Welcome to A2 Differentiation!
Welcome to one of the most powerful and rewarding topics in CCEA A2 1: Pure Mathematics. In AS Mathematics, you learned how to differentiate simple polynomials like \(y = x^3 - 4x\). In A2, we take those foundational ideas and unlock the full toolkit of calculus.
Differentiation allows us to analyse how quickly things change in the world around us—from the decay of medicine in the bloodstream to the flight path of a rocket. Don't worry if this topic seems daunting at first; we will break down every rule and concept step-by-step with clear examples, memory aids, and common pitfalls to watch out for.
1. Standard Derivatives: The New Toolkit
Before we can tackle complex problems, we need to know the derivatives of standard functions. In A2, you meet exponential, logarithmic, and trigonometric functions.
Exponential and Logarithmic Functions
1. Natural Exponential: If \(y = e^{kx}\), then \(\frac{dy}{dx} = k e^{kx}\).
Example: If \(y = e^{5x}\), then \(\frac{dy}{dx} = 5e^{5x}\).
Special Case: The derivative of \(e^x\) is simply \(e^x\). It is its own derivative!
2. General Exponential: If \(y = a^x\) (where \(a > 0\)), then \(\frac{dy}{dx} = a^x \ln a\).
Example: If \(y = 3^x\), then \(\frac{dy}{dx} = 3^x \ln 3\).
3. Natural Logarithm: If \(y = \ln(x)\), then \(\frac{dy}{dx} = \frac{1}{x}\).
Key Rule: If \(y = \ln(kx)\), using log laws \(\ln(kx) = \ln(k) + \ln(x)\). Since \(\ln(k)\) is a constant, the derivative is still \(\frac{dy}{dx} = \frac{1}{x}\).
Trigonometric Functions
Crucial Rule for Calculus: All angles must be in radians when differentiating trigonometric functions.
Here are the standard trigonometric derivatives you must know:
• If \(y = \sin(kx)\), then \(\frac{dy}{dx} = k\cos(kx)\)
• If \(y = \cos(kx)\), then \(\frac{dy}{dx} = -k\sin(kx)\)
• If \(y = \tan(kx)\), then \(\frac{dy}{dx} = k\sec^2(kx)\)
• If \(y = \sec(x)\), then \(\frac{dy}{dx} = \sec(x)\tan(x)\)
• If \(y = \operatorname{cosec}(x)\), then \(\frac{dy}{dx} = -\operatorname{cosec}(x)\cot(x)\)
• If \(y = \cot(x)\), then \(\frac{dy}{dx} = -\operatorname{cosec}^2(x)\)
Memory Trick: The "Co-" Rule
Notice a pattern? Every trig function starting with "co-" (\(\cos\), \(\cot\), \(\operatorname{cosec}\)) has a negative derivative! This simple check will save you valuable marks in exams.
Key Takeaway
Always ensure your calculator is in radians mode, and check your signs when differentiating "co-" trigonometric functions.
2. The Core Differentiation Rules
Most functions in exam questions are combinations of simpler functions. We use three fundamental rules to differentiate them: the Chain Rule, the Product Rule, and the Quotient Rule.
The Chain Rule (Function of a Function)
Use the Chain Rule when you have an "inside" function nested inside an "outside" function, such as \(y = (3x^2 + 1)^5\) or \(y = \sin(4x^3)\).
Formula: \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)
Analogy: Peeling an Onion
1. Differentiate the outer layer (leaving the inside unchanged).
2. Multiply by the derivative of the inside layer.
Generalised Power and Function Rules:
• If \(y = [f(x)]^n\), then \(\frac{dy}{dx} = n[f(x)]^{n-1} \times f'(x)\)
• If \(y = e^{f(x)}\), then \(\frac{dy}{dx} = f'(x) e^{f(x)}\)
• If \(y = \ln[f(x)]\), then \(\frac{dy}{dx} = \frac{f'(x)}{f(x)}\)
Example: Differentiate \(y = \ln(x^2 + 5x)\).
Here, \(f(x) = x^2 + 5x\), so \(f'(x) = 2x + 5\).
Therefore, \(\frac{dy}{dx} = \frac{2x + 5}{x^2 + 5x}\).
The Product Rule
Use the Product Rule when two variable expressions are multiplied together, such as \(y = x^2 \sin(x)\) or \(y = e^{3x}\ln(x)\).
Formula: If \(y = u v\), then \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\)
Memory Rhyme: "First times derivative of the second, plus second times derivative of the first."
Example: Differentiate \(y = x^3 e^{2x}\).
• Let \(u = x^3 \implies \frac{du}{dx} = 3x^2\)
• Let \(v = e^{2x} \implies \frac{dv}{dx} = 2e^{2x}\)
Applying the rule:
\(\frac{dy}{dx} = (x^3)(2e^{2x}) + (e^{2x})(3x^2) = x^2 e^{2x}(2x + 3)\).
The Quotient Rule
Use the Quotient Rule when one variable expression is divided by another: \(y = \frac{u}{v}\).
Formula: \(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)
Memory Rhyme: "Low d-high minus High d-low, all over Low squared."
