Differentiation Gradient Rate of Change

The derivative as a gradient

The derivative of $f(x)$ as the gradient of the tangent to the graph $y = f(x)$ at a point. In addition: interpretation of a derivative as a rate of change; second-order derivatives; knowledge of notation: $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$, $f'(x)$, and $f''(x)$. Differentiation from first principles is excluded.

In order to understand what derivative is, you need to have a good grasp of the notion of a 'rate of change'. We will explore this idea briefly. You will be used to ideas such as speed, and speed is a rate of change. Speed tells you how fast distance is changing compared to time: a speed of 3m/s means that distance is changing at a rate of 3m for every one second of time that elapses. So, rates of change tell you how fast one measure changes compared to another measure. Usually, we express rate of change as 'how many units of one thing change per single unit of another'. Speed is 'how much distance changes for every single unit change of time'. Acceleration is another example of a rate of change: it tells you how much the speed changes [m/s] for every one unit of time [s]: acceleration is usually measured in metres-per-second changed for every second of time and this is usually written [slightly confusingly] as metres per second per second or m/s/s and often the 'per second per second' is changed to $s^{-2}$.

Gradients are also rates of change. Recall we said earlier that 'we can think of the gradient as telling us how much we have to go vertically to get back on the line for every 1 unit we move horizontally from a point of the line'. In other words, gradient is the rate of change of $y$ compared to [or, with respect to] $x$. The gradient tells us how much $y$ changes for every one unit change in $x$. So gradient is just a rate of change.

You should be able to understand why the gradient of a distance-time graph will give you speed, and the gradient of a speed-time graph will give you acceleration [we have ignored the vector vs scalar issues here].

Now you have an idea that gradient is just a rate of change, we need to look at what a rate of change for a curve might mean. Actually, we will look at what we mean by the rate of change of a curve at a point on the curve; and recall we know what we mean by the rate of change for a straight line – it is its gradient.

We define the rate of change at a point on a curve as being the gradient of the tangent to the curve at that point. Intuitively this definition should make sense – it is worth your spending some time convincing yourself that this is the best definition. In fact, the definition gives rise to the interpretation of differentiation that we shall look at below – and if you have dealt with differentiation from first principles [which is not on the TMUA/ESAT specification] you will have seen how this leads to the interpretation of differentiation that we set out next.

Now we turn to look at how to find the differential of an expression and how we should understand what the differential of an expression means.

Let's start by thinking about what the differential mean. You already know that when you are given an expression such as $y = x^3 + 7x^2 - 3x + 11$ you can find what $y$ value corresponds to a given $x$ value by substituting the given $x$ value into the expression. What then does $\frac{dy}{dx}$ tell you? In this case we have $\frac{dy}{dx} = 3x^2 + 14x - 3$ and this tells you the gradient of the curve [its rate of change at a point] for any given $x$ value. And recall this is gradient of the tangent to the curve at that point.

So, the $y = \cdots$ gives you $y$ values once you have an $x$ value; and the $\frac{dy}{dx} = \cdots$ gives you gradients [or rates of change] of the curve [the tangent to the curve] at the point corresponding to the given $x$ value.

Below we will see why calculating the gradient of a curve at a point is so useful.

As well as knowing what differentiation tells us, you should make sure you are familiar with the various notation used for differentiation. We expect you to know $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$, $f'(x)$, and $f''(x)$ [note where the two 2s go in the notation: $\frac{d^2y}{dx^2}$].

Although we do not use it in the TMUA/ESAT, you should also be aware of the 'dot' notation for differentiation. The dot notion tends to be used in physics [and mechanics] when differentiating with respect to time [i.e. time is on the $x$-axis]: for instance $\frac{ds}{dt}$ might be written as $\dot{s}$, and $\frac{d^2s}{dt^2}$ as $\ddot{s}$.

You will note that the specification in this section says 'second-order derivatives'; we will explain what we expect you to know about these below.