Further Pure Mathematics | Calculus

Welcome to the World of Calculus!

Hello! Get ready to explore Calculus, arguably the most powerful tool in mathematics. Don't worry if it sounds complicated—at its core, Calculus is simply the mathematics of change. It helps us figure out how things are moving, growing, or shrinking at any exact moment in time.

In this chapter, we will master the two fundamental operations:

• Differentiation: Finding the instantaneous rate of change (like the exact speed of a car right now).
• Integration: Finding the accumulated total (like the distance travelled by a car over a period, or the area under a curve).

Let's dive in!

Section 1: The Foundations of Differentiation

1.1 The Power Rule: The Engine of Differentiation

Differentiation allows us to find the gradient function, or the derivative, of a curve, denoted by \(\frac{dy}{dx}\).

The Power Rule (for polynomials):

If \(y = ax^n\), then the derivative \(\frac{dy}{dx}\) is given by:

\(\frac{dy}{dx} = n a x^{n-1}\)

How it works (The Simple Trick):

1. Multiply the existing coefficient (\(a\)) by the power (\(n\)).
2. Reduce the power by 1 (\(n-1\)).

Example: If \(y = 5x^3\)

• Step 1: Multiply \(5 \times 3 = 15\).
• Step 2: Reduce the power \(3 - 1 = 2\).
• Result: \(\frac{dy}{dx} = 15x^2\).

Important Special Cases:

• If \(y = x\) (i.e., \(1x^1\)), then \(\frac{dy}{dx} = 1\).
• If \(y = c\) (a constant number, like 5), then \(\frac{dy}{dx} = 0\). (The gradient of a flat line is zero!)

Quick Review: Handling Roots and Fractions

Before differentiating, you must rewrite terms using index notation (\(x^n\)):

• \(\frac{1}{x^n} = x^{-n}\)
• \(\sqrt{x} = x^{1/2}\)
• \(\frac{5}{x^2} = 5x^{-2}\)

Key Takeaway: Differentiation is about finding the slope at any point. The Power Rule is the basic tool: multiply by the power, then subtract one from the power.

Section 2: Advanced Differentiation Techniques (The Rules)

When functions get complicated, we need specific rules to handle products, quotients (divisions), and functions nested inside other functions.

2.1 The Product Rule

We use the Product Rule when \(y\) is the product of two separate functions of \(x\).
Let \(y = uv\), where \(u\) and \(v\) are functions of \(x\).

The Formula:

\(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\)

Memory Aid (The UV trick):

If \(y = uv\), the derivative is: U V-dash plus V U-dash. (Where "dash" means differentiate.)

Step-by-Step Example: Differentiate \(y = (x^2 + 1)(x^3)\).

1. Identify \(u\) and \(v\):
   \(u = x^2 + 1\)
   \(v = x^3\)
2. Find their derivatives:
   \(\frac{du}{dx} = 2x\)
   \(\frac{dv}{dx} = 3x^2\)
3. Apply the rule (\(u \frac{dv}{dx} + v \frac{du}{dx}\)):
   \(\frac{dy}{dx} = (x^2 + 1)(3x^2) + (x^3)(2x)\)
4. Simplify:
   \(\frac{dy}{dx} = 3x^4 + 3x^2 + 2x^4 = 5x^4 + 3x^2\)

2.2 The Quotient Rule

We use the Quotient Rule when \(y\) is one function divided by another.
Let \(y = \frac{u}{v}\), where \(u\) and \(v\) are functions of \(x\).

The Formula:

\(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)

Memory Aid (The LO D HI trick):

Low D High minus High D Low, over Low Squared.

• Low = \(v\)
• D High = \(\frac{du}{dx}\)

COMMON MISTAKE: The order matters here because of the minus sign! Always start with \(v\) (the denominator) multiplying the derivative of the numerator (\(u\)).

2.3 The Chain Rule (Function of a Function)

The Chain Rule is used when you have a function nested inside another function, like differentiating "an onion." You peel the layers one by one.

If \(y\) is a function of \(u\), and \(u\) is a function of \(x\), the formula is:

\(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)

Analogy: Imagine a bracket raised to a power, like \(y = (x^3 + 5)^4\).

1. Outer Layer (\(dy/du\)): Treat the bracket as \(u\). Differentiate the power part first (bring the 4 down, reduce the power by 1).
2. Inner Layer (\(du/dx\)): Differentiate what is inside the bracket.
3. Multiply the results together.

Example: Differentiate \(y = (3x^2 - 7)^5\).

