and

These are derived from the formula for double cos as follows

by subsitituion of

and similarly for sin²x but using the substitution for cos²x instead of sin²x.

To use these identities we simply substitute them into the integral and find the integral as normal since we know

so we get

and

F=ma

where f= (resultant) force, m=mass,a=acceleration.

Example

So for example if gravity is pulling down on a body of mass 2kg and is causing it to accelerate at 9.81ms^-2 then we can work out that the force of gravity acting down on the body (what we refer to as its weight) is
2 x 9.81 = 19.62N

When more than one force is acting

If multiple forces are acting on a body along one line (ie we don’t need to worry about the angles between the forces) then each one is causing the body to accelerate at different rates but obviously the body can only have a single rate of acceleration. To understand this problem you need consider the resultant force and resultant acceleration. These are formed by summing all the forces and acceleration respectively. So

where the ’s are the different forces acting on the body, so gives the resultant force and the ’s are the different accelerations associated with these forces, so gives the resultant acceleration. Also note that both the forces and accelerations have signs, you must take one direction as positive and the other as negative and give signs accordingly.

Multiple Force Example

Suppose the the body from the above example (of mass 2kg) is again falling under gravity of 19.62 newtons but it is encountering air resistance on 2 newtons against its fall. Find the acceleration of the body.
SO applying the formula (taking downwards as positive)
19.62 – 2 = 2 x a
which gives

f(x)	f ‘(x)
c (constant)	0
xn	nxn-1
ln(x)	1/x
ex	ex
eax	aeax
sin(x)	cos(x)
cos(x)	-sin(x)
tan(x)	sec2(x)
cosec(x)	-cosec(x)cot(x)
sec(x)	sec(x)tan(x)
cot(x)	-cosec2(x)
sin-1(x)
cos-1(x)
tan-1
sinh(x)	cosh(x)
cosh(x)	sinh(x)
tanh(x)	sech2(x)
cosech(x)	-coth(x)cosech(x)
sech(x)	-tanh(x)sech(x)
coth(x)	-cosech2(x)
sinh-1(x)
cosh-1
tanh-1
a(x)b(x)	a’(x)b(x) + b’(x)a(x)
a(b(x))

If there any more standard derivatives that I have missed out please leave them in the comments section

These can be defined as follows,
if for all x₁ < x₂

• f(x₁) ≤ f(x₂) then f is increasing
• f(x₁ ) < f(x₁ ) then f is strictly increasing
• f(x₁ ) ≥ f(x₂) the f is decreasing
• f(x₁) < f(x₂) then f is strictly decreasing

Monotonicity and Derivatives

If a function f(x) is increasing then what we mean is that the slope is always positive, so if f is continuous we can relate the the properties of increasing and decreasing to the derivative as shown in the table below.

Its important to note that these rules only work if the function is continuous, for example consider f(x) =1/x, which is discontinuous at 0.
We can differentiate it to get f’(x) = -1/x² which we know is always negative (because the squared term is always positive) so we would expect it to be a decreasing function. However if we consider two point either side o, 1 and -1 say  we find that f is not a decreasing function because whilst -1 < 1, -1/x < 1/x contrary to our definition of decreasing

So for example f(x) = x+2 (from reals to reals) has an inverse as every element y is mapped to only by y-2 but f(x) = x² (from reals to reals) doesnt because both -1 and 1 map to 1 so how would we decide which of these would be the unique result of the inverse applied to 1 (we could of course define the domain to be the positive reals in which case it would have an inverse).

When finding the inverse of a function you are really looking to see what maps to each element of the range or codomain, to find the inverse of f you are looking for the element x in the domain for a given y in the range such that f(x) = y. This basically means you are reversing the process of the function.

Finding the Inverse when the function is a formula

When a function is given by a formula what you need to try and do is apply the operations of that formula backwards. This easiest way of doing this is to let the function f(x) = y. Now you know the formula to get from x to y so substitue this in for f(x). All that now needs to be done is to rearrange this equation so that x is the subject and the resulting rexpression on the otherside is only in terms of y. This function is the inverse of f, to show this lets denote it g so we g(y) = x.

Then g(f(x)) = g(y) = x since we started by letting f(x) = y and created g such that g(y) =x.
and f(g(y)) = f(x) = y

so g is indeed the inverse of f.

Example

This is most easily demostrated through an example.
Let
Then to find the inverse of f we first write

and rearrange as follows

So the inverse of f is given by the formula so we can write

Note that we have replaced the y’s with x’s, this doesnt matter as we can put any variable we like into the function but it important to make sure that you use the same variable as the parameter of the function and in the formula that defines it.

1. A domain
2. A co-domain or range
3. A “rule” for assigning each element of the domain to a unique element of the co-domain.

Important Points

There are some important things about functions, in particular part 3, which you need to remember.

Firstly the function must be able to be applied to every element of the domain so f(x) = 1/x isn’t a function from the real numbers to the real numbers since 0 cant be assigned a value, as 1/0 isn’t a real number. There are two ways round this problem, we can define the co-domain not to include 0 (ie ) or can give 1/0 a different rule, eg f(x) = 1/x for all and f(0) = 7.

