Welcome to the World of Differentiation!
Hello there, future mathematician! You are about to enter one of the most powerful and exciting areas of Pure Mathematics: Differentiation.
If algebra is about finding unknown quantities, differentiation is about finding rates of change. Essentially, it tells you how steep a line is, or how fast something is moving, at any precise moment. This knowledge is crucial for physics, economics, engineering, and much more.
Don't worry if the symbols look intimidating at first. We will break down this chapter step-by-step using clear language and practical examples. Let’s get started!
Section 1: The Idea of the Derivative (The Steepness Finder)
1.1 Gradient of a Curve vs. Gradient of a Line
You already know how to find the gradient (slope) of a straight line, \(m\). It’s constant everywhere! But what about a curve, like \(y = x^2\)?
The steepness of a curve is constantly changing. At one point, it might be flat; at another, it might be incredibly steep.
The Key Concept: When we differentiate a function, we find a new function called the Derivative. This derivative function allows us to calculate the exact gradient of the tangent line at any single point on the original curve.
Notation Check:
- If the original function is \(y\), the derivative is written as \(\mathbf{\frac{dy}{dx}}\).
- If the original function is \(f(x)\), the derivative is written as \(\mathbf{f'(x)}\) (pronounced "f prime of x").
1.2 Differentiation from First Principles (Optional Deep Dive)
Differentiation is fundamentally based on limits. This section explains *why* the rules work, by looking at the gradient of a very, very small interval.
Imagine you have a point P on a curve and a second point Q very close to P. The gradient of the chord PQ is an approximation of the gradient at P.
We let the horizontal distance between P and Q, often called \(\delta x\) or \(h\), shrink towards zero. The gradient of the chord then becomes the gradient of the tangent at P.
The formal definition of the derivative is:
$$\mathbf{f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}}$$
Don't worry! While you might be asked to prove the derivative of a simple function (like \(y=x^2\)) using this formula, most of the time you will use the much faster rules below!
Quick Review: What is Differentiation?
It is the process of finding the instantaneous rate of change (the exact steepness/gradient) of a function at any given point.
Section 2: The Rules of Differentiation (The Power Rule)
This is the core mechanic of P1 Differentiation. If you master this rule, you can differentiate any polynomial function!
2.1 The Power Rule: Differentiating \(x^n\)
The rule for differentiating a term where \(n\) is a constant power (positive, negative, whole number, or fraction) is:
$$\text{If } y = x^n, \text{ then } \mathbf{\frac{dy}{dx} = n x^{n-1}}$$
Memory Aid:
"Multiply by the power, then reduce the power by one."
Step-by-Step Example: Differentiate \(y = x^4\).
- Identify the power, \(n = 4\).
- Multiply the term by the power: \(4x^4\).
- Reduce the power by one: \(4 - 1 = 3\).
- Result: \(\frac{dy}{dx} = 4x^3\).
2.2 Differentiating \(ax^n\) and Constants
If there is a coefficient \(a\) in front of \(x^n\), we just keep multiplying it:
$$\text{If } y = ax^n, \text{ then } \mathbf{\frac{dy}{dx} = anx^{n-1}}$$
Did You Know?
If \(y = 5x\), this is the same as \(y = 5x^1\). Differentiating gives \(1 \times 5x^{1-1} = 5x^0\). Since \(x^0 = 1\), the derivative is just 5. This makes sense: the gradient of the straight line \(y=5x\) is 5!
The Constant Rule:
If \(y\) is just a constant (a number with no \(x\)), its gradient is zero.
$$\text{If } y = k, \text{ then } \mathbf{\frac{dy}{dx} = 0}$$
Analogy: A constant function, like \(y=7\), is a perfectly horizontal line. Horizontal lines have zero steepness.
2.3 Differentiating Sums and Differences (Polynomials)
We can differentiate polynomials term by term. If a function is made up of several terms added or subtracted, just apply the rule to each part individually.
Example: Differentiate \(y = 3x^5 - 2x^2 + 4x - 10\).
- Term 1 (\(3x^5\)): \(5 \times 3x^{5-1} = 15x^4\)
- Term 2 (\(-2x^2\)): \(2 \times (-2)x^{2-1} = -4x^1 = -4x\)
- Term 3 (\(4x\)): Differentiates to \(4\)
- Term 4 (\(-10\)): Differentiates to \(0\)
Result: \(\frac{dy}{dx} = 15x^4 - 4x + 4\)
2.4 Preparation is Key: Rewriting Terms
This is often the trickiest part for students! The Power Rule only works on terms of the form \(ax^n\). If you see roots or fractions, you must rewrite them using index laws first.
