MATH 1441 Technical Mathematics for Biological Sciences
By now, you are able to explain without hesitation that the derivative of a function is simply a formula for the slope of its graph. Of course, this means that the derivative of a function is itself a function -- this is clear from all of the examples you've studied so far. And so, if a derivative of a function is itself a function, there is nothing to stop us from producing the graph of that derivative, and that graph will have a slope, and the slope of that graph will be given by some formula -- nothing more than the derivative of the derivative. That is, to get a formula for the slope of the graph of a derivative, we simply apply the appropriate combination of methods presented so far to find the derivative of the expression representing the original derivative itself.
We'll restate this notion with more symbols. Suppose we have some function y = y(x). We use the symbol dy
 d dx dx to denote the formula for the slope of the graph of y plotted vs. x. This formula for the slope is d obtained by applying the operation dx
to the function y. Now, dy/dx is itself a function of x, and so we could plot a graph of the values of dy/dx vs x. That graph will have a slope at each point, and the formula for the d slope of that graph is obtained by applying the operation dx
to the function dy/dx. The result is another formula, which can be symbolized as d

 dy


 d dx
2
2
 dx
2
 y
 and is called the second derivative of y with respect to x . The various equivalent forms of notation just above are all intended to indicate that the operations of "find the derivative with respect to x" has been applied to the original function, y, twice in succession. Either append the "operator" d/dx twice to the left of the symbol, y, for the original function, or use a superscript "2", or place two primes as superscripts on the function symbol, y.
Once the language "second derivative of y with respect to x" was adopted to indicate the result of applying the differentiation operation twice in succession, people began to refer to the result of one step of differentiation, dy/dx, as the first derivative of y with respect to x .
Of course, this process can be continued. If is a function of x, its graph can be plotted, and the formula dx 2 for the slope of that graph is obtained by finding the derivative: d

 d y
2

 d
3
 dx
3
 dx
3
 y
  y called the third derivative of y with respect to x . The notational options remain the same, though for third and further derivatives, the use of repeated primes as superscripts becomes difficult to manage reliably, and so instead of a succession of primes, people will write a small number in brackets as a superscript. The brackets are intended to indicate that this is an order of derivative rather than a power.
Obviously, this process can be continued indefinitely, producing fourth derivatives, fifth derivatives, sixth derivatives, etc., each of which is just obtained by applying the "find the derivative" process to the previous member of the sequence. In general, these formulas are referred to as higher derivatives . From a technical point of view, there is nothing new here -- we are just applying the already well-known methods of differentiation to a succession of functions.

David W. Sabo (2000) Higher Derivatives Page 1 of 12

Example 1: Determine all of the derivatives with respect to x of the function y

5 x 4 
3 x 3 
7 x 2 
2 x

8
Solution
It may not be entirely clear what the "all the derivatives" means here, but the question is clearly asking us to find successive derivatives of this function. In the absence of a better idea, we might as well just begin determining derivatives and see if inspiration strikes.
So, dy    

  dx

2 20 x 3 
9 x 2 
14 x

2 and so d y
 d

 dy


 d dx
2 dx dx dx
20 x
3 
9 x
2 
14 x

2

 
   
14

60 x
2 
18 x

14 and so d y dx
3
 d

 d y
2


 d dx
 60 x
2 
18 x

14

120 x

18
Then d y dx
4
 d

 d y
3


 d dx

120 x

18


120 and thus d y dx
5
 d

 d y
4


 d dx
 

0
So, the fifth derivative is zero, and differentiating zero just gives zero again. Thus, the derivatives of order 5 and higher are all zero. With this statement, we have been able to specify all derivatives of the given function.

You can see from the pattern of results in Example 1 that polynomials containing just positive whole number powers of x will have just a finite number of nonzero higher derivatives. Each time a derivative is taken, the powers of x in each term are reduced by 1, until finally no terms containing x survive. After that, all further derivatives just produce zeros.
For functions that do not have this simple form, there is generally no stage at which all further higher derivatives are identically zero. You can see this sort of behavior from the next example.
Example 2: Determine the first four derivatives with respect to x of the function y
 x
5 x 2

3
.
Solution
We need to use the quotient formula here. Because you've seen many examples in which the full details of the quotient formula were demonstrated explicitly, we'll just show an outline of the work here. dy
 d


5 dx dx x x

2
3



 x

 x


 x

3

2
   
Page 2 of 12 Higher Derivatives

David W. Sabo (2000)


