A-level Mathematics/AQA/MPC3

Mappings and functions
We think of a function as an operation that takes one number and transforms it into another number. A mapping is a more general type of function. It is simply a way to relate a number in one set, to a number in another set. Let us look at three different types of mappings: Only the first two of these mappings are functions. An example of a mapping which is not a function is $$ f(x) = \pm \sqrt x $$
 * one-to-one - this mapping gives one unique output for each input.
 * many-to-one - this type of mapping will produce the same output for more than one value of $$ x $$.
 * one-to-many - this mapping produces more than one output for each input.

Domain and range of a function
In general:


 * $$ f(x) $$ is called the image of $$ x $$.
 * The set of permitted $$ x $$ values is called the domain of the function
 * The set of all images is called the range of the function

Modulus function
The modulus of $$ x $$, written $$ |x| $$, is defined as


 * $$ |x| = \begin{cases} x & \mbox{for } x \ge 0 \\ -x & \mbox{for } x < 0 \end{cases} $$

Chain rule
The chain rule states that:

If $$ y $$ is a function of $$ u $$, and $$ u $$ is a function of $$ x $$,


 * $$ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} $$

As you can see from above, the first step is to notice that we have a function that we can break down into two, each of which we know how to differentiate. Also, the function is of the form $$ f( g(x) ) $$. The process is then to assign a variable to the inner function, usually $$ u $$, and use the rule above;

Differentiate $$ y = 2(x - 1)^3 $$

We can see that this is of the correct form, and we know how to differentiate each bit.

Let $$ u = x - 1 $$

Now we can rewrite the original function, $$y = 2u^3$$

We can now differentiate each part;

$$ \frac{dy}{du} = 6u^2 $$ and $$ \frac{du}{dx} = 1 $$

Now applying the rule above; $$ \frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx} = 6u^2 * 1 = 6u^2 = 6(x - 1)^2 $$

Product rule
The product rule states that:

If $$ y = uv $$, where $$ u $$ and $$ v $$ are both functions of $$ x $$, then


 * $$ \frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx} $$

An alternative way of writing the product rule is:


 * $$ (uv)' = uv' + u'v \,\!$$

Or in Lagrange notation:

If $$k(x) = f(x)g(x)$$,

then $$k'(x) = f'(x)g(x) + f(x)g'(x)$$

Quotient rule
The quotient rule states that:

If $$ y = \frac{u}{v} $$, where $$ u $$ and $$ v $$ are functions of $$ x $$, then


 * $$ \frac{d}{dx} \left ( \frac{u}{v} \right ) = \frac{ v \frac{du}{dx} - u \frac{dv}{dx} }{v^2} $$

An alternative way of writing the quotient rule is:


 * $$ \left ( \frac{u}{v} \right )' = \frac{u'v - uv'}{v^2} $$

x as a function of y
In general,


 * $$ \frac{dy}{dx} = \frac{1}{ \frac{dx}{dy} } $$

The functions cosec θ, sec θ and cot θ
$$ \operatorname{cosec}{\theta} = \frac{1}{\sin{\theta}} $$

$$ \sec{\theta} = \frac{1}{\cos{\theta}} $$

$$ \cot{\theta} = \frac{1}{\tan{\theta}} $$

Standard trigonometric identities
$$ \cot{\theta} = \frac{ \cos{ \theta } } {\sin{ \theta } } $$

$$ \sec^2{\theta} = 1 + \tan^2{\theta} \,\!$$

$$ \operatorname{cosec}^2{\theta} = 1 + \cot^2{\theta} $$

Differentiation of sin x, cos x and tan x
$$ \frac{d}{dx} \left ( \sin{x} \right ) = \cos{x} $$

$$ \frac{d}{dx} \left ( \cos{x} \right ) = -\sin{x} $$

$$ \frac{d}{dx} \left ( \tan{x} \right ) = \sec^2{x} $$

Integration of sin(kx) and cos(kx)
In general,

$$ \int \cos{kx}\ dx = \frac{1}{k} \sin{kx} + c $$

$$ \int \sin{kx}\ dx = - \frac{1}{k} \cos{kx} + c $$

Differentiating exponentials and logarithms
In general,

$$ \mbox{when}\ y = e^{kx},\ \frac{dy}{dx} = ke^{kx} $$

$$ \int e^{kx}\ dx = \frac{1}{k} e^{kx} + c $$

Natural logarithms
If $$y=\ln{x}$$, then


 * $$ \frac{dy}{dx} = \frac{1}{x} $$

It follows from this result that


 * $$ \int \frac{1}{x}\ dx = \ln{x} + c $$

$$ \int \frac{f'(x)}{f(x)}\ dx = \ln{f(x)} + c,\ \mbox{provided}\ f(x) > 0 $$

Integration by parts
$$ \int u \frac{dv}{dx}\ dx = uv - \int v \frac{du}{dx}\ dx $$

Standard integrals
$$ \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}{ \left ( \frac{x}{a} \right ) } + c $$

$$ \int \frac{dx}{\sqrt{(a^2 - x^2)}} = \sin^{-1}{ \left ( \frac{x}{a} \right ) } + c $$

Volumes of revolution
The volume of the solid formed when the area under the curve $$ y = f(x) $$, between $$ x = a $$ and $$ x = b $$, is rotated through 360° about the $$ x $$-axis is given by:


 * $$ V = \pi \int_a^b y^2\ dx $$

The volume of the solid formed when the area under the curve $$ y = f(x) $$, between $$ y = a $$ and $$ y = b $$, is rotated through 360° about the $$ y $$-axis is given by:


 * $$ V = \pi \int_a^b x^2\ dy $$

Iterative methods
An iterative method is a process that is repeated to produce a sequence of approximations to the required solution.

Numerical integration
Mid ordinate rule


 * $$\int_a^b y\ dx \approx h \lbrack y_{\frac{1}{2}} + y_{\frac{3}{2}} + \ldots + y_{n-\frac{3}{2}} +  y_{n-\frac{1}{2}} \rbrack $$


 * $$ \mbox{where}\ h = \frac{b-a}{n} $$

Simpson's rule


 * $$\int_a^b y\ dx \approx \frac{h}{3} \lbrack \left ( y_0 + y_n \right ) + 4\left ( y_1 + y_3 \ldots + y_{n-1} \right ) + 2\left ( y_2 + y_4 + \ldots + y_{n-2} \right ) \rbrack$$


 * $$\mbox{where}\ h = \frac{b-a}{n}\ \mbox{and}\ n\ \mbox{is even}$$