Welcome to Differentiation!
Welcome to one of the most exciting and powerful chapters in Pure Mathematics. If you’ve ever wondered how we can measure exactly how fast something is changing at a single moment—like a car’s speed at exactly 2.5 seconds—then Differentiation is the answer.
Don't worry if this seems a bit abstract at first. We are essentially just finding the gradient (the steepness) of a curve. Once you master a few simple "shortcuts," you’ll find this is one of the most rewarding parts of the course!
1. The Big Idea: Gradient as a Rate of Change
In previous chapters, you learned how to find the gradient of a straight line using the formula \(\frac{y_2 - y_1}{x_2 - x_1}\). But what if the line is curved? The steepness changes at every single point!
Differentiation is the process of finding an equation (called the derivative) that tells us the gradient of a curve at any point \(x\).
- Notation: If our curve is \(y = f(x)\), the derivative is written as \(\frac{dy}{dx}\) or \(f'(x)\).
- Meaning: \(\frac{dy}{dx}\) literally means "the change in \(y\) for a tiny change in \(x\)."
Analogy: Imagine you are driving up a hill. At any specific moment, the "steepness" of the road under your tires is the derivative.
Key Takeaway:
The derivative \(\frac{dy}{dx}\) represents the gradient of the tangent to the curve at a specific point.
2. Differentiation from First Principles
Before we use shortcuts, we need to prove how differentiation works. This is called differentiation from first principles. For AS Level, you only need to know how to do this for small powers of \(x\) (like \(x^2\) or \(x^3\)).
We imagine two points on a curve that are very close together. The distance between them on the x-axis is a tiny amount called \(h\).
The formula is:
\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)
Step-by-Step Example: Differentiating \(f(x) = x^2\)
- Write out the expression: \(\frac{(x+h)^2 - x^2}{h}\)
- Expand the brackets: \(\frac{x^2 + 2xh + h^2 - x^2}{h}\)
- Simplify the top: \(\frac{2xh + h^2}{h}\)
- Divide everything by \(h\): \(2x + h\)
- As \(h\) gets closer and closer to 0 (the limit), we are left with \(2x\).
Quick Tip: In the exam, if a question asks you to differentiate "from first principles," you must show these steps. You cannot just use the shortcut!
3. The Power Rule: The Ultimate Shortcut
Most of the time, you won't need first principles. There is a simple rule for differentiating any term in the form \(ax^n\):
Multiply by the power, then subtract one from the power.
If \(y = x^n\), then \(\frac{dy}{dx} = nx^{n-1}\)
Examples:
- If \(y = x^5\), then \(\frac{dy}{dx} = 5x^4\)
- If \(y = 3x^2\), then \(\frac{dy}{dx} = 6x\) (because \(2 \times 3 = 6\))
- If \(y = 7x\), then \(\frac{dy}{dx} = 7\) (because \(x\) is \(x^1\), and \(x^0 = 1\))
- If \(y = 10\), then \(\frac{dy}{dx} = 0\) (The gradient of a flat constant line is zero!)
Dealing with Fractions and Negatives:
You must be comfortable with your laws of indices here. Always rewrite the expression as \(x^n\) before differentiating.
- Fractions: \(\sqrt{x}\) becomes \(x^{1/2}\). The derivative is \(\frac{1}{2}x^{-1/2}\).
- Denominators: \(\frac{1}{x^2}\) becomes \(x^{-2}\). The derivative is \(-2x^{-3}\).
Common Mistake to Avoid:
When you subtract 1 from a negative power, it goes "further" from zero. For example, the derivative of \(x^{-3}\) is \(-3x^{-4}\), not \(-3x^{-2}\)!
4. Tangents and Normals
Now that we can find the gradient (\(m\)), we can find the equations of specific lines.
- The Tangent: A line that just touches the curve. It has the same gradient as the curve.
- The Normal: A line perpendicular (at 90 degrees) to the tangent. Its gradient is the negative reciprocal: \(-\frac{1}{m}\).
How to find the equation:
- Differentiate the function to get \(\frac{dy}{dx}\).
- Substitute the \(x\) value of your point to find the gradient \(m\).
- Use the straight-line formula: \(y - y_1 = m(x - x_1)\).
5. Stationary Points: Maxima and Minima
A stationary point occurs when the gradient of a curve is zero (\(\frac{dy}{dx} = 0\)). This is where the curve is perfectly flat for a split second—usually at the "top" of a hill or the "bottom" of a valley.
The Second Derivative Test
To find out if a point is a Maximum or a Minimum, we differentiate a second time. We call this \(\frac{d^2y}{dx^2}\) or \(f''(x)\).
- If \(\frac{d^2y}{dx^2} > 0\) (positive), it is a Minimum point (it's turning upwards like a smile).
- If \(\frac{d^2y}{dx^2} < 0\) (negative), it is a Maximum point (it's turning downwards like a frown).
Memory Aid: Positive = Happy face (Minimum point at the bottom). Negative = Sad face (Maximum point at the top).
6. Increasing and Decreasing Functions
Sometimes you just need to know if a function is going up or down.
- A function is increasing when \(f'(x) \geq 0\).
- A function is decreasing when \(f'(x) \leq 0\).
To prove a function is always increasing, you differentiate it and show that the resulting expression is always positive for the given range of \(x\).
7. Sketching Gradient Functions
The exam might ask you to sketch the graph of \(y = f'(x)\) based on the graph of \(y = f(x)\).
Step-by-step logic:
- Find where the original graph is flat (stationary points). On your new graph, these points will be on the x-axis (gradient = 0).
- Find where the original graph is going up. On your new graph, the line will be above the x-axis (positive gradient).
- Find where the original graph is going down. On your new graph, the line will be below the x-axis (negative gradient).
Quick Review Checklist
[ ] Can I differentiate \(x^n\) including negative and fractional powers?
[ ] Do I know the formula for first principles by heart?
[ ] Can I find the equation of a tangent and a normal?
[ ] Do I remember that \(\frac{dy}{dx} = 0\) at a stationary point?
[ ] Can I use the second derivative to identify Maxima and Minima?
Final Encouragement: Differentiation is like a new language. At first, you have to think about the rules, but with practice, "multiplying by the power and subtracting one" will become second nature!