10 x
2 
30 x

5 x
2
 x

3

2

5 x
2 
30 x
 x

3

2

5


6

 x

3

2
Then dx
2

5 d dx


 x x
2 

6 x
3

2

 
5


 x

3
 
2 x

6


 x
2 
 x

3

4
6 x

2
 x

  


5




  x

3



2
 x

3

2 
2
 x

3

4 3
 x 2 
6 x







 
10



 x 2 
6 x
  x 2
 x

3

3

6 x





90
 x

3

2
Now y
 d y
 d dx 3 dx


90
 x

3
 
3

 
90
  x

3
  

 270
 x

3

4
In fact, it is clear that from this stage on, all higher derivatives will just amount to derivatives of a negative power of (x + 3), so the process can be carried forward very easily. We are just required to find one more derivative: y
 d y
 d dx
4 dx


270
 x

3
 
4  
270
  x

3
  

1080
 x

3

5
It is also clear that this process can go on forever, with the result never being identically zero, because each successive derivative increases the numerical value of the numerator, and increases the power of (x + 3) in the denominator.

Of course, any function that begins with a non-integer power will have successive derivatives that never terminate in an identically-zero function because reducing an initial power in steps of 1 never reaches the power zero (but bypasses it). This is illustrated by the next example.
Example 3: Determine the first five derivatives with respect to x of the function y
  x
2
.
Solution
We have y

5 3 x
2 

5 3 x
2
 1
2
Thus y
  d dx




 x 2
 1
2




2
1 
 x 2
  1
2
 
6 x
  
3 x

 x 2
  1
2 

3 x
 x 2
This is a quotient, and so to obtain the second derivative, we need to employ the quotient formula, which gives
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David W. Sabo (2000) Higher Derivatives Page 3 of 12

y
  x
2
  
3 x
 1 

2 x
2
2
 x
2
  1
2


6 x

 x
2 x
2

9 x
2



 x
2



 x

2 x
2


 3
2
9 x
2



15 x
2
 3
2
The third derivative is now obtained with relative ease, because this second derivative involves just the power of a function: y
  d dx






15 x
2
 3
2




 d dx




 x
2
  3
2


 
15 d dx


 x
2
  3
2


 
15



3
2
 x
2
  5
2


6 x






135 x x
2
 5
2
Now we're back to a full-fledged quotient, so to get the fourth derivative, we need to apply the quotient formula again. y
 
 x
2
 5
2

135


135 x
 x
2


5
2

5

 x
2
 3
2


6 x



 

5 3 x
2
 3
2

 x
2
 
135
 
135 x
 
15 x
 x
2

5
7
2




 x 2 
15 x 2 x
2
 7
2




 x
2
 7
2 x
2

In the first two lines of work above, we factored the minus sign out of the expression as a whole to avoid having to keep track of it term-by-term. It looks like the successive derivatives are getting more complicated again (which is true), but we have just one more to determine to satisfy the requirements of the example, and again, it is necessary to begin with the quotient formula. The work is eased a bit by removing the factor
(-135) from the term-by-term operation. y
 
135 



 x
2
 7
2

24 x



5 12
 x
2 x
2

7
 
2 x
2
 5
2


6 x





 
135 



 x
2
 5
2
 x
2
 
24 x



5 12 x
2
 
21 x

 x
2

7
9
2




Page 4 of 12 Higher Derivatives

David W. Sabo (2000)

 
135



 120 x

72 x

3 
105 x

252 x
3 x
2
 9
2




 
135



 225 x

180 x
3
 x
2
 9
2







   x

4 x
3 x
2
 9
2



6075 x

5

4 x
2
 x
2
 9
2


The next example involves a product of two functions to begin with. The first derivative ends up being a product of three functions, and so to determine the second derivative we need to go somewhat beyond the basic formulas for finding derivatives presented so far. Fortunately, it is easy to do so, because products of more than two functions can be handled by applying the product-of-two-functions formula successively. For instance, for a product of three functions, y = u
 v
 w, we can start by considering y to be the product of two functions, y = f
 g = (u
 v)
 w. Then, apply the standard product formula for derivatives with f = u
 v and g = w: dy
 d dx dx
 
 g df dx
 f dg dx
 w d dx
 u v
 u v dw dx

 w v du dx
 u dv x

 dw dx du dv dw dx dx dx
The pattern of terms in the formula for a derivative of a product of two or more factors is quite obvious from this result. There will be the same number of terms in the derivative as there is factors in the original function. The terms are each formed as the product of the derivative of one factor and the product of the other factors. In the above formula for the derivative of a product of three factors, you can see that there are three terms. The first term contains du/dx and the product of the other two factors, v and w. The second term contains dv/dx, and the product of the other two factors, u and w. Finally, the third term contains dw/dx, and the product of the other two factors, u and v. Generalizing this to a product of four or more factors may give a result that is tedious to write down, but is not difficult to devise.
Example 4: Determine the first three derivatives with respect to x of the function y