However not every element in the co-domain needs to be assigned to something so is a valid function even though nothing goes to -1

Every element of the domain is assigned a unique element of the c0-domain. This uniqueness is often a source of confusion – here what we mean is that the function only assigns a single element of the co-domain to each element of the domain, but more than one element of the domain can be assigned the same element of the co-domain.

Simple examples

Here are some simple examples of functions from the real numbers to the real numbers.

– This simply applys this operation to any number and give you another number.

for $ x \not 0 $ and – This takes the reciprocal of a number unless that number is zero where the reciprocal is undefined so gives 7.

-Note that  f(a) = f(-a) and nothing gives f(x) = -1 however this is still a function as f(x) is defined for all x and f(x) gives only one answer.

Composing Functions

You can compose two or more functions to form a new function. Consider two functions f and g both from the reals to the reals, f composed with g, written, fg or is given by

so if f = x+1 and g=2x fg(x) = f(2x) = 2x+1

Note that in order to do this the domain of g has to be the same as the co-domain of f.

Inverses and the Identity map

The identity map or function is the function which does nothing. If I is the identity on the reals the I(x) = x.

The inverse of a function , written , is the function which when composed with f gives the identity. There are two types of inverse, a right sided inverse and a left sided inverse, depending upon which side you compose them with f.

g is the left sided inverse of f is

gf(x) = I(x) = x

and h is the right sided inverse of f if

fh(x) = I(x) = x

If there is a function h such that

hf = fh = I

then h is said to be a two sided inverse of f and is often simply refereed to as the inverse of f.

Note that not all functions have inverses, for example x² has no inverse, but when a function does have one it is unique.

A point of inflection in the graph of a polynomial of degree 5

Note
While points of inflection are often found at stationary points, particularly when trying to find the nature of the stationary point, it is not a requirement that they are at one.

The number of points of infection in a graph is equal to the number of distinct real roots of the equation formed by equating the second derivative to zero.

Finding the Stationary Points

We know that stationary point occur when the gradient is 0 so when the derivative of the graph is 0, so in order to find the stationary points we but first differentiate the curve.

For example lets consider the graph . We cab differentiate this to find

We must then equate the derivative to 0 and solve the resulting equation. This is because we are trying to find the points where the gradient is zero and these point occur exactly at the solutions of the equation we have formed.

So in our example we form the equation

by equating our expression for , , to 0
Solving this equation we find that stationary points occur exactly when

Note that there can be more than solution to this equation, each of which is a valid stationary point.

Finally we should also find the y co-ordinate for the stationary point by putting this value of x into the initial equation. So for this example
So the only stationary point is at

Nature of Stationary Points

The nature of a stationary point simply means what the graph is doing around it and are characterised by the second derivative, (found by differentiating the derivative). There are three types of stationary point:

1. Maximum Points: These are stationary points where the graph is sloping down on either side of the stationary point (a sad face type of curve).
   Here
2. Minimum Points: These are stationary where the graph is sloping upwards on either side of the point (a happy face)

   Here

3. Point of Inflection: Here the direction of the slope of the graph is the same either side of the stationary point, it can be in either direction.

   At a point of inflection but isn’t enough to ensure that a point really is a point of inflection as it could still be a maximum or minimum point

Checking the nature of a Stationary Point when
In this case the easiest thing to do is look a small distance either side of the point and see whether the y value is greater than or less than that of the stationary point. You can then draw yourself a picture to see what it is. For example if they are both greater than the stationary point you know it is a minimum point, but if one is greater and one is less than it is a point of inflection

Warning: checking points either side does not guarantee the correct result as there may be another stationary point or a break in the graph between where you are checking and the stationary point so you should always check using the derivatives if possible

Differentiating brackets raised to a power

The chain rule can be a great short cut to differentiating brackets raised to a power as it doesn’t require you to multiply them all out, it also enables you to differentiate brackets raised to an unknown power.
Consider
This is the composite of the functions ax+b and tⁿ. So we differentiate them both to get a and nt^n-1 and then apply the formula to get

Notice how we multiplied the derivative of the inner function, a, by the derivative of the outer function nt^n-1 but substituted ax+b back in for t.

To generalise we can replace the ax+b with f(x) and by applying the above get

Differentiating Trigonometric functions

We can also use the chain rule when differentiating sin(f(x)) and cos(f(x)) since we know how to differentiate sin(x) and cos(x).
Using the chain rule we get

and

Example y=x²
Find the equation of the tangent to the curve when x=4?

To do this we first need to find the gradient of the curve which we can do by differentiating it.

so at the point x=t the gradient is 2t.

From this we can get a general equation for the tangent using the equation for the gradient of a straight line
grad =
to get the general equation for the tangent at the point x=t by substituting x₁=t, y₁=t² and m=2t

Then we can substitute in t=4 to find the equation of the tangent when x=4 to get

which is our final answer.