("Low" is the denominator \(v\), "High" is the numerator \(u\), and "d-high" means the derivative of \(u\)).
Example: Differentiate \(y = \frac{\sin x}{x^2}\).
• Let \(u = \sin x \implies \frac{du}{dx} = \cos x\)
• Let \(v = x^2 \implies \frac{dv}{dx} = 2x\)
Applying the formula:
\(\frac{dy}{dx} = \frac{x^2(\cos x) - (\sin x)(2x)}{(x^2)^2} = \frac{x^2 \cos x - 2x \sin x}{x^4} = \frac{x \cos x - 2 \sin x}{x^3}\).
Common Pitfall
Unlike the product rule, the quotient rule has a minus sign in the numerator. Order matters! You must start with the denominator multiplied by the derivative of the numerator (\(v \frac{du}{dx}\)).
Key Takeaway
Always identify the algebraic structure first: Is it a single composite function (Chain), a product of two terms (Product), or a fraction (Quotient)? Label \(u\) and \(v\) clearly before starting.
3. Implicit Differentiation
So far, equations have been written explicitly in the form \(y = f(x)\). But what happens when \(x\) and \(y\) are tangled together, like in the equation of a circle \(x^2 + y^2 = 25\) or a curve \(x^3 + y^3 = 3xy\)?
Instead of trying to rearrange for \(y\) (which is often impossible), we differentiate every term with respect to \(x\) directly. This is called implicit differentiation.
The Core Principle
Because \(y\) is a function of \(x\), any term containing \(y\) must be differentiated using the Chain Rule:
\(\frac{d}{dx}[g(y)] = \frac{d}{dy}[g(y)] \times \frac{dy}{dx}\)
Quick examples:
• \(\frac{d}{dx}[y^2] = 2y \frac{dy}{dx}\)
• \(\frac{d}{dx}[\sin y] = \cos(y) \frac{dy}{dx}\)
• \(\frac{d}{dx}[e^{3y}] = 3e^{3y} \frac{dy}{dx}\)
Handling Product Terms with \(x\) and \(y\)
When you see a mixed term like \(xy\) or \(x^2 y^3\), use the Product Rule!
Example: Differentiate \(x^2 y\) with respect to \(x\):
• Let \(u = x^2 \implies \frac{du}{dx} = 2x\)
• Let \(v = y \implies \frac{dv}{dx} = 1 \times \frac{dy}{dx} = \frac{dy}{dx}\)
• \(\frac{d}{dx}[x^2 y] = x^2 \frac{dy}{dx} + 2xy\)
Step-by-Step Method for Implicit Equations
Step 1: Differentiate every term on both sides with respect to \(x\). (Remember: constant terms differentiate to \(0\)).
Step 2: Whenever you differentiate a term with \(y\), attach \(\frac{dy}{dx}\).
Step 3: Collect all terms containing \(\frac{dy}{dx}\) on one side of the equation.
Step 4: Move all other terms to the opposite side.
Step 5: Factor out \(\frac{dy}{dx}\) and divide to solve for it.
Full Example: Find \(\frac{dy}{dx}\) for the curve \(x^2 + y^2 - 4x + 6y = 12\).
1. Differentiating each term: \(2x + 2y\frac{dy}{dx} - 4 + 6\frac{dy}{dx} = 0\)
2. Group \(\frac{dy}{dx}\) terms: \(2y\frac{dy}{dx} + 6\frac{dy}{dx} = 4 - 2x\)
3. Factorise: \(\frac{dy}{dx}(2y + 6) = 4 - 2x\)
4. Solve: \(\frac{dy}{dx} = \frac{4 - 2x}{2y + 6} = \frac{2 - x}{y + 3}\).
Key Takeaway
Every time you differentiate a \(y\) variable with respect to \(x\), simply differentiate normally with respect to \(y\) and multiply by \(\frac{dy}{dx}\).
4. Parametric Differentiation
Sometimes the coordinates \(x\) and \(y\) are given in terms of a third variable, called a parameter (typically \(t\) for time or \(\theta\) for an angle):
\(x = f(t)\) and \(y = g(t)\)
Finding the First Derivative: \(\frac{dy}{dx}\)
By the Chain Rule, we can connect the rates of change:
\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} \quad \text{or} \quad \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\)
Example: A curve is defined parametrically by \(x = 2t^3\) and \(y = 4t^2 - 5\). Find \(\frac{dy}{dx}\) in terms of \(t\).
• \(\frac{dx}{dt} = 6t^2\)
• \(\frac{dy}{dt} = 8t\)
• \(\frac{dy}{dx} = \frac{8t}{6t^2} = \frac{4}{3t}\)
Finding the Second Derivative: \(\frac{d^2y}{dx^2}\) (Crucial Exam Skill!)
Warning: One of the most common mistakes at A2 is thinking that \(\frac{d^2y}{dx^2} = \frac{d^2y/dt^2}{d^2x/dt^2}\). This is incorrect!