1. Outer derivative (Power Rule on the whole bracket):
   \(5(3x^2 - 7)^4\)
2. Inner derivative (Derivative of \(3x^2 - 7\)):
   \(6x\)
3. Multiply them together:
   \(\frac{dy}{dx} = 5(3x^2 - 7)^4 \times (6x)\)
   \(\frac{dy}{dx} = 30x(3x^2 - 7)^4\)

Section 3: Differentiation of Trigonometric Functions

You must know and memorise the derivatives of the three standard trigonometric functions. These are fundamental results and often appear in conjunction with the Chain, Product, or Quotient Rules.

Let \(y\) be a function of \(x\):

• If \(y = \sin x\), then \(\frac{dy}{dx} = \cos x\).
• If \(y = \cos x\), then \(\frac{dy}{dx} = -\sin x\).
• If \(y = \tan x\), then \(\frac{dy}{dx} = \sec^2 x\).

Watch the signs! Differentiating \(\sin x\) gives positive \(\cos x\). Differentiating \(\cos x\) gives negative \(\sin x\).

Applying the Chain Rule to Trig Functions

If the function inside the sine or cosine is not just \(x\) (e.g., \(\sin(3x)\)), we must use the Chain Rule.

If \(y = \sin(ax + b)\), where \(a\) and \(b\) are constants, then:

1. Differentiate the function itself: \(\cos(ax + b)\).
2. Multiply by the derivative of the inner part (\(ax+b\)): \(\times a\).
3. Result: \(\frac{dy}{dx} = a \cos(ax + b)\).

Example: Differentiate \(y = 4 \cos(5x)\).

\(\frac{dy}{dx} = 4 \times (-\sin(5x)) \times 5 = -20 \sin(5x)\).

Section 4: Applications of Differentiation

The derivative \(\frac{dy}{dx}\) tells us the gradient (slope) of the curve at any point. We use this to find tangents, normals, and turning points.

4.1 Tangents and Normals

If we want to find the equation of a line (tangent or normal) at a point \((x_1, y_1)\) on the curve, we need its gradient, \(m\).

The equation of a straight line is \(y - y_1 = m(x - x_1)\).

1. Finding the Gradient of the Tangent (\(m_T\))

The gradient of the tangent is simply the value of the derivative at that specific point:

\(m_T = \left. \frac{dy}{dx} \right|_{x=x_1}\)

2. Finding the Gradient of the Normal (\(m_N\))

The normal is the line perpendicular (at 90 degrees) to the tangent. If \(m_T\) is the gradient of the tangent, the gradient of the normal is the negative reciprocal:

\(m_N = - \frac{1}{m_T}\)

Key Takeaway: Differentiation is essential for finding the slope needed to define these straight lines.

4.2 Stationary Points (Turning Points) and Optimisation

Stationary Points (or Turning Points) are points on the curve where the gradient is exactly zero—the curve is momentarily flat. This occurs at local maximums, local minimums, or points of inflection.

Finding Stationary Points:

1. Set the derivative equal to zero: \(\frac{dy}{dx} = 0\).
2. Solve the resulting equation for \(x\) to find the x-coordinates of the stationary points.
3. Substitute these \(x\) values back into the original equation \(y\) to find the corresponding y-coordinates.

Determining the Nature (Max or Min): The Second Derivative Test

To determine if a stationary point is a maximum or a minimum, we use the Second Derivative, denoted by \(\frac{d^2y}{dx^2}\).

The Test:

• If \(\frac{d^2y}{dx^2} > 0\) (Positive), the point is a Local Minimum. (Think of a smiling face, concave up.)
• If \(\frac{d^2y}{dx^2} < 0\) (Negative), the point is a Local Maximum. (Think of a frowning face, concave down.)
• If \(\frac{d^2y}{dx^2} = 0\), the test fails. (It could be a point of inflection, and you need to check the sign of the gradient on either side.)

Optimisation Problems:

Differentiation is used to solve practical problems where we need to find the largest (maximum) or smallest (minimum) value of a quantity (e.g., maximum area, minimum cost). The process is identical to finding stationary points:

1. Formulate an equation for the quantity to be maximised/minimised (e.g., Area \(A\)).
2. Differentiate the equation (\(\frac{dA}{dx}\)).
3. Set the derivative to zero and solve.
4. Use the Second Derivative Test to confirm if it yields a maximum or minimum.

