Welcome to FP1 Calculus!
In your standard Mathematics course, you’ve already learned how to differentiate and integrate functions. In Further Mathematics, we take those skills and look "under the hood" to see how they actually work. We will explore how to find gradients from scratch, how tiny changes in one variable affect another, and what happens when we try to find the area under a curve that goes on forever!
Don't worry if some of these ideas seem a bit "infinite" at first—we’ll break them down step-by-step.
1. Differentiation from First Principles
Usually, when you see \(x^2\), you just say the derivative is \(2x\). But where does that come from? Differentiation from first principles is the formal way to find the gradient of a tangent to a curve by looking at a chord that gets shorter and shorter.
How it works:
Imagine two points on a curve, \(P\) and \(Q\).
Point \(P\) is at \((x, f(x))\).
Point \(Q\) is a tiny bit further along, at \((x+h, f(x+h))\).
The distance between them on the x-axis is just \(h\). If we find the gradient of the straight line (the chord) joining \(P\) and \(Q\), and then make \(h\) so small that it basically becomes zero, we find the exact gradient at point \(P\).
The Formula:
\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)
Step-by-Step Example:
Let's find the gradient of \(f(x) = x^2 - 2x\).
1. Find \(f(x+h)\):
\((x+h)^2 - 2(x+h) = x^2 + 2xh + h^2 - 2x - 2h\)
2. Subtract \(f(x)\):
\((x^2 + 2xh + h^2 - 2x - 2h) - (x^2 - 2x) = 2xh + h^2 - 2h\)
3. Divide by \(h\):
\(\frac{2xh + h^2 - 2h}{h} = 2x + h - 2\)
4. Take the limit as \(h \to 0\):
As \(h\) disappears, we are left with \(2x - 2\). Success!
Quick Review: Always remember to expand your brackets carefully! The \(x^2\) terms should always cancel out, leaving you with only terms containing \(h\).
Key Takeaway: Differentiation is just finding the gradient of a chord as the distance between the two points (\(h\)) shrinks to zero.
2. Connected Rates of Change
Sometimes, variables are linked like a chain. For example, if you pump air into a balloon, the volume increases, which causes the radius to increase, which depends on time.
We use the Chain Rule to link these changes together:
\(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\)
Analogy:
Think of it like currency conversion. If 1 GBP = 1.20 EUR, and 1 EUR = 1.10 USD, then to find how many USD you get for 1 GBP, you multiply the rates together! Calculus does the same with rates of change.
Example:
If \(p = kv^{4/3}\), then \(\frac{dp}{dv} = \frac{4}{3}kv^{1/3}\).
If we know how fast \(v\) is changing (\(\frac{dv}{dt}\)), we can find how fast \(p\) is changing (\(\frac{dp}{dt}\)) by multiplying them.
Common Mistake: Make sure you have your "fractions" the right way up so the terms seem to "cancel out" diagonally!
3. Small Changes (Approximations)
If we make a tiny change in \(x\) (called \(\delta x\)), we can estimate the tiny change in \(y\) (called \(\delta y\)) without doing a full calculation.
The Formula:
\(\delta y \approx \frac{dy}{dx} \times \delta x\)
This works because, for a very small distance, the curve is almost a straight line (the tangent).
Example:
Given \(h = 20x^{-2}\), find the approximate change in \(h\) when \(x\) changes by a small amount \(\delta x\).
1. Differentiate: \(\frac{dh}{dx} = -40x^{-3}\).
2. Apply formula: \(\delta h \approx -40x^{-3} \times \delta x\).
Did you know? This technique is used by engineers to calculate "tolerances"—for example, how much a metal beam might expand or contract if the temperature changes slightly.
Key Takeaway: \(\delta y\) is the approximate change. The smaller the \(\delta x\), the more accurate your estimate will be.
4. Improper Integrals
In standard maths, integrals have clear start and end points. An improper integral is one where "infinity" gets involved. There are two main types you need to know:
Type 1: Infinite Limits
This is when the area goes on forever along the x-axis, like \(\int_{4}^{\infty} x^{-3/2} dx\).
To solve this, we replace the \(\infty\) with a letter (like \(t\)), integrate normally, and then see what happens as \(t\) gets huge.
Example: \(\int_{4}^{\infty} x^{-3/2} dx = [-2x^{-1/2}]_{4}^{\infty}\)
As \(x \to \infty\), \(\frac{-2}{\sqrt{x}}\) becomes \(0\).
So the area is \(0 - (-2/\sqrt{4}) = 1\).
Type 2: Undefined at a Limit
This is when the function itself shoots up to infinity at one of the boundaries, like \(\int_{0}^{4} \frac{1}{\sqrt{x}} dx\).
Here, the function is "broken" at \(x = 0\) (you can't divide by zero!). We integrate and then evaluate the limit as we get closer and closer to \(0\).
Don't worry if this seems tricky! Just treat the limit like a normal number during the integration phase. The magic happens at the very last step when you ask: "What happens to this value as the number gets infinitely large (or small)?"
Key Takeaway: If an improper integral results in a finite number, we say it converges. If it results in infinity, we say it diverges.
Summary Checklist
1. First Principles: Can you use \(\lim_{h \to 0}\) to differentiate \(x^2\) or \(x^4\)?
2. Connected Rates: Can you link \(\frac{dy}{dx}\) and \(\frac{dx}{dt}\) to find \(\frac{dy}{dt}\)?
3. Small Changes: Can you use \(\delta y \approx f'(x) \delta x\)?
4. Improper Integrals: Can you identify if a limit is infinite or makes the function undefined?