Calculus (Differentiation)

### Introduction to Differentiation

Calculus is a branch of mathematics that deals with rates of change. Differentiation is the process of finding the exact rate of change of one quantity with respect to another. In the context of graphs, this means finding the gradient (or steepness) of a curve at a specific point. While you can find the gradient of a straight line using `(y2 - y1) / (x2 - x1)`, a curve's gradient is constantly changing. Differentiation gives us a tool, the gradient function, which allows us to calculate this gradient at any point on the curve.

The gradient function is denoted as dy/dx (read as 'dee-why by dee-ex'), which represents the rate of change of `y` with respect to `x`.

### The Power Rule for Differentiation

The fundamental rule for differentiating polynomial functions at the O Level is the power rule. For any function of the form y = axⁿ, where `a` and `n` are constants, its derivative is given by the formula:

dy/dx = anxⁿ⁻¹

To apply this rule, you bring the power (`n`) down to multiply with the coefficient (`a`) and then reduce the original power by one (`n-1`).

Let's break this down with examples:

* If `y = x³`, then `a=1` and `n=3`. So, `dy/dx = (1)(3)x³⁻¹ = 3x²`.

* If `y = 5x⁴`, then `a=5` and `n=4`. So, `dy/dx = (5)(4)x⁴⁻¹ = 20x³`.

This rule applies to each term in a polynomial. Two special cases are:

* Differentiating a constant: If `y = c` (e.g., `y = 7`), the line is horizontal, so its gradient is zero. d/dx (c) = 0.

* Differentiating x: If `y = ax` (e.g., `y = 6x`), this is `y = 6x¹`. Applying the rule, `dy/dx = (6)(1)x¹⁻¹ = 6x⁰ = 6(1) = 6`.

Example Process: Differentiate `y = 2x³ - 4x² + 7x - 5`.

We differentiate term by term:

`dy/dx = d/dx(2x³) - d/dx(4x²) + d/dx(7x) - d/dx(5)`

`dy/dx = 6x² - 8x + 7 - 0`

`dy/dx = 6x² - 8x + 7`

### Finding Stationary Points

Stationary points (or turning points) are points on a curve where the gradient is zero. At these points, the curve is momentarily flat, neither rising nor falling. These can be a local maximum (the peak of a hill) or a local minimum (the bottom of a valley).

The process to find stationary points is:

To determine the nature of these stationary points (whether they are a maximum or a minimum), we use the second derivative test.

The second derivative, written as d²y/dx², is found by differentiating `dy/dx` a second time.

The Second Derivative Test:

* If at the stationary point, d²y/dx² > 0 (positive), the point is a local minimum.

* If at the stationary point, d²y/dx² < 0 (negative), the point is a local maximum.

Example: Find and determine the nature of the stationary points of `y = x³ - 12x + 1`.

* When `x = 2`, `y = (2)³ - 12(2) + 1 = 8 - 24 + 1 = -15`. Point is (2, -15).

* When `x = -2`, `y = (-2)³ - 12(-2) + 1 = -8 + 24 + 1 = 17`. Point is (-2, 17).

* At `x = 2`: `d²y/dx² = 6(2) = 12`. Since `12 > 0`, the point (2, -15) is a local minimum.

* At `x = -2`: `d²y/dx² = 6(-2) = -12`. Since `-12 < 0`, the point (-2, 17) is a local maximum.

### Application to Kinematics

Differentiation is essential for describing motion (kinematics). We relate displacement (s), velocity (v), and acceleration (a) with respect to time (`t`).

* Velocity (v) is the rate of change of displacement. Therefore, v = ds/dt.

* Acceleration (a) is the rate of change of velocity. Therefore, a = dv/dt.

This also means that acceleration is the second derivative of displacement: a = d²s/dt².

Example: A particle's displacement `s` in metres from a point O after `t` seconds is given by `s = t³ - 6t² + 9t`.

(a) Find its velocity at `t = 4`.

(b) Find the time when the particle is momentarily at rest.

(a) Find the velocity function: `v = ds/dt = 3t² - 12t + 9`.

At `t = 4`, `v = 3(4)² - 12(4) + 9 = 3(16) - 48 + 9 = 48 - 48 + 9 = 9` m/s.

(b) The particle is at rest when v = 0:

`3t² - 12t + 9 = 0`

Divide by 3: `t² - 4t + 3 = 0`

Factorise: `(t - 1)(t - 3) = 0`

So, the particle is at rest at t = 1 second and t = 3 seconds.