Welcome to Differentiation
Welcome to one of the most exciting and powerful areas of Pure Mathematics! If you have ever wondered how a speedometer calculates your exact speed at a single split-second, or how engineers design roller coasters with perfectly smooth drops, the answer is differentiation.
At its heart, differentiation is all about measuring rates of change. In GCSE mathematics, you found the gradient of straight lines. In AS Level Mathematics, differentiation gives you the superpower to find the exact slope or steepness of any curved graph at any given point.
Don't worry if this seems tricky at first! We will break every concept down into small, bite-sized steps with clear examples, handy memory tricks, and common pitfalls to watch out for.
1. The Foundation: Differentiation from First Principles
To understand where the rules of calculus come from, we look at first principles. Let's look at how we move from finding the slope of a straight chord to the slope of a curve at a single point.
The Big Idea: From Chord to Tangent
Imagine picking a point \(P(x, f(x))\) on a curve. Now pick a second point slightly further along the curve, \(Q(x+h, f(x+h))\), where \(h\) represents a tiny step in the horizontal direction.
The gradient of the straight line joining \(P\) and \(Q\) (called a secant line or chord) is:
\(\text{Gradient of chord } PQ = \frac{\text{Change in } y}{\text{Change in } x} = \frac{f(x+h) - f(x)}{(x+h) - x} = \frac{f(x+h) - f(x)}{h}\)
Now imagine letting that tiny step \(h\) shrink down towards zero (\(h \to 0\)). As point \(Q\) slides closer and closer to point \(P\), the chord turns into the exact tangent to the curve at \(P\).
The Formal Definition
The derivative of \(f(x)\), denoted as \(f'(x)\) or \(\frac{\text{d}y}{\text{d}x}\), is formally defined as:
\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)
The notation \(\lim_{h \to 0}\) means "the limit as \(h\) tends to zero".
Step-by-Step Example: Differentiating from First Principles
Problem: Prove from first principles that the derivative of \(f(x) = x^2\) is \(2x\).
Step 1: Write down the expression for \(f(x+h)\):
\(f(x+h) = (x+h)^2 = x^2 + 2xh + h^2\)
Step 2: Substitute \(f(x+h)\) and \(f(x)\) into the definition formula:
\(f'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2) - x^2}{h}\)
Step 3: Simplify the numerator:
\(f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h}\)
Step 4: Divide every term in the numerator by \(h\):
\(f'(x) = \lim_{h \to 0} (2x + h)\)
Step 5: Take the limit by setting \(h = 0\):
\(f'(x) = 2x + 0 = 2x\)
Key Takeaway: Differentiation from first principles finds the gradient of a tangent by shrinking the horizontal gap \(h\) between two points to zero.
2. The Core Rules of Differentiation
While first principles shows us why differentiation works, we usually use shortcut rules to differentiate quickly and efficiently.
The Power Rule
If \(y = x^n\), where \(n\) is any real number, then:
\(\frac{\text{d}y}{\text{d}x} = n x^{n-1}\)
Memory Trick: "Bring the power to the front, and knock one off the power."
Multiplying by a Constant and Sum Rule
• If \(y = a x^n\), then \(\frac{\text{d}y}{\text{d}x} = a \cdot n x^{n-1}\)
• If \(y = f(x) \pm g(x)\), differentiate each term separately: \(\frac{\text{d}y}{\text{d}x} = f'(x) \pm g'(x)\)
• The derivative of a constant number alone is always \(0\) (e.g., if \(y = 7\), then \(\frac{\text{d}y}{\text{d}x} = 0\)).
Crucial Skill: Rewriting Terms into Index Form
Before differentiating, you must write all terms in the form \(a x^n\). Do not try to differentiate fractions or roots directly at AS Level!
• Fractions: \(\frac{1}{x} = x^{-1}\), \(\frac{3}{x^2} = 3x^{-2}\), \(\frac{5}{2x^3} = \frac{5}{2}x^{-3}\)
• Roots: \(\sqrt{x} = x^{\frac{1}{2}}\), \(\sqrt[3]{x} = x^{\frac{1}{3}}\), \(\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}\)
• Products / Quotients: Expand brackets and split single-denominator fractions before differentiating.
Worked Example: Differentiating Mixed Expressions
Problem: Differentiate \(y = 4x^3 - \frac{2}{x^2} + 6\sqrt{x} - 5\) with respect to \(x\).
Step 1: Rewrite in power form:
\(y = 4x^3 - 2x^{-2} + 6x^{\frac{1}{2}} - 5\)
Step 2: Apply the power rule to each term individually:
\(\frac{\text{d}y}{\text{d}x} = 4(3)x^{3-1} - 2(-2)x^{-2-1} + 6\left(\frac{1}{2}\right)x^{\frac{1}{2}-1} - 0\)
Step 3: Simplify the powers and coefficients:
\(\frac{\text{d}y}{\text{d}x} = 12x^2 + 4x^{-3} + 3x^{-\frac{1}{2}}\)
Step 4: (Optional) Rewrite with positive powers and roots:
\(\frac{\text{d}y}{\text{d}x} = 12x^2 + \frac{4}{x^3} + \frac{3}{\sqrt{x}}\)
Common Mistake to Avoid: When subtracting \(1\) from a negative power, remember it becomes more negative! For example, \(-2 - 1 = -3\), not \(-1\).
