined by \(t\) (e.g., \(x = 2t^3, y = 3t^2\)).
To find \(\frac{dy}{dx}\), we use a rearrangement of the Chain Rule:

The Parametric Formula

Formula: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\)

Step-by-Step Parametric Differentiation

Example: A curve is defined by \(x = t - e^{2t}\) and \(y = t + e^{2t}\). Find \(\frac{dy}{dx}\). (Syllabus Example)

  1. Find \(\frac{dx}{dt}\):
    \(\frac{dx}{dt} = 1 - 2e^{2t}\) (Remember Chain Rule on \(e^{2t}\))
  2. Find \(\frac{dy}{dt}\):
    \(\frac{dy}{dt} = 1 + 2e^{2t}\)
  3. Apply the formula:
    \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1 + 2e^{2t}}{1 - 2e^{2t}}\)

Did you know? Parametric equations are perfect for describing motion, where \(t\) represents time, giving us the exact position \((x, y)\) at any moment.

Key Takeaway: Differentiate \(x\) and \(y\) separately with respect to the parameter (\(t\)), then divide \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\).

5. Applications of Differentiation

5.1. Tangents and Normals

The core application remains the same as P1, but now you apply it to the complex P2 functions (including implicit and parametric forms).

  • The gradient of the Tangent at a point \((x_0, y_0)\) is \(m_T = \frac{dy}{dx}\) evaluated at that point.
  • The gradient of the Normal (perpendicular to the tangent) is \(m_N = -\frac{1}{m_T}\).
  • Use the straight line equation \(y - y_0 = m(x - x_0)\) to find the equation of the line.
5.2. Connected Rates of Change

This application links different variables changing over time (e.g., how fast the volume of a sphere is increasing if its radius is increasing at a certain rate).

We use the Chain Rule structure to connect rates:

If \(A\) depends on \(B\), and \(B\) depends on time \(t\), then the rate of change of \(A\) with respect to \(t\) is:
$$\frac{dA}{dt} = \frac{dA}{dB} \times \frac{dB}{dt}$$

Example: The radius \(r\) of a circular oil slick is increasing at a rate of \(0.5 \text{ m/s}\) (\(\frac{dr}{dt} = 0.5\)). Find the rate at which the Area \(A\) is increasing when \(r=10 \text{ m}\) (\(\frac{dA}{dt}\)).

  1. Identify the relationship: \(A = \pi r^2\).
  2. Find the link derivative: \(\frac{dA}{dr} = 2\pi r\).
  3. Use the Chain Rule: \(\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}\).
  4. Substitute and calculate: \(\frac{dA}{dt} = (2\pi r) \times (0.5)\).
    When \(r=10\): \(\frac{dA}{dt} = (2\pi \cdot 10) \times 0.5 = 10\pi \text{ m}^2/\text{s}\).

Key Takeaway: Connected Rates problems always rely on linking three variables using two known rates and one derivative calculated from a formula (like Area or Volume).

Quick Review: Key P2 Differentiation Techniques
  • New Derivatives: \(e^x\), \(\ln x\), \(\sin x\), \(\cos x\), \(\tan x\).
  • Product Rule (UV): \(u v' + v u'\).
  • Quotient Rule (U/V): \(\frac{v u' - u v'}{v^2}\).
  • Implicit: Differentiate \(y\) terms, then multiply by \(\frac{dy}{dx}\).
  • Parametric: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).