Product, quotient and function of a function rules

Introduction to Advanced Differentiation

Welcome to one of the most powerful chapters in Further Pure Mathematics! So far, you have likely learned how to differentiate simple terms like \(x^n\), \(\sin ax\), or \(e^{ax}\) one by one. But what happens when these functions are multiplied together, divided, or tucked inside one another? In this chapter, we will master three essential tools: the Chain Rule, the Product Rule, and the Quotient Rule. These rules allow you to break down complex expressions into smaller, manageable pieces. Think of these as the "Swiss Army Knife" of calculus—once you know how to use them, no function is too difficult to differentiate!

1. The Function of a Function Rule (The Chain Rule)

We use this rule when one function is sitting "inside" another function. For example, in \(y = (3x + 2)^{10}\), the "inner" function is \(3x + 2\) and the "outer" function is "something to the power of 10". The Analogy: The Onion Differentiating a composite function is like peeling an onion. You start with the outer layer, and then move to the inner layer. The Formula: 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 Example: Differentiate \(y = \sin(5x^2)\).
1. Identify the "inside": Let \(u = 5x^2\). This means \(y = \sin u\). 2. Differentiate both: \(\frac{du}{dx} = 10x\) and \(\frac{dy}{du} = \cos u\). 3. Multiply them: \(\frac{dy}{dx} = (\cos u) \times (10x)\). 4. Substitute back: Replace \(u\) with \(5x^2\) to get the final answer: \(\frac{dy}{dx} = 10x \cos(5x^2)\). Common Mistake to Avoid: Students often forget to differentiate the "inside" part. For \(y = (4x + 1)^3\), the answer is NOT just \(3(4x + 1)^2\). You must multiply by the derivative of \(4x + 1\), which is \(4\), giving \(12(4x + 1)^2\).

2. The Product Rule

We use this rule when two functions of \(x\) are multiplied together. For example: \(y = x^2 e^{3x}\). Here, \(x^2\) is one function and \(e^{3x}\) is another. The Formula: If \(y = uv\), where \(u\) and \(v\) are functions of \(x\), then:

\(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\)

Memory Aid: "Left d-Right + Right d-Left" Keep the first part (Left) and multiply by the derivative of the second (d-Right), then add the second part (Right) multiplied by the derivative of the first (d-Left). Step-by-Step Example: Differentiate \(y = x^3 \cos(2x)\).
1. Label your parts: \(u = x^3\) and \(v = \cos(2x)\). 2. Differentiate each: \(\frac{du}{dx} = 3x^2\) and \(\frac{dv}{dx} = -2\sin(2x)\). 3. Plug into the formula:

\(\frac{dy}{dx} = (x^3)(-2\sin(2x)) + (\cos(2x))(3x^2)\)

4. Tidy up: \(\frac{dy}{dx} = -2x^3 \sin(2x) + 3x^2 \cos(2x)\). Key Takeaway: Always list \(u, v, \frac{du}{dx},\) and \(\frac{dv}{dx}\) clearly in a small table before you start. It prevents simple mistakes!

3. The Quotient Rule

We use this rule when one function is divided by another. For example: \(y = \frac{\sin x}{x^2}\). The Formula: If \(y = \frac{u}{v}\), then:

\(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)

Note: This formula is provided on your exam formula sheet, so you don't need to stress about memorizing it perfectly—but you must know how to use it! Memory Aid: "Low d-High minus High d-Low, square the bottom and away we go!" "Low" is the denominator (\(v\)) and "High" is the numerator (\(u\)). Step-by-Step Example: Differentiate \(y = \frac{e^{4x}}{x + 3}\).
1. Label: \(u = e^{4x}\) and \(v = x + 3\). 2. Differentiate: \(\frac{du}{dx} = 4e^{4x}\) and \(\frac{dv}{dx} = 1\). 3. Apply Formula:

\(\frac{dy}{dx} = \frac{(x+3)(4e^{4x}) - (e^{4x})(1)}{(x+3)^2}\)

4. Simplify: Factor out \(e^{4x}\) from the top:

\(\frac{dy}{dx} = \frac{e^{4x}(4x + 12 - 1)}{(x+3)^2} = \frac{e^{4x}(4x + 11)}{(x+3)^2}\)

Common Mistake to Avoid: In the Quotient Rule, the order on top matters because there is a minus sign. Always start with \(v\frac{du}{dx}\) (the bottom times the derivative of the top).

Summary Checklist

When do I use which rule?

• Chain Rule: Use if there is a "bracket" or a function "inside" another (e.g., \(\sin(x^2)\) or \((ax+b)^n\)).
• Product Rule: Use if two different expressions of \(x\) are being multiplied (e.g., \(x^2 \times e^x\)).
• Quotient Rule: Use if one expression of \(x\) is being divided by another (e.g., \(\frac{\sin x}{x}\)).

Quick Review Tips:

• Did you know? The Chain Rule is often used inside the Product or Quotient rules. If you have \(y = x^2 \sin(5x)\), you use the Product Rule first, but when you differentiate \(\sin(5x)\), you use the Chain Rule!
• Simplify at the end: Examiners love it when you factorise your final answer (like taking out common factors of \(e^{ax}\) or \(x^n\)).
• Don't rush: Write down your \(u\) and \(v\) values explicitly. Most marks are lost by trying to do the whole calculation in one head-step.

For more details on differentiating basic functions like \(\sin ax\) or \(e^{ax}\), see the chapter on "Differentiating powers of x, sin ax, cos ax and e^(ax)".