3 x

5
 
2 x

7

8
.
Solution
Initially, this is the product of just two functions. Thus, applying the standard product formula for derivatives, we get y
  d dx

3 x

5
 
2 x

7

8 
 x


8
 x
 5
2 7 6 3 5 3 3 x

5

6
 x

7
  
 
3 x

5
 
2 x

7
 
2 x
   
3 x
  


3 x

5
 
2 x

7
 
36 x

126

48 x

80



3 x

5
 
2 x

7
 
84 x

46


David W. Sabo (2000) Higher Derivatives Page 5 of 12


 x

5
 
2 x

7
 
42 x

23

Now, to determine the second derivative, we are dealing with a product of three functions. (The constant factor, 2, out front here can just be held as a separate constant factor throughout our work.) Applying the three factor formula described on the previous page, we get y
  d
2 dx

3 x

5
 
2 x

7
 
42 x

23

 d
2 dx
 

 x

  x

  x
 4
2[ 2 7 42 23 5 3 5 3 3 x

5
 
42 x

23
  x

7
  


3 x

5
 
2 x

7
  

 x

5
 
2 x

  x

 x

  
3 x

 x

 


3 x

 x

 

 x

5
 
2 x

7

6

3276 x
2 
3588 x

665

The third derivative is found in much the same way, though the algebra is even a bit longer because the third factor here is now a trinomial. Nevertheless, if you slog your way through the work, you should end up with the simplified result: y
   x

5
 
2 x

7

5

19656 x
3 
32292 x
2 
11970 x

11795

Notice that the powers of (3x+5) and (2x-7) are dropping by one with each successive derivative, so eventually, these factors will disappear from the expression of the higher derivatives. Although the third factor appears to be increasing in complexity, it is just a polynomial itself. Had we expanded the original function, we would have obtained a polynomial of order 14, so we know that only the first fourteen derivatives will nonzero.

What are Higher Derivatives Good For?
You may be thinking: So, we can calculate higher derivatives … so what? Well, it turns out that second derivatives arise in some very important applications in science and technology. It is not so common to encounter the need for third and higher order derivatives, though such applications do exist. As well, there are some very important mathematical techniques that in principle require derivatives of all orders. This course does not allow time to explore any of these sorts of applications in the sort of detail that even introducing them warrants. However, we will give a brief description of two such areas of application here.
Equations of Motion in Physics
Think of an object that is restricted to moving along a straight line path (for example, a train on a track or a vehicle on a straight line stretch of road, or an elevator in a vertical shaft). Let x = x(t) be a formula for its position, x, as a function of time. As the value of t varies, the value of x varies, resulting in motion along the line (which we are calling the x-axis here).
In this situation, the first derivative, dx/dt, represents the instantaneous rate of change of position x with respect to time, t. But this is nothing more than what you are more familiar with calling the velocity , v(t), of the object. If this velocity changes from instant to instant (either in magnitude or direction or both), then v is itself a function of t, as indicated by the notation v = v(t). Thus,
 d dt
Now, if v(t) is a function of t -- that is, v changes with time -- then its derivative, dv/dt, gives the instantaneous rate of change of v with respect to time. This is called the acceleration of an object, a(t):
Page 6 of 12 Higher Derivatives

David W. Sabo (2000)

 d dt

 
However, since v can be written as the first derivative of the position, x(t), of the object, we can also write the acceleration as
 d dt

 
 d
 dx


  d
2 dt
2
That is, the acceleration of the object is the first derivative of its velocity, but is also the second derivative of its position.
Now, you know from high school physics that Newton's Law of Motion, the fundamental description of how everyday things in the world move, relates the acceleration of an object to the applied force through the formula F = ma. Thus, Newton's Law of Motion is a statement about the acceleration of an object, which you now see is really a statement about the second derivative with respect to time of the position of an object. So, you've really been familiar with a second derivative for a while, even if you didn't know that it was a second derivative. (Just think -- instead of calling that pedal on the floor of your car that makes it speed up an accelerator pedal, you can now call it a "second derivative pedal"! Will your friends be impressed or what?)
Example 5: The formula for the height, y, in meters, above the ground of an object thrown straight up is given by y
 
