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
yk+1 = a.b! . (b-k).xb-(k+1)
(b-k)!
= a.b! . xb-(k+1)
(b-(k+1))!
Therefore if equation true for n=k also true for n=(k+1). Since true for n=1, then by Mathematical Induction also true for all positive integer values of n.
Next I will look the at Sine and Cosine functions. Sine and cosine functions differentiate in a cycle: the (n+4) th differential will always contain the same trigonometric function as the nth.
start: y = sin(x)
yI = cos(x)
yII = - sin(x)
yIII = - cos(x)
yIV = sin(x)
...
This is a preview of the whole essay
yk+1 = a.b! . (b-k).xb-(k+1)
(b-k)!
= a.b! . xb-(k+1)
(b-(k+1))!
Therefore if equation true for n=k also true for n=(k+1). Since true for n=1, then by Mathematical Induction also true for all positive integer values of n.
Next I will look the at Sine and Cosine functions. Sine and cosine functions differentiate in a cycle: the (n+4) th differential will always contain the same trigonometric function as the nth.
start: y = sin(x)
yI = cos(x)
yII = - sin(x)
yIII = - cos(x)
yIV = sin(x)
From the rules of differentiation we know that to differentiate a function such as sin(3x2 + 4) we simply differentiate sin to cos of the bracket, and multiply by the differential of the bracket. In this case the differential would be 6x.cos(3x + 4).
For now I'm only going to use functions of the form sin(bx)...
start: y = sin(bx)
yI = b.cos(bx)
yII = - b2.sin(bx)
yIII = - b3.cos(bx)
yIV = b4.sin(bx)
It can be observed straight away that b is always to the power n. The next stage makes it difficult at first to see a general rule. It would be easy to find a rule for the (n + 1) th differential (given the nth differential), I'd just have to put the above paragraph into algebraic terms, but to find the nth differential itself is not as obvious.
Consider the graphs of sin(x) and cos(x)...
sin(x)
cos(x)
These two graphs make it immediately clear that a sine graph and a cosine graph are simply translations of each other. A cosine is effectively a sine wave but shifted backwards (in the negative direction) by ?/2 radians.
In general: f(x) ? f(x+a)
ie. a translation by ??-a ?
???0 ?
sin(x + ?/2)
Therefore, for example if y = cos(?/6)
then y = sin(?/6 + ?/2)
or y = sin(2?/3)
This can easily be checked on a scientific calculator.
So, using this method if I now differentiate the same function I can convert each answer into terms of sine rather than cosine. And therefore each time I differentiate I'll be differentiating sin, and never -cos, cos or -sin.
start: y = sin(bx)
yI = b.cos(bx) = b.sin(bx + ?/2)
yII = b2.cos(bx + ?/2) = b2.sin(bx + 2?/2)
yIII = b3.cos(bx + 2?/2) = b3.sin(bx + 3?/2)
yIV = b4.cos(bx + 3?/2) = b4.sin(bx + 4?/2)
It is now clear that n appears as a multiplying factor to ?/2 as well as being the power of b. The n as a multiplying factor serves to shift the graph n times ?/2 backwards. The general rule is now easily derived...
dn ( sin(bx) ) = bn sin(bx + n?/2 )
dxn
* This can be proved by induction for y = sin(bx)
[1] Test for n=1
yI = b.cos(bx)
= b.sin(bx + ?/2 )
equation correct for n=1
[2] Assume true for n=k
yk = bk sin(bx + (k.?)/2??
[3] Differentiate both sides w.r.t. x
yk+1 = bk b.cos(bx + (k.?)/2 )
= bk+1 sin(bx + (k.?)/2 + ?/2 )
= bk+1 sin(bx + ((k+1).?)/2 )
Therefore if equation true for n=k also true for n=(k+1). Since true for n=1, then by Mathematical Induction also true for all positive integer values of n.
A general rule for the nth differential of y = cos(bx) would be very similar to the rule I have proved above for y = sin(bx) . As I have already demonstrated, the sine graph is simply the cosine graph translated back ?/2 radians (ie. cos(x) = sin(x + ?/2 ) ). As the 1st differential of cos(x) is -sin(x) I need to find what -sin(x) is in terms of the cosine graph.