Key Takeaway: Always rewrite roots and algebraic fractions into the form \(a x^n\) before differentiating.
3. Tangents and Normals to Curves
One of the most practical uses of the derivative \(\frac{\text{d}y}{\text{d}x}\) is finding the equation of straight lines that touch or intersect curves at specific points.
Definitions
• Tangent: A straight line that touches the curve at a single point and has the exact same gradient as the curve at that point (\(m_T = \frac{\text{d}y}{\text{d}x}\)).
• Normal: A straight line that is perpendicular (\(90^\circ\)) to the tangent at that specific point. Its gradient is the negative reciprocal: \(m_N = -\frac{1}{m_T}\).
Straight Line Equation Formula
To find the equation of either line, use the straight-line equation:
\(y - y_1 = m(x - x_1)\)
where \((x_1, y_1)\) is the point of contact and \(m\) is the appropriate gradient.
Step-by-Step Example: Finding Tangents and Normals
Problem: Find the equation of the normal to the curve \(y = x^2 - 4x + 3\) at the point where \(x = 4\).
Step 1: Find the \(y\)-coordinate:
Substitute \(x = 4\) into the original curve equation:
\(y = (4)^2 - 4(4) + 3 = 16 - 16 + 3 = 3\)
So the point is \((4, 3)\).
Step 2: Find the gradient function \(\frac{\text{d}y}{\text{d}x}\):
\(\frac{\text{d}y}{\text{d}x} = 2x - 4\)
Step 3: Calculate the gradient of the tangent at \(x = 4\):
\(m_T = 2(4) - 4 = 4\)
Step 4: Find the gradient of the normal:
\(m_N = -\frac{1}{m_T} = -\frac{1}{4}\)
Step 5: Write down the equation using \(y - y_1 = m(x - x_1)\):
\(y - 3 = -\frac{1}{4}(x - 4)\)
Multiply through by \(4\):
\(4(y - 3) = -(x - 4)\)
\(4y - 12 = -x + 4\)
\(x + 4y - 16 = 0\)
Key Takeaway: Differentiate to find the gradient of the tangent. Flip and change the sign to get the gradient of the normal.
4. Increasing and Decreasing Functions
We can use the derivative to describe whether a graph is going "uphill" or "downhill" across different intervals.
• Increasing Function: The curve is sloping upwards from left to right. Here, \(\frac{\text{d}y}{\text{d}x} > 0\) (or \(f'(x) \ge 0\)).
• Decreasing Function: The curve is sloping downwards from left to right. Here, \(\frac{\text{d}y}{\text{d}x} < 0\) (or \(f'(x) \le 0\)).
Worked Example
Problem: Find the range of values of \(x\) for which \(f(x) = 2x^3 - 9x^2 + 12x - 5\) is decreasing.
Step 1: Find the derivative \(f'(x)\):
\(f'(x) = 6x^2 - 18x + 12\)
Step 2: Set up the inequality for a decreasing function (\(f'(x) < 0\)):
\(6x^2 - 18x + 12 < 0\)
Step 3: Divide by \(6\):
\(x^2 - 3x + 2 < 0\)
Step 4: Factorise to find the critical values:
\((x - 1)(x - 2) < 0\)
Critical values are \(x = 1\) and \(x = 2\).
Step 5: Determine the region where the quadratic is negative (below the \(x\)-axis):
The parabola opens upwards, so it is negative between the roots:
\(1 < x < 2\)
Key Takeaway: To find where a function is increasing or decreasing, find \(\frac{\text{d}y}{\text{d}x}\) and solve the corresponding inequality.
5. Stationary Points and the Second Derivative
A stationary point (or turning point) occurs on a curve where the gradient is flat, meaning:
\(\frac{\text{d}y}{\text{d}x} = 0\)
Types of Stationary Points
1. Local Maximum: A "peak" where the curve stops rising and starts falling.
2. Local Minimum: A "valley" where the curve stops falling and starts rising.
3. Stationary Point of Inflection: The curve flattens out momentarily but continues in the same direction (e.g., \(y = x^3\) at \(x = 0\)).
The Second Derivative Test
To find how the gradient is changing, we differentiate a second time to find \(\frac{\text{d}^2y}{\text{d}x^2}\) (read as "d-two-y by d-x-squared") or \(f''(x)\).
• If \(\frac{\text{d}^2y}{\text{d}x^2} < 0\) at the stationary point, the point is a Local Maximum.
• If \(\frac{\text{d}^2y}{\text{d}x^2} > 0\) at the stationary point, the point is a Local Minimum.
• If \(\frac{\text{d}^2y}{\text{d}x^2} = 0\), the test is inconclusive! You must test the sign of \(\frac{\text{d}y}{\text{d}x}\) slightly to the left and right of the point.
Memory Trick:
• Negative second derivative (\(< 0\)): Think of a frown \(\cap\) \(\implies\) Maximum (highest point).
