If we can differentiate two functions f and g the derivative is

This rule can be repeated so that we can differentiate the product of more than two functions for exampe

Example

Suppose we want to differentiate

First we notice that f is the product of two functions we know how to differentiate, and sin(x).

The derivative of x² is 2x and the derivative of sin(x) is cos(x) so putting these into the rule we get

So putting in the derivative for sin and x²; we get

Proof of the Product Rule

Completely rigourus proofs of most rules with differentiation can be quite complicated and I don’t this site covers some of the concepts in enough detail, but we can give a basic outline of a proof which you can expand as you learn higher levels of maths (mainly analysis).

So as we did when deriving from first principles we want to the limit of the gradient between two point on the curve as they get closer together.

We now add and subtract f(x+t)g(x) because we can cunningly use it to get the derivatives work and because it is the same as adding 0 it is perfectly valid.

Now we bracket in such a way that we can see the derivatives we want

Spotting the limits that make dg/dx and df/dx(and making use of the fact that we can multiply limits) we get

and finally taking the limit of f(x+t) to be f(x) (since f is differentiable)

which is the required result!

Also see

• Integration by Parts (0)
• Points of Inflection (0)
• Stationary Points (Maximum and Minimums) and Differentiation (1)
• Fundamental Theorem of Calculus (0)
• Why the proof 2=1 is wrong (0)
• Differentiate Logs with Proof (0)

log_ab^c = c log_ab

For example,

log(9) = log(3²) = 2 log(3)

This can be used to either take a power outside of a log so it becomes a multiplication that is easier to work with or take a multiplication inside a log so you are only working with logs.

Proof when c is an integer

When c in the above equation is a positive integer it is simple. Since b^c means b multiplied by itself c times we can use the rules of log addition to get
log(b^c) = log(b….b) = log(b) + log(b) + … + log(b) = clog(b)

Similarly when c is negative we divide 1 by b c times so with using the rules of log subtraction we get
log(b^-c = log(1/b …. 1/b) = -log(b) – … -log(b) = -clog(b)

Proof when c is a real number

Noticing that b = a^log_ab
We can write
b^c = (a^log_ab)^c=a^clog_ab
So taking log a of both sides gives us
log(b^c) = c log_ab as required

Notice that this uses the fact that a log and a power cancel.
Ie) log_aa^b = b
and a^log_ab = b

Also see

• Binomial Theorem (Expansion) for positive integer powers (0)
• Differentiate Logs with Proof (0)
• Why anything to the power 0 is eqaul to 1 (2)
• Log Laws: Adding and Subtracting Logs (0)
• Matrix Multiplication (1)
• Matrix Calculator, C++ Program by David Woodford (5)

The Unit Components

Usually, when working in 2 or 3 dimensions we use the standard unit component vectors i,j,k. These are vectors of length 1 with i in the direction of the x axis, j in the direction of the y axis and k in the direction of the z axis as shown below.

Writing vectors in component form

If we have a given vector v imagine the start of v being at the origin. Then suppose the other end has co-ordinates (a,b,c). That is if we travel distance a along the x axis, b along the y axis and c along the z axis we will have travelled along v. However, if you remember vector addition from the introduction to vectors you will notice that moving along a certain distance in one direction and then a certain distance in another is really just adding vectors.

So moving a along the x axis is moving a in the direction of i and since i is of length 1 we have really done ai. If we now move b along the y axis (the direction of j) we have relay just added bj to ai. Again if we move along the z axis by c we add ck. So we can now write

v = ai + bj + ck

mple of the vector 3j+2i in component form

Column and Row Vectors

When we have a vector in component form we can then write it as a column or a row vector. To do this we just write

or

Why have we done this? (addition)

Writing in component form may seen an odd thing to do but it is very useful. We now have some real numbers to work with (a,b,c) so we will find working with vectors simpler. For example addition is made easier as we can just add the components.

Example of addition

If v = 2i + 3j + k and u = -i + 3j then

v + u = (2-1)i + (3+3)j + (1+0)k= i +  6j + k

Components are also useful because they have some meaning. The tell us what the vector looks like in our normal co-ordinate system making it easier to understand what a vector “looks like”.

Finding the Norm or Size of a vector in component form

One of the benefits of the component form is it allows us to work out the norm or size of the vector. This is done using Pythagoras theorem since the components form a right angled triangle.

