2 Notation for derivatives

Just as there is a notation for the first derivative so there is a similar notation for higher derivatives. Consider the function, y ( x ) . We know that the first derivative is d y d x or d d x ( y ) which is the instruction to differentiate the function y ( x ) . The second derivative is calculated by differentiating the first derivative, that is

second derivative =  d d x d y d x

So, using a fairly obvious adaptation of our derivative notation, the second derivative is denoted by d 2 y d x 2 and is read as ‘dee two y by dee x squared’. This is often written more concisely as y .

In similar manner, the third derivative is denoted by d 3 y d x 3 or y and so on. So, referring to Example 6 we could have written

first derivative = d y d x = 4 x 3 + 12 x second derivative = d 2 y d x 2 = 12 x 2 + 12 third derivative = d 3 y d x 3 = 24 x
Key Point 7

If y = y ( x ) then its first, second and third derivatives are denoted by:

d y d x d 2 y d x 2 d 3 y d x 3
or y y y

In most examples we use x to denote the independent variable and y the dependent variable. However, in many applications, time t is the independent variable. In this case a special notation is used for derivatives. Derivatives with respect to t are often indicated using a dot notation, so d y d t can be written as , pronounced ‘ y dot’. Similarly, a second derivative with respect to t can be written as ÿ , pronounced ‘ y double dot’.

Key Point 8

If y = y ( t ) then

stands for d y d t , ÿ stands for d 2 y d t 2 etc
Task!

Calculate d 2 y d t 2 and d 3 y d t 3 given y = e 2 t + cos t .

First find d y d x :

d y d t = 2 e 2 t sin t

Now obtain the second derivative:

4 e 2 t cos t

Finally, obtain the third derivative:

8 e 2 t + sin t

Note that in the last Task we could have used the dot notation and written = 2 e 2 t sin t , ÿ = 4 e 2 t cos t and ÿ ̇ = 8 e 2 t + sin t

We may need to evaluate higher derivatives at specific points. We use an obvious notation.

The second derivative of y ( x ) , evaluated at say, x = 2 , is written as d 2 y d x 2 ( 2 ) , or more simply as y ( 2 ) . The third derivative evaluated at x = 1 is written as d 3 y d x 3 ( 1 ) or y ( 1 ) .

Task!

Given y ( x ) = 2 sin x + 3 x 2 find

  1.   y ( 1 )  
  2. y ( 1 )  
  3. y ( 0 )

First find y ( x ) , y ( x ) and y ( x ) :

y ( x ) = 2 cos x + 6 x y ( x ) = 2 sin x + 6 y ( x ) = 2 cos x

Now substitute x = 1 in y ( x ) to obtain y ( 1 ) :

y ( 1 ) = 2 cos 1 + 6 ( 1 ) = 7.0806 . Remember, in cos 1 the ‘ 1 ’ is 1 radian.

Now find y ( 1 ) :

y ( 1 ) = 2 sin ( 1 ) + 6 = 7.6829 Finally, find y ( 0 ) :

  y ( 0 ) = 2 cos 0 = 2 .

Exercises
  1. Find d 2 y d x 2 where y ( x ) is defined by:
    1. 3 x 2 e 2 x   
    2. sin 3 x + cos x   
    3.   x   
    4. e x + e x   
    5. 1 + x + x 2 + ln x
  2. Find d 3 y d x 3 where y is given in Exercise 1.
  3. Calculate ÿ ( 1 ) where y ( t ) is given by:
    1.   t ( t 2 + 1 )   
    2.   sin ( 2 t )   
    3. 2 e t + e 2 t   
    4. 1 t   
    5. cos t 2
  4. Calculate y ... ( 1 ) for the functions given in Exercise 3.
    1. 6 4 e 2 x   
    2. 9 sin 3 x cos x   
    3. 1 4 x 3 2   
    4. e x + e x   
    5. 2 1 x 2
    1. 8 e 2 x   
    2. 27 cos 3 x + sin x   
    3. 3 8 x 5 2   
    4. e x e x   
    5. 2 x 3
    1. 6   
    2. 3.6372   
    3. 34.9927   
    4. 2   
    5. 0.2194
    1. 6   
    2. 3.3292   
    3. 1.8184   
    4. 6   
    5. 0.0599