Section 5: Integration: The Accumulation

Integration is the reverse process of differentiation. It is often called the antiderivative. If differentiation finds the rate of change, integration finds the original function.

5.1 Indefinite Integration

When integrating a function \(f(x)\) with respect to \(x\), we write \(\int f(x) \, dx\).

The Reverse Power Rule:

If \(f(x) = ax^n\), then the integral is:

\(\int ax^n \, dx = \frac{a x^{n+1}}{n+1} + C\) (where \(n \ne -1\))

How it works (The Simple Trick):

1. Increase the power by 1 (\(n+1\)).
2. Divide the coefficient by the new power (\(n+1\)).
3. Crucially, always add the Constant of Integration, \(+ C\).

Why \(+ C\)?

When we differentiated, constants vanished (\(C\)). Since integration reverses this, we must include the arbitrary constant \(C\) to represent any constant that might have been there. This is why it is called an indefinite integral.

Example: Integrate \(y = 6x^2 + 5x - 3\).

\(\int (6x^2 + 5x^1 - 3x^0) \, dx = \frac{6x^{2+1}}{3} + \frac{5x^{1+1}}{2} - \frac{3x^{0+1}}{1} + C\)

Result: \(2x^3 + \frac{5}{2}x^2 - 3x + C\)

5.2 Integration of Trigonometric Functions

Since integration reverses differentiation, the rules are straightforward:

• \(\int \cos x \, dx = \sin x + C\)
• \(\int \sin x \, dx = -\cos x + C\)
• \(\int \sec^2 x \, dx = \tan x + C\)

Dealing with the Inner Function (Reverse Chain Rule):

When integrating functions like \(\cos(ax + b)\), we integrate the function normally and then divide the entire result by the coefficient of \(x\).

\(\int \cos(ax + b) \, dx = \frac{1}{a} \sin(ax + b) + C\)

Example: \(\int 7 \sin(2x) \, dx\)

The integral of \(\sin(2x)\) is \(-\cos(2x)\), divided by the coefficient 2.

\(\int 7 \sin(2x) \, dx = 7 \left( - \frac{\cos(2x)}{2} \right) + C = - \frac{7}{2} \cos(2x) + C\).

Section 6: Applications of Integration (Definite Integrals)

When integration has limits (numbers at the top and bottom of the integral sign), it is a definite integral. It no longer has a \(+C\) because we are calculating a specific value.

\(\int_{a}^{b} f(x) \, dx = [F(x)]_{a}^{b} = F(b) - F(a)\)

Where \(F(x)\) is the integrated function.

6.1 Area Under a Curve

The primary use of definite integration is finding the area bounded by a curve \(y = f(x)\), the x-axis, and two vertical lines \(x=a\) and \(x=b\).

Area \(A = \int_{a}^{b} y \, dx\)

Important Rule: Areas Below the Axis

If the curve is below the x-axis within the interval \([a, b]\), the integral will result in a negative value. Since area must be positive, you must calculate the integral and then take the absolute value (make it positive).

If a region spans both above and below the axis, you must split the integral at the x-intercept and calculate the positive area for each part separately, then add the results.

Area Between Two Curves

If a region is bounded by two curves, \(y_1\) (the upper curve) and \(y_2\) (the lower curve), the area between them is found by integrating the difference:

Area \(A = \int_{a}^{b} (y_{upper} - y_{lower}) \, dx\)

6.2 Volume of Revolution (Around the x-axis)

In Further Pure Mathematics, you are required to find the volume of a solid generated when an area bounded by a curve and the x-axis is rotated 360 degrees around the x-axis.

Imagine spinning the graph of the function around the x-axis—this creates a 3D solid (like a vase or a spinning top).

The Formula for Volume of Revolution (around the x-axis):

Volume \(V = \pi \int_{a}^{b} y^2 \, dx\)

Step-by-Step Process:

1. Identify the function \(y\) and the limits \(a\) and \(b\).
2. Square the function: calculate \(y^2\). (Make sure to expand brackets if necessary!)
3. Integrate the resulting \(y^2\) expression with respect to \(x\).
4. Evaluate the definite integral using \(F(b) - F(a)\), and multiply the final answer by \(\pi\).

Did you know? The formula for the volume of revolution comes from summing up infinitely many thin cylindrical discs (like slices of salami) along the x-axis. Each disc has an area of \(\pi r^2\), where \(r\) is the height \(y\).

Key Takeaway: Definite integration gives you a numerical result, representing accumulated quantity like area or volume. Remember the \(\pi\) and the \(y^2\) for volume!