If v = ai + bj + ck then the norm of v is

Also see

• An introduction to vectors (0)
• Vectors: Dot or Scalar Product (0)
• Log Laws: Adding and Subtracting Logs (0)
• Basic Arithmatic in Binary (1)

Note// Whilst when first being introduced to vectors we usually are dealing with some sort of physical space, such as a plane, vectors can actually be used much more generally for example functions actually form and infinite dimensional vector space and colours are really just an example of another 3D vector space.

To start with vectors can be thought of as “arrows” on either a plane or in space (from the more general definition of vectors you might use later you’ll find that you can use any set of “axis” or bases  such as functions). The magnitude of a vector is represented by the length of the line whilst the direction is by the direction of the arrow.

Lets consider the 2D plane.

Drawing Vectors

The vector a is drawn an arrow in order that we can show the direction. The length of the arrow indicated the magnitude. By reversing the the direction of the arrow we can get -a.

Equality of Vectors

Two vectors are equal if they have the same direction and size or magnitude. This means that when we represent vectors using arrows two vectors can be equal even if they are drawn in different places provided they are parallel, pointing the same way and of the same length.

Addition of Vectors in the Plane

We can also add vectors together. To do this we “join” them so that we can see the combination of where they are heading. This means that adding vectors changes both the direction and the magnitude of the vector.

Multiplication of a Vector by a Scalar

And finally we can multiply a vector by a scaler.

This only changes the magnitude of the vector. In the case in the diagram multiplying a vector by 2 doubles its magnitude but keeps the direction the same.

More complex things can be done with vectors, such as the dot product, and they can also be written down numerically so that computations can be done with them which will be shown in later lessons.

Also see

• Writing Vectors in component form (0)
• Vectors: Dot or Scalar Product (0)
• Log Laws: Adding and Subtracting Logs (0)
• Matrix Calculator, C++ Program by David Woodford (5)
• Basic Arithmatic in Binary (1)

f(x)	f ‘(x)
c	0
	nxn-1
ln(x)	1/x
	ex
	aeax
sin(x)	cos(x)
-cos(x)	sin(x)
ln|sec(x)|	tan(x)
ln|sin(x)|	cot(x)
sinh(x)	cosh(x)
cosh(x)	sinh(x)

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

Also see

• Common Derivatives (0)
• Fundamental Theorem of Calculus (0)
• Differentiation From First Principles (with example) (1)
• Integration by Parts (0)
• Implicit Differentiation (1)
• First Order Differential Equations (6)

When multiply out this bracket we get all the combinations of powers for a and b such that the sum of their powers is n. So there is a term of aⁿwhere the power of b is 0 and one of bⁿ and then there are terms in between with both a and b raised to powers. Then all that is left is to find the co-efficents of each of these terms. For the case that n is a positive integer there are two ways of doing this, one is to use the Binomial Theorem or the other is to use Pascal’s triangle.

Binomial Theorem

The binomial Theorem is:

Example

If we know calculate the factorials this becomes

Which you can check since we know 3+1 = 4 and 4⁴ = 256.

However there is another simpler way of finding the coefficients for the power which is to use a Pascal’s triangle. This is an quick to calculate way for moderately large powers (the triangle can get quite large for big powers above say 10) by hand. A Pascal’s triangle with 5 rows is below.

Each item in the triangle is the sum of the two numbers above it and there is a 1 at the end of each row. When the binomial terms are written in order of powers of x (either way as it is symmetric) each number in the nth row is the coefficient of x in corresponding term (check it with the example above)

More complex sums

We can also use the binomial theorem on sums where the first term is something other than 1. In general when we want to find (a+b)ⁿ.
To do this we first factor out a within the sum to get

We then treat as x and use the binomial theorem to expand to get

we can then multiply through the sum by aⁿ to get

Example

Expand (3 + 2x)⁴.

Firstly factorise the 3 to get

Using Pascal’s triangle or our previous calculations we can find the coefficients and write out the sum

and now working out the arithmetic of each term we get our final answer

The binomial theorem can also be used to calculated sums raised to non integer or negative powers but that is for another post.

