Repeated Differentiation

I am going to investigate what effect differentiation has on several different functions of x and try to find general rules for the nth differential for each function. I'm interested in investigating this particular topic as I enjoy differentiation and integration in the A-Level class work I do and would like to take it further to see I can find any patterns and hopefully find general formulae for nth differentials.

Repeated differentiation to A-Level standard is really only used up to the second differential of a function, as this gives the turning points on the graph of the function. In this piece of coursework I will look into what happens when functions are differentiated n times and the results achieved when different functions are multiplied together.

I am going to first study repeated differentiation on a simple function of x, for example a.xb , and then look at increasingly more complicated functions including sine and cosine. I could use the " Differentiation of products " technique to incorporate two different functions and I could may be find a way of using the " R, ??formula " and looking to see if there is a rule for the nth differential of a function of x in that form. I could prove any general rules I might find using the method known as "Proof by induction".

Notation in use throughout coursework:

yI = dy/dx

yII = d2y/dx2

yIII = d3y/dx3

... etc.

yn = dny/dxn

nCr = n! / [ r! (n-r)! ]

In general, unless otherwise stated, for an equation to be true both n and r must be a positive non-zero integers where n is the differential number and r is the term number.

The simplest function of x to differentiate is xb . This is an example of what happens to such a function upon differentiation...

start: y = x11

yI = 11x10 = 11x10

yII = 11.10x9 = 110x9

yIII = 11.10.9x8 = 990x8

yIV = 11.10.9.8x7 = 7920x7

yV = 11.10.9.8.7x6 = 55440x6

... etc.

We know from the rules of differentiation for such a function that for the (n+1) th differential you multiply the nth function by its power and lower its power by one. However, if we don't know the nth differential and we know only the original function this simple method is obviously not possible.

Looking at the above example suggests that a solution would have to have something to do with factorials. The way the power and the differential number add together to equal b is of particular interest.

The rule for the nth differential of xb can easily be deduced...

b! x(b-n)

(b-n)!

If a constant multiplying factor is introduced to y (call this a, ie. axb) , then this is what happens to the same function...

start: y = a.x11

yI = a.11x10

yII = a.11.10x9

yIII = a.11.10.9x8

yIV = a.11.10.9.8x7

yV = a.11.10.9.8.7x6

We can see that if a constant multiplying factor is inserted into the equation then this constant is unaffected by differentiation. From observing what happens to this function we can simply multiply the above rule by a.

dn ( axb ) = a.b! x(b-n)

dxn (b-n)!

N.B. the above rule is only true when (b-n) is not negative,

ie. true when (b-n)???0

* I can prove this rule by induction for y = axb

[1] Test for n=1

yI = a.b! x(b-1) = a.b x(b-1)

(b-1)!

equation correct for n=1 (see previous work)

[2] Assume true for n=k

yk = a.b! x(b-k)

(b-k)!

[3] Differentiate both sides w.r.t. x

I am going to investigate what effect differentiation has on several different functions of x and try to find general rules for the nth differential for each function. I'm interested in investigating this particular topic as I enjoy differentiation and integration in the A-Level class work I do and would like to take it further to see I can find any patterns and hopefully find general formulae for nth differentials.

Repeated differentiation to A-Level standard is really only used up to the second differential of a function, as this gives the turning points on the graph of the function. In this piece of coursework I will look into what happens when functions are differentiated n times and the results achieved when different functions are multiplied together.

I am going to first study repeated differentiation on a simple function of x, for example a.xb , and then look at increasingly more complicated functions including sine and cosine. I could use the " Differentiation of products " technique to incorporate two different functions and I could may be find a way of using the " R, ??formula " and looking to see if there is a rule for the nth differential of a function of x in that form. I could prove any general rules I might find using the method known as "Proof by induction".

Notation in use throughout coursework:

yI = dy/dx

yII = d2y/dx2

yIII = d3y/dx3

... etc.

yn = dny/dxn

nCr = n! / [ r! (n-r)! ]

In general, unless otherwise stated, for an equation to be true both n and r must be a positive non-zero integers where n is the differential number and r is the term number.

The simplest function of x to differentiate is xb . This is an example of what happens to such a function upon differentiation...

start: y = x11

yI = 11x10 = 11x10

yII = 11.10x9 = 110x9

yIII = 11.10.9x8 = 990x8

yIV = 11.10.9.8x7 = 7920x7

yV = 11.10.9.8.7x6 = 55440x6

... etc.

We know from the rules of differentiation for such a function that for the (n+1) th differential you multiply the nth function by its power and lower its power by one. However, if we don't know the nth differential and we know only the original function this simple method is obviously not possible.

Looking at the above example suggests that a solution would have to have something to do with factorials. The way the power and the differential number add together to equal b is of particular interest.

The rule for the nth differential of xb can easily be deduced...

b! x(b-n)

(b-n)!

If a constant multiplying factor is introduced to y (call this a, ie. axb) , then this is what happens to the same function...

start: y = a.x11

yI = a.11x10

yII = a.11.10x9

yIII = a.11.10.9x8

yIV = a.11.10.9.8x7

yV = a.11.10.9.8.7x6

We can see that if a constant multiplying factor is inserted into the equation then this constant is unaffected by differentiation. From observing what happens to this function we can simply multiply the above rule by a.

dn ( axb ) = a.b! x(b-n)

dxn (b-n)!

N.B. the above rule is only true when (b-n) is not negative,

ie. true when (b-n)???0

* I can prove this rule by induction for y = axb

[1] Test for n=1

yI = a.b! x(b-1) = a.b x(b-1)

(b-1)!

equation correct for n=1 (see previous work)

[2] Assume true for n=k

yk = a.b! x(b-k)

(b-k)!

[3] Differentiate both sides w.r.t. x