To find the second derivative, remember that \(\frac{d^2y}{dx^2}\) means differentiating \(\frac{dy}{dx}\) with respect to \(x\). Using the chain rule:
\(\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \times \frac{dt}{dx} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}\)
Continuing the example above:
1. We have \(\frac{dy}{dx} = \frac{4}{3}t^{-1}\).
2. Differentiate this with respect to \(t\): \(\frac{d}{dt}\left(\frac{4}{3}t^{-1}\right) = -\frac{4}{3}t^{-2} = -\frac{4}{3t^2}\).
3. Divide by \(\frac{dx}{dt} = 6t^2\):
\(\frac{d^2y}{dx^2} = \frac{-\frac{4}{3t^2}}{6t^2} = -\frac{4}{18t^4} = -\frac{2}{9t^4}\).
Key Takeaway
To get the second derivative of a parametric curve, differentiate your first derivative with respect to \(t\), then divide by \(\frac{dx}{dt}\).
5. Connected Rates of Change
In real life, changing one quantity causes related quantities to change simultaneously. For example, pumping air into a spherical balloon increases its volume, which in turn causes its radius and surface area to increase over time.
We use the Chain Rule to link rates of change with respect to time (\(t\)).
Step-by-Step Strategy
Step 1: Identify the given rate: Look for phrases like "increasing at a rate of \(5\text{ cm}^3/\text{s}\)". (This is \(\frac{dV}{dt} = 5\)).
Step 2: Identify the required rate: E.g., "Find the rate of increase of the radius" (\(\frac{dr}{dt} = ?\)).
Step 3: Find a geometric or physical formula connecting the variables: E.g., Volume of a sphere \(V = \frac{4}{3}\pi r^3\).
Step 4: Differentiate the formula: E.g., \(\frac{dV}{dr} = 4\pi r^2\).
Step 5: Form a chain rule equation and solve:
\(\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{dV/dt}{dV/dr}\)
Example: Oil is leaking into a circular puddle such that the area \(A\) is increasing at a constant rate of \(12\text{ cm}^2/\text{s}\). Find the rate at which the radius \(r\) is increasing when \(r = 6\text{ cm}\).
1. Given: \(\frac{dA}{dt} = 12\)
2. Required: \(\frac{dr}{dt}\) when \(r = 6\)
3. Formula: \(A = \pi r^2 \implies \frac{dA}{dr} = 2\pi r\)
4. Chain Rule: \(\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} \implies 12 = (2\pi r) \times \frac{dr}{dt}\)
5. Substitute \(r = 6\): \(12 = 12\pi \times \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{12}{12\pi} = \frac{1}{\pi}\text{ cm/s}\).
Key Takeaway
Let the units guide you! For instance, \(\text{cm}^3/\text{s}\) must represent \(\frac{dV}{dt}\), while \(\text{cm}/\text{s}\) represents \(\frac{dr}{dt}\) or \(\frac{dx}{dt}\).
6. Applications: Tangents, Normals, and Points of Inflection
Tangents and Normals
• The gradient of the tangent to a curve at point \((x_1, y_1)\) is \(m_T = \frac{dy}{dx}\).
• The normal is perpendicular to the tangent, so its gradient is \(m_N = -\frac{1}{m_T}\) (provided \(m_T \neq 0\)).
• Use the straight-line equation: \(y - y_1 = m(x - x_1)\).
Stationary Points and Their Nature
Stationary points occur where the gradient is zero: \(\frac{dy}{dx} = 0\).
We classify stationary points using the second derivative \(\frac{d^2y}{dx^2}\):
• If \(\frac{d^2y}{dx^2} > 0\), the curve is convex (smiling) \(\implies\) Local Minimum.
• If \(\frac{d^2y}{dx^2} < 0\), the curve is concave (frowning) \(\implies\) Local Maximum.
• If \(\frac{d^2y}{dx^2} = 0\), the test is inconclusive! You must examine the sign of \(\frac{dy}{dx}\) just before and just after the point.
Points of Inflection and Concavity
A point of inflection is a point on a curve where the concavity changes (from concave to convex, or from convex to concave).
Conditions for a Point of Inflection:
1. \(\frac{d^2y}{dx^2} = 0\) (or is undefined).
2. \(\frac{d^2y}{dx^2}\) changes sign (from positive to negative or negative to positive) as it passes through the point.
Note: A point of inflection does not need to be stationary. If \(\frac{dy}{dx} = 0\) at the inflection point, it is called a stationary point of inflection. If \(\frac{dy}{dx} \neq 0\), it is a non-stationary point of inflection.
Key Takeaway
\(\frac{d^2y}{dx^2} = 0\) is a necessary condition for a point of inflection, but it is not sufficient on its own. Always verify that \(\frac{d^2y}{dx^2}\) changes sign on either side of the point.
Quick Reference Summary
• Product Rule: \(\frac{d}{dx}(uv) = u v' + v u'\)
• Quotient Rule: \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}\)
• Chain Rule: \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)
• Implicit Differentiation: Differentiate \(y\)-terms normally and multiply by \(\frac{dy}{dx}\)
• Parametric First Derivative: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\)
• Parametric Second Derivative: \(\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}\)