4.905
t
2 
22.50
t

1.90
Determine the position, velocity, and acceleration of the object when t = 0 seconds and when t = 3.0 seconds. Also, what can you say about these quantities when t = 10.0 s?
Solution
We get values of the position of the object by just substituting the value of t of interest into the formula for y.
To get a formula for the velocity of the object, we need to take the first derivative of y(t), getting
 d dt


4.905
t
2 
22.50
t

1.90
 
9.81
t

22.50
and the formula for the acceleration will result from differentiating the formula for v(t):
 d dt


9.81
t

22.50

 
9.81
You see immediately that the acceleration is a constant -- in fact, it's the familiar acceleration due to gravity at the surface of the earth, 9.81 m/s 2 , with the minus sign because y is measured in an upwards direction, but the acceleration is downwards.
Now, when t = 0, we get y(0) = 1.90 m v(0) = 22.50 m/s a(0) = -9.81 m/s 2
Notice that the constant term in the formula for y(t) gives the position at t = 0, the initial position of the object.
Similarly, the coefficient of t in the formula for y(t) gives the velocity at t = 0, the initial velocity. At t = 0, the velocity is +22.50 m/s, indicating that the object is moving upwards. The acceleration is -9.81 m/s 2 , indicating that the velocity will be continuously decreasing with time, which corresponds with the way we know objects thrown upwards behave. Their speed slows rather quickly to zero, and then they begin accelerating towards the ground.

David W. Sabo (2000) Higher Derivatives Page 7 of 12

When t = 3.0 s, we get y(3) = 25.255 m v(3) = -6.93 m/s a(3) = -9.81 m/s 2
Here, we see that the object is 25.255 meters above the ground, but its velocity is now -6.93 m/s, a negative value, indicating that the object is moving groundwards.
When t = 10.0 s, we get y(10) = -263.6 m
We needn't do any further calculations, because this is clearly an unphysical result. This says that the object is now 263.6 m below the surface of the ground. This is not an uncommon situation in applied mathematics. You can substitute any value of t into the original formula for y(t) and get a numerical result, but the result will only apply to the physical situation for a limited interval of values of t, corresponding to the time interval between when the object is thrown (t = 0) and the instant when it subsequently lands on the ground (here at t

4.67 seconds). Thus, when t = 10.0 seconds, we can say that the object is lying on the ground (y = 0) and has velocity equal to 0.0 m/s. Although the object would still be "feeling" the force of gravity, the ground is resisting its movement sufficiently to reduce the net force it experiences to zero, so it really has an acceleration of 0.0 m/s 2 as well at that time. If its acceleration were not zero, then it would begin moving.

Computing Values for Mathematical Functions: Power Series Expansions
We will show you a simple example of one other more mathematical application that requires higher derivatives. This is a trivial example from a very deep and complex branch of mathematics. It involves the development of something called a power series , which amounts to a polynomial with a potentially infinite number of terms. Because of infinity, the method about to be illustrated for one simple case must be applied with great caution. So, you could think of this example as falling into the category of "don't try this at home without a good bit more study of the issues". We present this application both to illustrate how important the ability to work with higher derivatives is in something as technical as designing the e x key on your calculator, and to give you a hint of a very powerful technique of advanced mathematics.
Earlier in the course, we stated without proof the formula for the derivative of the exponential function, y = e x : d e x e x dx
Since the derivative of e x with respect to x is e x , this means that all higher derivatives of e x are also equal to e x . This makes e x a very nice function to find derivatives for. (The method presented below will work for other functions that don't have this nice property, but then you really do have to do all the work necessary to calculate as many derivatives as you eventually require.)
Now, the goal here is to attempt to develop a formula for e x in terms of x that we can use to calculate the value of e x using just simple arithmetic operations. We start by making the following crucial assumption: that y = e x can be expanded in the form of a power series: y
 e x  a
0
 a x
1
 a x
2
2  a x
3
3  a x
4
4 
(i) where a
0
, a
1
, a
2
, a
3
, a
4
, and so on are at present undetermined numerical coefficients. The subscripts on the symbols 'a' are labels to identify with which power of x that particular coefficient is associated. The form of equation (i) is left indefinite because at this stage, we don't know how many terms may be necessary in this formula. In fact (with good reason, you'll see before we're done), we leave open the possibility that the number of terms in (i) may actually be infinite!
It is the ability to find successive derivatives of formula (i) that gives us a tool for determining the values of these coefficients, a
0
, a
1
, a
2
, a
3
, a
4
, … .
Page 8 of 12 Higher Derivatives