Again using the graphs I can see that -sin(x) ? cos(x + ?/2 ) . This identity can be used to finding a general rule for the nth differential of y = cos(x) .
start: y = cos(x)
yI = -sin(x) = cos(x + ?/2 )
yII = -sin(x + ?/2 ) = cos(x + 2?/2 )
yIII = -sin(x + 2?/2 ) = cos(x + 3?/2 )
Once again, as with the pattern for progressive differentiations of sine functions, the nth differential for y = cos(x) appears to be cos(x + n.?/2 ). This is exactly the same as what happened to the sine functions.
Because of this important observation anything I prove for progressive differentiations of functions containing sine will also be true for the same function containing cosine, but with the sin replaced by a cos . A quick proof for the above hypothesis will be enough to obtain the link between the two trigonometric functions that makes any general rule proved for sine also true for cosine after the rule is modified with cos replacing sin .
If y=cos(x)
To prove... yn = cos(x+ n.?/2 )
[1] Test for n=1
yI = -sin(x)
= cos(x + ?/2 )
equation correct for n=1 (see previous work)
[2] Assume true for n=k
yk = cos(x + k.?/2??
[3] Differentiate both sides w.r.t. x
yk+1 = -sin(x + k.?/2 )
= cos(x + (k+1)??/2 )
Therefore if equation true for n=k also true for n=(k+1). Since true for n=1, then by Mathematical Induction also true for all positive integer values of n.
Differentiation of products centres on a very simple equation: dy/dx = uv' + vu' where y = v.u . One of the simplest examples of product differentiation is y = ex .sin(x), (because ex isn't changed by differentiation). For example...
start: y = ex sin(x)
yI = ex (sin(x) + cos(x) )
= ex (sin(x) + sin(x + ?/2 ) )
yII = ex (sin(x) + sin(x + ?/2 ) ) + ex (cos(x) + cos(x +?/2 ) )
= ex (sin(x) + 2.sin(x + ?/2 ) + sin(x + 2?/2 ) )
yIII = ex (sin(x) + 2.sin(x + ?/2 ) + sin(x + 2?/2 ) ) + ex (cos(x) + 2.cos(x + ?/2 ) + cos(x + 2?/2 ) )
= ex (sin(x) + 3.sin(x + ?/2 ) + 3.sin(x + 2?/2 ) + sin(x + 3?/2 ) )
As you can see I have simplified each term in cos into a term in sin and grouped all of the same terms together. The multiplying factors of each term belong to a pattern know as Pascal's Triangle:
,2,1
,3,3,1
,4,6,4,1
,5,10,5,1 ...etc.
The numbers in this pattern can be written algebraically as nCr . For example in the 5th row the 3rd number is 5C3 = 5! / (3! 2!) = 10 . The rest of the general solution is easily derived using the fact that ex differentiates to ex (ie. it isn't changed by differentiation), as well as the rule obtained above from a simplified version of the formula for the nth differential of sin(bx) .
dn ( ex sin(x) ) = ex ( nC0 sin(x) + nC1 sin(x + ?/2 ) + ... + nCr-1 sin (x + (r-1).?/2?????????nCn sin (x + n.?/2??
dxn
* The above rule can be proved by induction for y = ex sin(x)
[1] Test for n=1
yI = ex (sin(x) + sin(x + ?/2 ) )
equation correct for n=1 (see previous work)
[2] Assume true for n=k
yk = ex (sin(x) + sin(x + ?/2 ) + ... + kCr sin(x + (r-1).?/2???????????sin(x + k.?/2????
[3] Differentiate both sides w.r.t. x
yk+1 = ex (sin(x) + kC1 sin(x + ?/2 ) + ... + kCr-1 sin(x + (r-1).?/2???????????sin(x + k.?/2????
+ ex (cos(x) + kC1 cos(x + ?/2 ) + ... + kCr-1 cos(x + (r-1).?/2???????????cos(x + k.?/2????
= ex (sin(x) + [ kC1 + kC0 ] sin(x + ?/2 ) + ... + [ kCr + kCr-1 ] sin(x + (r-1).?/2?????????
? sin(x + (k+1).?/2????