Important Prerequisite: Index Laws
Do NOT differentiate until the function looks like \(ax^n\).
- Roots: \(\mathbf{\sqrt{x} = x^{1/2}}\) and \(\mathbf{\sqrt[3]{x^2} = x^{2/3}}\)
- Fractions: \(\mathbf{\frac{1}{x^n} = x^{-n}}\) (Move the term up and change the sign of the power)
- Example: \(\mathbf{\frac{5}{x^3} = 5x^{-3}}\)
- Example: \(\mathbf{\frac{1}{2\sqrt{x}} = \frac{1}{2} x^{-1/2}}\)
Full Example of Rewriting: Differentiate \(y = \frac{4}{x^2} + 3\sqrt{x}\).
- Rewrite: \(y = 4x^{-2} + 3x^{1/2}\) (Notice we haven't differentiated yet!)
- Differentiate Term 1: \(\frac{d}{dx}(4x^{-2}) = (-2) \times 4 x^{-2-1} = -8x^{-3}\)
- Differentiate Term 2: \(\frac{d}{dx}(3x^{1/2}) = (1/2) \times 3 x^{1/2 - 1} = \frac{3}{2} x^{-1/2}\)
- Result: \(\frac{dy}{dx} = -8x^{-3} + \frac{3}{2} x^{-1/2}\)
- (Optional cleanup, return to original notation): \(\frac{dy}{dx} = -\frac{8}{x^3} + \frac{3}{2\sqrt{x}}\)
Common Mistake Alert!
If you have expressions like \(\frac{x^3 + 5}{x}\), you must split the fraction first before differentiating:
$$y = \frac{x^3}{x} + \frac{5}{x} = x^2 + 5x^{-1}$$
THEN differentiate: \(\frac{dy}{dx} = 2x - 5x^{-2}\).
Section 3: Applications of Differentiation – Gradients, Tangents, and Normals
3.1 Finding the Gradient at a Specific Point
Once you have the derivative, \(f'(x)\), finding the gradient at a point \(x=a\) is simple: substitute \(a\) into \(f'(x)\).
Example: Find the gradient of \(y = x^3 - 4x\) at the point where \(x=2\).
- Differentiate: \(\frac{dy}{dx} = 3x^2 - 4\).
- Substitute \(x=2\): \(\frac{dy}{dx}\bigg|_{x=2} = 3(2)^2 - 4 = 3(4) - 4 = 12 - 4 = 8\).
The gradient of the curve at \(x=2\) is 8.
3.2 The Tangent Equation
The tangent is a straight line that touches the curve at exactly one point and has the exact same gradient as the curve at that point.
We use the standard equation of a straight line: \(\mathbf{y - y_1 = m(x - x_1)}\).
Step-by-Step: Finding the Equation of the Tangent
- Find the Coordinates \((x_1, y_1)\): Use the original equation \(y = f(x)\) to find \(y_1\) for the given \(x_1\).
- Find the Gradient \(m\): Differentiate the function to find \(\frac{dy}{dx}\), then substitute \(x_1\) to find the numerical gradient \(m\).
- Substitute and Solve: Plug \(m\), \(x_1\), and \(y_1\) into \(y - y_1 = m(x - x_1)\).
3.3 The Normal Equation
The normal is a straight line that is perpendicular (at 90 degrees) to the tangent at the point of contact.
If the gradient of the tangent is \(m_{tangent}\), the gradient of the normal, \(m_{normal}\), is the negative reciprocal.
$$\mathbf{m_{normal} = -\frac{1}{m_{tangent}}}$$
Example: If the tangent gradient is \(m_T = 5\), then the normal gradient is \(m_N = -\frac{1}{5}\). If the tangent gradient is \(m_T = -\frac{2}{3}\), then \(m_N = +\frac{3}{2}\).
To find the Normal equation, follow the three steps above, but use \(m_{normal}\) in the final step instead of \(m_{tangent}\).
Section 4: Stationary Points (Turning Points)
4.1 What are Stationary Points?
Stationary points (or turning points) are crucial features on a curve. They occur where the function momentarily stops increasing or decreasing.