David W. Sabo (2000)

To begin with, substitute x = 0 into formula (i). Since e 0 = 1 and all of the terms containing x on the righthand side are then equal to zero, we get a
0
= 1 and so we've determined the value of the first unknown constant.
Now, take the derivative of both sides of formula (1): d dx
 d dx a
0
 a x
1
 a x
2
2  a x
3
3  a x
4
4  or e x  a
1

2 a x
2

3 a x
3
2 
4 a x
4
3 
5 a x
5
4 
(ii)
Now, substitute x = 0 into this formula (ii). Again, the left-hand side becomes equal to 1, and all of the righthand side terms containing x become zero, leaving a
1
= 1
A second coefficient has now been determined. Now, repeat this last process, taking the derivatives of both sides of equation (ii): e x 
2 a
2
  a x
3
  a x
4
2   a x
5
3   a x
6
4 
(iii)
When x = 0 is substituted into formula (iii), we get
1

2 a
2
 a
2

1
2
Repeat the process again. Take the derivative of formula (iii) e x   a
3
2 3 
(iv)
Setting x = 0 in formula (iv) gives
1
  a
3
 a
3


1

1
3 2 3!
In this last form, we've used the factorial notation . The factorial of a positive whole number is the product of that whole number and all positive whole numbers less than it. Thus 3! = 3 x 2 x 1 = 6, and
5! = 5 x 4 x 3 x 2 x 1 = 120, and so on. You can see that as we continue to differentiate the right-hand sides of this series of formulas, the factorial of the original power of x in each term will appear as a numerical coefficient. Just to be sure, repeat the process once more. The derivative of formula (iv) is e x    a
4
    a x
5
    a x
6
2 
(v) and when x = 0 is substituted into this formula, we get
1 4 3 2 a
4
 a
4

1
 

1
4 3 2 4!
Two things are now obvious. First, the pattern is clear enough that we can write that in general, for this series expansion of e x , the coefficient, a n
, of x n is given by a n

1 n !
so that the working formula for e x can be written

David W. Sabo (2000) Higher Derivatives Page 9 of 12

e x x
1
2!
x
2 
1
3!
x
3 
1
4!
x
4 
1
5!
x
5 
(vi)
The other observation is that this formula has no end -- the repetitive process illustrated by the first few steps above will never terminate. Thus, the right-hand side of formula (vi) continues endlessly with the pattern showing.
So, what's the use of an infinitely long formula. In this case, the formula turns out to be remarkably useful, because the factorial denominators in the coefficients increase in size so rapidly that all but the first few terms of the right-hand side contribute a negligible amount to its overall value. A series that has this property is said to be convergent , and even though convergent power series formulas have an infinite number of terms, accuracy of any specific degree can be achieved using just a finite number of terms.
Before you can use a power series formula of this type, you need to establish that it is convergent using one of several available test procedures. To use a power series expansion of a function without being able to demonstrate that it satisfies convergency conditions is a serious error in mathematics.
Anyway, the series in formula (4) can be shown to be convergent. We can also see how well it works by trying it out for some specific value of x. For example, take x = 1. Using your calculator, you can probably determine that e 1 = 2.718282, rounded to six decimal places. The following table lists the first twelve terms of formula (vi) with x = 1, along with the partial sums through each term, with values rounded to seven decimal places.
Cumulative
Sum
Cumulative
Sum a
0
= 1 a
1
= 1 a
2
= 0.5 a
3
= 0.1666667 a
4
= 0.0416667 a
5
= 0.0083333
1
2
2.5
2.6666667
2.7083333
2.7166667 a
6
= 0.0013889 a
7
= 0.0001984 a
8
= 0.0000248 a
9
= 0.0000028 a
10
= 0.0000003 a
11
= 0.0000000
2.7180556
2.7182540
2.7182788
2.7182815
2.7182818
2.7182818
You see that by the twelfth term (a
11
1 11 ), the value of the term doesn't show up in the first seven decimal places, and so, do not contribute to the sum in formula (vi) in those first seven decimal places. Thus, although formula (vi) indicates you need to sum an infinite number of terms to get e 1 exactly, we find that summing the first twelve is adequate to get e 1 accurate to apparently seven decimal places. (You may not be impressed with even a twelve term formula for e 1 , but it is possible to set up this calculation very efficiently for an electronic device such as your electronic calculator or a microcomputer. The point of this example is to demonstrate that the ability to determine higher derivatives actually gives us a way to obtain formulas of this type.)