This can be rearranged to...
= ex (sin(x) + [ k+1C0 ] sin(x + ?/2 ) + ... + [ kCr-1 + kCr-2 ] sin(x + (r-1).??????????? sin(x + k.?????
The multiplying factor for the r th term: [ kCr-1 + kCr-2 ] is the same as [ kCr + kCr-1 ] which can also be simplified...
[ kCr + kCr-1 ] = k! / ( r! (k-r)! ) + k! / ( (r-1)! (k-(r-1))! )
= ( k! (k-r+1) + k! r ) / ( r! ((k+1) - r)! )
= ( k! (k-r+1+r) ) / ( r! ((k+1) - r)! )
= ( k! (k+1) ) / ( r! ((k+1) - r)! )
= (k+1)! / (r! ((k+1) - r)! )
= k+1Cr
Because [ kCr + kCr-1 ] = k+1Cr
then [ kCr-1 + kCr-2 ] = k+1Cr-1
Therefore if I replace [ kCr-1 + kCr-2 ] with k+1Cr-1 then...
y(k+1) = ex (sin(x) + (k+1)C1 sin(x + ?/2 ) + ... + (k+1)Cr-1 sin(x + (r-1).?/2?????????? sin(x + (k+1).?/2????
Therefore if equation true for n=k also true for n=(k+1). Since true for n=1, then by Mathematical Induction also true for all positive integer values of n.
If I now insert multiplying factors to both the x 's, this is what happens...
start: y = eax sin(bx)
yI = eax b.sin(bx + ?/2 ) + a.eax sin(bx)
yII = eax b2.sin(bx + ? ) + 2a.eax b.sin(bx + ?/2 ) + a2.eax sin(bx)
yIII = eax b3.sin(bx + 3?/2 ) + 3a.eax b2.sin(bx + ? ) + 3a2.eax b.sin(bx+ ?/2 ) + a3.eax sin(bx)
As you can see from the example above the pattern for the numeric multiplying factors (ie. not a or b ) is the same as the pattern for the previous function ( ex sin(x) ). This is expected as the effect of the insertion of the multiplying factors a and b is just to multiply each term by a and/or b to a power. Therefore any general rule derived from the pattern observed above will contain the general rule proved previously for ex sin(x) .
The a and the b belong to a different pattern, namely the binomial theorem. It is clear from the example above that the power of b and the power of a add together to give n . As the value of b increases the value of a decreases, and vice versa. This pattern can be expressed algebraically as a(n-r) br . I can now use this equation in my general formula as it is simply the multiplying factor of the general formula for ex sin(x) .
dn ( eax sin(bx) ) = eax ( an sin(bx) + nC1 a(n-1) b.sin(bx + ?/2 ) + ... + nC(r-1) an-(r-1) br-1 sin(bx + (r-1).?/2 ) + ...
dxn
+ bn sin(bx + n.?/2?) )
* To prove this is true I can use Mathematical Induction...
[1] Test for n=1
yI = eax ( a.sin(bx) + b.sin(bx + ?/2 ) )
equation correct for n=1 (see previous work)
[2] Assume true for n=k
yk = eax ( ak sin(bx) + kC1 a(k-1) b.sin(bx + ?/2 ) + ... + kCr-1 ak-(r-1) br-1 sin(bx + (r-1).?/2 ) + ...
+ bk sin(bx + k.?/2?) )
[3] Differentiate both sides w.r.t. x
yk+1 = eax ( ak+1-0 sin(bx) + kC1 ak+1-1 b.sin(bx + ?/2 ) + ... + kCr-1 ak-(r-1+1) br-1 sin(bx + (r-1).?/2?????????
??bk sin(bx + k.?/2???
+ eax ( ak b.cos(bx) + kC1 a(k-1) b2 cos(bx + ?/2 ) + ... + kCr-2 ak-(r-1+1) br-2+1 cos(bx + (r-2).?/2 ????????
??bk+1?cos(bx + k.?/2????
yk+1 = eax (ak+1 sin(bx) + [ kC1 a(k+1)-(2-1) b + kC0 ak ] sin(bx + ?/2 ) + ...