Also see

• Log Laws: Taking logs of powers (0)
• Why anything to the power 0 is eqaul to 1 (2)

1 + 2 + 4 + 8 + 16 + …       each term is twice the previous one
4 + 12 + 36 + 108 + …       each term is three times the previous one

In general the nth term of a geometric series is written as where a is the first term and r is the factor we multiply each term by. If a is 1 then the terms are just the powers of r.

We are often interested at find the value of a geometric sum, that is the sum of all the values of a geometric series to a certain point. We can also find the value of the sum of infinitely many terms if . To find both of these first let the sum of the first n terms be denoted S_n.

So

The formula for a finite sum that we will try to prove is

Proof of finite sum

Consider
from the definition of S_n


We can now factorise the left-hand side and tidy up the right multiply the r through the first sum and change the index down on the second,





We can now break up the sums into parts in order that we cancel bits of them in the next step





So cancelling out the two sums and factorising the a noting that anything the the power of 0 is 1 gives





and then dividing through by (r-1) gives us our required result


.

Infinite Sums

As noted earlier is we can also find the value of an infinite sum. It may seem like a strange idea to be able to find the value of the sum of an infinite number of things but if the size of each term keeps getting smaller so the amount the sum increases by each time gets smaller and smaller.

A simple way of visualising how this can work is to imagine a piece of rope two meters long. We are going to cut this rope up into an infinite number of pieces with the length of each piece representing a term in the sum. Firstly cut the rope in two. Put one piece to the side as this is our first term so a=1 as this piece is 1 meter long. r=1/2 because each piece of rope will be half the length of the previous one. Now cut the remaining piece of the rope in half and again put one piece to the side as the second term and cut the remainder in half. Theoretically we will always be able to repeat this process because you can always halve something. This will mean that we will now have an infinite number of pieces of rope with lengths 1m, 1/2m, 1/4m, 1/8m, … so the lengths of the pieces of rope form a geometric progression and we know that the sum of their lengths must be less than two as we only started out with two meters of rope. In fact it is exactly two which we can prove.

In general,


if


To show this simply note that rⁿ tends to 0 as n tends to infinity. Then using the algebra of limits and multiplying the top and bottom y -1 we find the above expression to be true.

Also see

• Arithmetic Series and Progression (2)
• Proof there are infinitley many prime numbers (0)

The easiest way to deal with this is to resolve forces into perpendicular components. For example if we are working in a 2D plane it might make sense to consider the horizontal and vertical parts of a force separately. We can then do two sets of calculations, one vertical and the other horizontal, to find out the affect of the force.

Consider the example below,

It shows how we can split a force into its two components and what they’re relative sizes would be. To work out the magnitude of the components we must know the magnitude of the force we are resolving and the angle between it and one of the two perpendicular component. We can then use trigonometry.

If we know the angle between the force and its component then the magnitude of the component is the magnitude of the force multiplied by the cos of the angle and the magnitude of the other, perpendicular component, is the magnitude of the force multiplied by the sin of the angle.

So in the above example where the magnitude of the force is M and that of the horizontal and vertical components is H and V:

V = Msin(a) and H = Mcos(a)
or
V = Mcos(b) and H = Msin(b)

Finding a resultant force from components

Once you have done calculations in two perpendicular directions you are likely to end up with two perpendicular forces which you want to combine to find the resultant force. To do this we must use pythagoras’s theorem and more trigonometry.

Again consider the example above. Suppose you know H and V and want to find M and the angles a and b. To find M we use Pythagoras so we get

and to find the angles a and b we must use the tan function to get
and

Example – finding the resultant force

Find the direction and magnitude of the resultant force below.

First find all the horizontal components (taking forces acting right as positive).
The horizontal component of the:
3N force is 3cos(30) = 2.598
2N force is 2cos(90) = 0 — note that this is expected because a vertical force has no horizontal component as they are perpendicular
4N force is -4cos(45) = -2.828

Then to find the horizontal component H of the resultant force we add these up to get
H = 2.598 + 0 – 2.828 = -0.23

And then we do the same in the vertical direction. The vertical components of the:
3N force is 3sin(30) = 1.5
2N force is 2sin(90) = -2
4N force is 4sin(45) = 2.828

And adding these up we find that the vertical component of the resultant force is
V = 1.5 -2 + 2.828 = 1.328

We now combine these to find the magnitude and direction of the resultant force.
The magnitude M is found using Pythagoras’s to get

and the angle, measured above the right horizontal, is
or 80.17 above the left horizontal

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1. The magnitude of the force in the direction parallel to the surface
2. The magnitude of the force acting on the object perpendicular to the surface (the normal force N)
3. The coefficient of friction acting between the object and the surface (this  is a measure of the “roughness” of the surface between 0 and 1 where 0 is a completely smooth surface and with coefficient 1 the object couldn’t be moved however large the force μ.