• Positive second derivative (\(> 0\)): Think of a smile \(\cup\) \(\implies\) Minimum (lowest point).
Worked Example: Finding and Classifying Stationary Points
Problem: Find the coordinates of the stationary points on the curve \(y = 2x^3 - 3x^2 - 12x + 8\) and determine their nature.
Step 1: Differentiate to find \(\frac{\text{d}y}{\text{d}x}\):
\(\frac{\text{d}y}{\text{d}x} = 6x^2 - 6x - 12\)
Step 2: Set \(\frac{\text{d}y}{\text{d}x} = 0\) and solve for \(x\):
\(6x^2 - 6x - 12 = 0\)
Divide by \(6\):
\(x^2 - x - 2 = 0\)
\((x - 2)(x + 1) = 0\)
So, \(x = 2\) and \(x = -1\).
Step 3: Find corresponding \(y\)-coordinates:
• When \(x = 2\): \(y = 2(2)^3 - 3(2)^2 - 12(2) + 8 = 16 - 12 - 24 + 8 = -12\)
Point \(A = (2, -12)\)
• When \(x = -1\): \(y = 2(-1)^3 - 3(-1)^2 - 12(-1) + 8 = -2 - 3 + 12 + 8 = 15\)
Point \(B = (-1, 15)\)
Step 4: Find the second derivative:
\(\frac{\text{d}^2y}{\text{d}x^2} = 12x - 6\)
Step 5: Test each stationary point:
• At \(x = 2\): \(\frac{\text{d}^2y}{\text{d}x^2} = 12(2) - 6 = 18 > 0 \implies\) Local Minimum at \((2, -12)\).
• At \(x = -1\): \(\frac{\text{d}^2y}{\text{d}x^2} = 12(-1) - 6 = -18 < 0 \implies\) Local Maximum at \((-1, 15)\).
Key Takeaway: Stationary points occur where \(\frac{\text{d}y}{\text{d}x} = 0\). Use the sign of \(\frac{\text{d}^2y}{\text{d}x^2}\) to determine if each point is a maximum or minimum.
6. Practical Optimisation Problems
Optimisation means finding the maximum or minimum value in a real-world scenario (for example, maximizing the volume of a box or minimizing the cost of materials).
General Strategy for Optimisation Problems
1. Read and sketch: Draw a diagram and label dimensions (\(x\), \(y\), \(h\), etc.).
2. Write an equation: Write an expression for the quantity you want to maximize or minimize (e.g., Volume \(V\)).
3. Eliminate extra variables: Use a given constraint (e.g., total surface area or total wire length) to express everything in terms of a single variable.
4. Differentiate: Find the derivative with respect to that variable.
5. Set to zero: Solve \(\frac{\text{d}V}{\text{d}x} = 0\) to find the optimal dimension.
6. Justify: Use the second derivative to confirm whether it is a maximum or minimum.
Worked Example: Maximising Area
Problem: A farmer has \(40\text{ m}\) of fencing to create a rectangular enclosure against an existing straight brick wall. Find the dimensions of the enclosure that maximize the enclosed area.
Step 1: Set up the variables:
Let the width perpendicular to the wall be \(x\) meters, and the length parallel to the wall be \(y\) meters.
Since the wall forms one side, the total fence used is:
\(2x + y = 40 \implies y = 40 - 2x\)
Step 2: Express Area \(A\) in terms of \(x\):
\(A = x \cdot y = x(40 - 2x) = 40x - 2x^2\)
Step 3: Differentiate \(A\) with respect to \(x\):
\(\frac{\text{d}A}{\text{d}x} = 40 - 4x\)
Step 4: Set to zero for maximum area:
\(40 - 4x = 0 \implies 4x = 40 \implies x = 10\text{ m}\)
Step 5: Find the corresponding length and maximum area:
\(y = 40 - 2(10) = 20\text{ m}\)
\(\text{Maximum Area} = 10 \times 20 = 200\text{ m}^2\)
Step 6: Justify that it is indeed a maximum:
\(\frac{\text{d}^2A}{\text{d}x^2} = -4 < 0\)
Since the second derivative is negative, \(x = 10\text{ m}\) gives the maximum area.
Key Takeaway: In word problems, rewrite the formula in terms of a single variable before differentiating and setting to zero.
Quick Reference Summary
• Definition from First Principles: \(f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\)
• Power Rule: \(\frac{\text{d}}{\text{d}x}(a x^n) = a n x^{n-1}\)
• Tangent Gradient: \(m_T = \frac{\text{d}y}{\text{d}x}\)
• Normal Gradient: \(m_N = -\frac{1}{m_T}\)
• Increasing Function: \(\frac{\text{d}y}{\text{d}x} > 0\)
• Decreasing Function: \(\frac{\text{d}y}{\text{d}x} < 0\)
• Stationary Points: \(\frac{\text{d}y}{\text{d}x} = 0\)
• Classification: \(\frac{\text{d}^2y}{\text{d}x^2} > 0 \implies \text{Min}\); \(\frac{\text{d}^2y}{\text{d}x^2} < 0 \implies \text{Max}\)