+ [ kCr-1 ak-(r-1+1) br-1 + kCr-2 ak-r br-2+1 ] sin(bx + (r-1).?/2 ) + ... + kCk bk+1 sin(bx+ (k+1).?/2?????
Incorporating the identity [ kCr-1 + kCr-2 ] ? k+1Cr-1 the above equation can be simplified to...
yk+1 = eax ( ak+1 sin(bx) + k+1C1 a(k+1)-(2-1) b.sin(bx + ?/2 ) + ...
+ k+1Cr-1 a(k+1)-(r-1) br-1 sin(bx + (r-1).?/2?????????? bk+1 sin(bx + (k+1).?/2????
Therefore if equation true for n=k also true for n=(k+1). Since true for n=1, then by Mathematical Induction also true for all positive integer values of n.
Using the rules I have proved in the preceding pages I can now find general rules for nth differentials of other similar functions.
For example I can look into what happens to the function eax (sin(bx) + cos(bx) ) when differentiated.
Consider:
y = sin(bx) + cos(bx)
Using the " R, ??formula " I know that
a.sin(x) + b.cos(x) ? R.sin(x + ??)
where R = ?(a2+ b2 ) and ??= tan-1 (b/a)
In this case:
a=1
b=1
therefore R=?2
??= tan-1 (1/1)
hence ??= ?/4
Because x is not changed / is not involved in the above manipulation, inserting a multiplying factor to x will not affect any of the calculations. Therefore if I change x to bx the below is true...
sin(bx) + cos(bx) = ?2.sin(bx + ?/4 )
Then multiplying by eax ...
eax (sin(bx) + cos(bx) ) = eax (?2.sin(bx + ?/4 ) ) = ?2.eax (sin(bx + ?/4 ) )
I can now use my general rule for the nth differential of eax (sin(bx) to find the nth differential of eax (sin(bx) + cos(bx) ) by simply multiplying everything by ?2 and adding ?/4 to bx in every term. So, the r th term would be:
?2.eax ( ... + nC(r-1) an-(r-1) br-1 sin(bx + (r-1).?/2 + ?/4 ) + ... )
Summary
These are the general formulae for the nth differentials of the five functions of x which I have proved so far.
dn ( axb ) = a.b! x(b-n)
dxn (b-n)!
dn ( sin(bx) ) = bn sin(bx + n?/2 )
dxn
dn ( ex sin(x) ) = ex ( nC0 sin(x) + nC1 sin(x + ?/2 ) + ... + nCr-1 sin (x + (r-1).?/2?????????nCn sin (x + n.?/2??
dxn
dn ( eax sin(bx) ) = eax ( an sin(bx) + nC1 a(n-1) b.sin(bx + ?/2 ) + ... + nC(r-1) an-(r-1) br-1 sin(bx + (r-1).?/2 ) + ...
dxn
+ bn sin(bx + n.?/2?) )
dn ( eax (sin(bx) + cos(bx) ) = ?2.eax ( an sin(bx + ????) + nC1 a(n-1) b.sin(bx + 3?/4 ) + ...
dxn + nC(r-1) an-(r-1) br-1 sin(bx + (r-1).?/2 + ?/4 ) + ... + bn sin(bx + n.?/2 + ?/4?) )
Further Investigation
Using the rules above I could investigate repeated differentiation further. I could may be try to find a rule for the nth differential of functions like eax (b.sin(x) ??d.cos(x) ) possibly by using the " R, ??formula ". I could investigate more simple functions of the form ( f(x) )-1 and see what happens to them. Or I could even bring in other trigonometric functions like sec(x) , cosec(x) or cot(x) and try and find some sort of general rule for those too. A general formula for this last example could be a combination of a rule for the nth differential of ( f(x)-1 ) and the rule for the nth differential of cosine or sine, so this is what I'd try first after finding a rule for inverse functions.
I could also look into the differentiation of functions like sin-1(x) - which, despite it immediately differentiating away from a function that includes a trigonometric function like sin-1(x) or even sin(x) , there might possibly be some sort of pattern which could be interesting to investigate. Other examples of this are ln(f(x)) or tan-1(f(x)) .
Repeated Differentiation - Winter 2001
Page 1 of 13
Ben Leavett
3E