Firction acts parallel to the surface opposing motion and the normal is perpendicular to the surface

We know that if the object is in equlibrium then the resultant force in the direction parallel to the surface must equal zero and hence the firctional force must be equal to the resultant force acting on the object parallel to the plane. However, what we are interested in finding is the minimum force that must be applied in order that the object begins to move, that is the maximum value that the firction can take before it moves.

To find this maximum we simply multiply the normal force by the coefficient of friction. That is the firctional force R on a static object on a surface being acted upon by a normal force N and a force parallel to the surface F with coefficent of friction between the object and the surface μ is

F ≤ Nμ

and if the object is on the point of moving (that is the we applying the minimum force required to make the object move)

F = Nμ

Example

An object of mass 5kg is on a rough slope at 30 degrees to the horizontal. The acceleration due to gravity is g and the object is on the point of slipping down the slope. Find the coefficient of friction between the slope and the object.

We know from Newtons laws that the force of gravity acting on the object will have a equal force opposing it since the object is at rest. To use the formulea above we want to break this into two components parallel and perpendicular to the slope. The component perpendicular to the slope is the normal force N and the component parallel to the slope is the force R which is trying to move the object down the slope.

Firstly it is always useful to draw a diagram of what we are interested in.

The forces acting upon the object on a slope

Since we know that the object is on the point of slipping down the slope we can the use the eqaulity from the equations above

F = Nμ

So to find μ we need to first calulate F and N. Because the object is in equilibrium F = R .

So we can resolve the weight into directions parallel and perpendicular to the slope to get

N = Wcos(30) = (5g√3)/2

and

F = Wsin(30) = 5g/2

so we can re arrange to get

Also see

• Will an object ever stop because of friction (11)

Example 1: simplifying single fractions

Consider

We want to find things that divide both the numerator (top) and denominator (bottom). In this case 2x does. We can then divide both the top and bottom by 2x (known as cancelling out the 2x) to gain the fraction

.

Note that you can only cancel things that divide the whole of the top and bottom not part of the top and bottom or things that are added to the top and bottom. For example

Sometimes the common factor that we are cancelling isn’t immediately obvious, for example if we have quadratic expressions. In these cases you should try to factorise the quadratic to see if there is anything you should cancel.

Note on quadratics: If a fraction has quadratics that can be factorised you should always leave the denominator factorised whilst it is common to multiply out the numerator (unless there is a need to leave it factorised or it is very complex)

Example 2: adding fractions

Consider the algebraic expression

Before we consider adding them we should make each of the fractions in the sum into its simplest form.

Firstly we want to have one denominator. To do this we need to find the simplest common denominator and then make the denominator of both these fractions this. This is done in a similar way in which we find the lowest common denominator when we are adding normal fractions. The easiest way to do this is to multiply numerator and the denominator of both factions by the denominator of the other so the expression becomes:

Now, as when adding normal fractions, because the denominator of both fractions is the same we can add their numerators together and put them over a single denominator to get

Example 3: Solving Equations

To solve an equation involving an algebraic fraction you should first multiply out the fractions as explained above.

Consider the equation

which is already simplified. To solve this we need to cross multiply, this involves multiplying both sides of the equation by the both the denominators and then simplifying both sides. However, because after the multiplication the denominator of each fraction will be multiplied by the numerator they will cancel out with the result that we have multiplied each side of the equation by the denominator of the other side and have removed the denominators from the fractions. This is most easily show by example

ie)

becomes

with intermediate step

We can then solve the equation using normal methods (note that if the result is a quadratic you may have to use the quadratic formula)

Also see

• First Order Differential Equations (6)
• Finding the Inverse of a Function (0)
• Find equation of tangent to a curve (2)
• Inequalities (0)
• Why the proof 2=1 is wrong (0)
• Simultaneous Equations (0)