At a stationary point, the tangent line is perfectly horizontal, meaning the gradient is zero.
$$\mathbf{\text{At a stationary point, } \frac{dy}{dx} = 0}$$
Stationary points can be classified into three types:
- Local Maximum: A peak (the curve goes up, stops, then goes down).
- Local Minimum: A trough (the curve goes down, stops, then goes up).
- Point of Inflection: Stops changing direction, but continues in the same general direction (less common in P1 basic examples).
Step-by-Step: Finding Stationary Points
- Differentiate: Find the derivative \(\frac{dy}{dx}\).
- Set to Zero: Set \(\frac{dy}{dx} = 0\) and solve the resulting equation for \(x\). These are the critical \(x\)-values.
- Find Coordinates: Substitute the \(x\)-values back into the original function \(y\) to find the corresponding \(y\)-coordinates.
4.2 Determining the Nature of Stationary Points
After finding the coordinates, we need to know if they are maxima or minima. In P1, we usually use the Second Derivative Test.
A) The Second Derivative (\(\frac{d^2y}{dx^2}\) or \(f''(x)\))
The second derivative is simply the derivative of the first derivative.
$$\mathbf{\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right)}$$
This tells us about the *rate of change of the gradient*.
B) The Second Derivative Test
Substitute the \(x\)-coordinate of the stationary point into \(\frac{d^2y}{dx^2}\).
| Condition | Result | Nature |
|---|---|---|
| \(\frac{d^2y}{dx^2} > 0\) (Positive) | The curve is smiling (concave up) | Local Minimum |
| \(\frac{d^2y}{dx^2} < 0\) (Negative) | The curve is frowning (concave down) | Local Maximum |
Memory Trick:
If the result is Positive (\(+\)), it looks like the bottom of a bowl: Minimum.
If the result is Negative (\(-\)), it looks like the top of a hill: Maximum.
Example: Determine the nature of the stationary points of \(y = x^3 - 3x\).
- Find \(\frac{dy}{dx}\): \(\frac{dy}{dx} = 3x^2 - 3\).
- Find Stationary Points (\(\frac{dy}{dx} = 0\)): $$3x^2 - 3 = 0 \implies 3x^2 = 3 \implies x^2 = 1$$ Stationary points occur at \(x = 1\) and \(x = -1\).
- Find \(\frac{d^2y}{dx^2}\): Differentiate \(3x^2 - 3\). $$\frac{d^2y}{dx^2} = 6x$$
- Test the Points:
- At \(x = 1\): \(\frac{d^2y}{dx^2} = 6(1) = 6\). Since \(6 > 0\) (Positive), this is a Local Minimum.
- At \(x = -1\): \(\frac{d^2y}{dx^2} = 6(-1) = -6\). Since \(-6 < 0\) (Negative), this is a Local Maximum.
- Final Coordinates (Optional but good practice): \(y(1) = 1^3 - 3(1) = -2\). Minimum at \((1, -2)\). \(y(-1) = (-1)^3 - 3(-1) = 2\). Maximum at \((-1, 2)\).
Alternative P1 Method (Sign Change Test):
If the second derivative test gives 0 (or if you prefer), you can check the sign of the gradient \(\frac{dy}{dx}\) just before and just after the stationary point \(x=a\):
- Maximum: Gradient goes from Positive \((+)\) to Zero \((0)\) to Negative \((\mathbf{-})\).
- Minimum: Gradient goes from Negative \((-\)) to Zero \((0)\) to Positive \((\mathbf{+})\).
Summary: Your Differentiation Checklist
You now have all the tools necessary for the Differentiation chapter in P1! Remember these key takeaways:
- Prepare the Function: Convert all roots and fractions to \(ax^n\) form (e.g., \(\frac{1}{x} = x^{-1}\)).
- Power Rule: Multiply by the power, reduce the power by one. \(\frac{d}{dx}(ax^n) = anx^{n-1}\).
- Tangents: Use \(\frac{dy}{dx}\) to find the gradient \(m\), then use \(y - y_1 = m(x - x_1)\).
- Normals: The gradient is the negative reciprocal: \(m_N = -1/m_T\).
- Stationary Points: Set \(\frac{dy}{dx} = 0\) to find the \(x\)-coordinates.
- Nature: Use \(\frac{d^2y}{dx^2}\). Positive result = Minimum. Negative result = Maximum.
Keep practicing those index laws, and differentiation will become second nature! Good luck!