Differentiation and its Applications for the ESAT

Understanding Rates of Change and Gradients

To understand differentiation, you must grasp the concept of a rate of change. A common example is speed, which tells us how fast distance changes compared to time. A speed of $3\text{ m/s}$ means distance changes at a rate of $3$ metres for every one second that passes. Similarly, acceleration is the rate of change of speed, measured in metres per second changed per second, often written as $m/s^2$ or $s^{-2}$.

In mathematics, the gradient of a line is also a rate of change. It tells us how much the $y$ coordinate changes for every $1$ unit move in the $x$ direction. This means the gradient of a distance-time graph represents speed, and the gradient of a speed-time graph represents acceleration. For a curve, the rate of change at a specific point is defined as the gradient of the tangent to the curve at that point.

Differentiation Notation and the Power Rule

When we have an expression for $y$ in terms of $x$, such as $y = x^3 + 7x^2 - 3x + 11$, the derivative $\frac{dy}{dx}$ provides the gradient of the tangent at any given $x$. In this case, $\frac{dy}{dx} = 3x^2 + 14x - 3$. You should be familiar with the following notation:

1. First derivative: $\frac{dy}{dx}$ or $f'(x)$
2. Second derivative: $\frac{d^2y}{dx^2}$ or $f''(x)$
3. Physics dot notation: $\dot{s}$ for $\frac{ds}{dt}$ and $\ddot{s}$ for $\frac{d^2s}{dt^2}$

The fundamental rule for differentiating a power of $x$ is $\frac{d}{dx} x^n = nx^{n-1}$. This applies to any rational $n$, including fractions and negative numbers.

Simplification and Term-by-Term Differentiation

For the ESAT, you are expected to differentiate simple expressions involving sums of powers. If an expression is not already in this form, you must simplify it first. Advanced rules like the chain, product, or quotient rules are not required: instead, expand brackets and divide through by denominators.

Worked Example: Differentiate the expression $y = \frac{(3x+2)^2}{x}$.

1. Expand the numerator: $(3x+2)^2 = 9x^2 + 12x + 4$.
2. Divide each term by $x$: $y = \frac{9x^2}{x} + \frac{12x}{x} + \frac{4}{x} = 9x + 12 + 4x^{-1}$.
3. Differentiate term by term: $\frac{dy}{dx} = 9(1)x^0 + 0 + 4(-1)x^{-2}$.
4. Final result: $\frac{dy}{dx} = 9 - 4x^{-2}$ or $9 - \frac{4}{x^2}$.

Stationary Points and the Second Derivative

Stationary points occur where the tangent to a curve is horizontal, meaning the gradient $\frac{dy}{dx} = 0$. To classify these points as local maxima or local minima, we use the second derivative, $\frac{d^2y}{dx^2}$, which measures how the gradient itself is changing.

1. Local Minimum: If $\frac{dy}{dx} = 0$ and $\frac{d^2y}{dx^2} > 0$ at a point, it is a minimum. This is because the gradient is increasing (changing from negative to positive).

2. Local Maximum: If $\frac{dy}{dx} = 0$ and $\frac{d^2y}{dx^2} < 0$ at a point, it is a maximum. Here, the gradient is decreasing (changing from positive to negative).

Note on Logic: These conditions are sufficient but not necessary. For example, if $y = x^4$, there is a minimum at $x = 0$, even though $\frac{d^2y}{dx^2} = 0$. If both the first and second derivatives are zero, the point could be a maximum, a minimum, or a point of inflexion.

Worked Example: Classifying Stationary Points

Consider the cubic $y = 2x^3 - 3x^2 - 12x + 6$.

1. Find the first derivative: $\frac{dy}{dx} = 6x^2 - 6x - 12$.
2. Set to zero to find stationary points: $6(x^2 - x - 2) = 0 \implies 6(x-2)(x+1) = 0$. Thus $x = 2$ and $x = -1$.
3. Find the second derivative: $\frac{d^2y}{dx^2} = 12x - 6$.
4. Evaluate at $x = 2$: $12(2) - 6 = 18$. Since $18 > 0$, this is a local minimum.
5. Evaluate at $x = -1$: $12(-1) - 6 = -18$. Since $-18 < 0$, this is a local maximum.

Strictly Increasing and Decreasing Functions

A strictly increasing function always slopes upwards: its $y$ value increases as $x$ increases. A strictly decreasing function always slopes downwards.

The ESAT uses the following sufficient conditions:

• If $f'(x) > 0$ for all $x$ in an interval, the function is strictly increasing in that interval.
• If $f'(x) < 0$ for all $x$ in an interval, the function is strictly decreasing in that interval.

Note that some functions may be strictly increasing even if the derivative is not always greater than zero (for example, at points where the derivative is zero or undefined due to a corner). However, for the simple polynomials tested in the ESAT, you will primarily use the $f'(x) > 0$ and $f'(x) < 0$ tests.

Points of Inflexion

You are expected to have a qualitative understanding of points of inflexion, such as the one found at the origin for $y = x^3$. While you should be able to recognise their shape when sketching polynomial graphs, you are not required to use advanced differentiation techniques to identify them formally.