Before leaving this topic of higher derivatives, we need to address one more issue briefly, more to make you aware of the problem than to provide formulas that you will use routinely.
The Chain Rule and Higher Derivatives
The so-called chain rule for first derivatives allows us to find the derivatives of functions of functions. At its simplest, we have used it to calculate derivatives of powers of functions: d dx u n n

1 du dx
(i) but it can be stated much more generally as d dx

   d du

   du dx
(ii)
Page 10 of 12 Higher Derivatives

David W. Sabo (2000)

In this course, we've really only had to use formula (i) and so we'll restrict our discussion here to that special form of the chain rule. This formula is a great convenience when determining the derivative of a simple whole number power of a binomial in x, say, but it is absolutely essential when the function involved is something like a square root which cannot be expanded into a polynomial with a few terms.
The question is: how do you determine second or higher derivatives of such functions. And the answer is: with great care. Simply taking the derivative of formula (i) gets us initially to d dx
2
2 u n d dx

 n

1  du dx


 d n dx
 u n

1  du dx


(iii)
We can factor the 'n' out of the derivative since it is just a constant. What is left is the derivative of a product, since in general, both u n-1 and du/dx will be functions of x. So, we need to apply the product formula: d dx
2
2 u n n d dx
 u n

1  du dx


 n 




 du dx

 n

1
 u n

2 du dx



 u n

1 d  du 
 
 





1
 u n

2
 du 

 dx


2
 nu n

1 dx 2
That is d 2 dx 2 u n
 
1
 u n

2
 du
 2

 dx


 nu n

1 dx 2
(iv)
This is quite a bit more complicated than formula (i), but was obtained using the basic rules for finding derivatives with which you are now very familiar. For the record, the more general case building on formula
(ii) is d
2 dx
2

 

 d f du
2


 du dx

2

 df dx


 d u dx
2


(v)
Formulas for third and higher order derivatives get much more complicated and really go beyond the scope of this course and typical applications. You need to be aware that in generalizing the chain rule to higher order derivatives, you cannot just "guess" a higher order form of formulas (i) and (ii), but rather, you need to work out the required formulas very cautiously using the other basic formulas for finding derivatives.
Example: Determine the first two derivatives with respect to x of the function y


5 x 2 
3

12
.
Solution
This will require use of formula (i) for the first derivative, and formula (iv) for the second derivative, since we can write y = u 12 with u = 5x 2 + 3. In this case du

10 x dx
Then, formula (I) gives dy dx and


12 u
11
  du

 dx

 dx 2

10 x
2 
3

11 
10 x


120 x

5 x
2 
3

11

David W. Sabo (2000) Higher Derivatives Page 11 of 12

Applying formula (iv) to get the second derivative gives d 2 dx 2
 u 12
 
  u 10
 du  2

 dx



12 u 11 dx 2
    x 2 
3

10 
10 x

2 
 x 2 
3

11  


5 x
2 

10
3 13200 x
2 
 x
2 
3


 x 2 
 x 2 
3

10
We've skipped several minor steps of algebraic manipulation between the second last and the last line here.
Note that we could have easily gotten the same result without using formula (iv) by simply applying our basic rules for derivatives to the first derivative. That is d y
 d dx 2 dx



120 x

5 x 2 
3

11



 d
120 dx

 x

5 x
2 
3

11


120 d dx
  where df f
 x
 
1 dx and g


5 x 2 
3

11
 dg dx

 x 2 
3

10 
10 x


110 x

5 x 2 
3

10 and so dx
2
 d
120 dx
 

120
 g df dx
 f dg dx



120
 
5 x
2 
3

11     

110 x

5 x
2 
3

10





 x 2 

10
3 5 x 2   x 2 
 x 2 
3
 
115 x 2 
3
 which is the same result as we obtained before when we used formula (iv).
Perhaps the best advice arising out of this example is that you don't really need to know formula (iv) to determine the second derivative when the chain rule was involved in finding the first derivative. Just apply the basic rules for finding derivatives to the first derivative expression directly. At the same time, it is important for you to be aware that there is no simple single term generalization of the chain rule for first derivatives to second and higher order derivatives.

Page 12 of 12 Higher Derivatives

David W. Sabo (2000)