# MATHEMATICS

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```					         C4
MATHEMATICS
SUPPORT CENTRE

Title: Chain Rule: further examples

Target: On completion of this worksheet you should be able to use the chain rule to
differentiate further examples of functions of a function.

The chain rule                                              Exercise
dy dy du                Differentiate the following:
If y  f (u ) and u  g ( x) then                         1. y  cos(2 x  3)
dx du dx
2. y  sin(4 x  1)
Examples                                                    3. y  e ( 5 x  4 )
Differentiate the following:                                4. y  tan( x 3  x)
1. y  sin( x  3)
2
5. y  sin(4 x 5  3 x)
let u  x  3 so y  sin u
2

6. y  e 2 x
2

du                dy
 2x             cosu                                 7. y  ln(3 x  2)
dx               du
dy                                                          8. y  cos(e x )
 cosu  2 x
dx                                                          9. y  ln(cos 4 x)
dy                                                          10. y  ln( x 2  3 x  4)
 2 x cos(x 2  3)
dx                                                          11. y  ln( 4 x 5  3)
12. y  e sin 5 x
2. y  ln(3 x  4 x  2)
3

u  3x 3  4 x  2         y  ln u                         Answers:
du                         dy 1                             1.  2 sin(2 x  3)
 9x 2  4                
dx                         du u                             2. 4 cos(4 x  1)
dy 1                                                        3. 5e ( 5 x  4 )
  (9 x 2  4)
dx u                                                        4. (3 x 2  1) sec2 ( x 3  x)
dy       (9 x 2  4)                                        5. (20 x 4  3) cos(4 x 5  3 x)

dx (3 x 3  4 x  2)
2             3
6. 4 xe 2 x             7.
3x  2
dy u '
                                                                      4 sin 4 x
Note that                                                   8.  e x sin(e x ) 9.
dx u                                                                       cos 4 x
In general if                                                      2x  3                        20 x 4
dy f ' ( x)                               10. 2                          11. 5
y  ln f ( x) then                                             x  3x  4                      4x  3
dx     f ( x)                             12. (5 cos5 x)e   sin 5 x

This is a useful result to remember.

Mathematics Support Centre,Coventry University, 2001
Examples                                                     Exercise
Differentiate the following:                                 1. y  cos3 x
1. y  sin 2 4 x                                             2. y  3 tan 2 2 x
Now sin 4 x  (sin 4 x)
2                 2
1
3. y 
let u  sin 4 x y  u 2                                             sin(2 x  3)
du
 4 cos 4 x
dy
 2u                                   4. y  (e13 x  3 x) 4
dx               du                                          5. s  5 sin 2 (t  7)  3 cos2 2t
dy
 2u  4 cos 4 x                                          6. p  2 ln(3  4q ) 5
dx
dy                                                           7. x  ln(sin t 2 )
 8 sin 4 x cos 4 x
dx                                                           8. x  ln(sin 2 t )
9. y  ln(5 x  3)
2. y  2 cos(3 x  7)   4
4
10. y 
u  (3 x  7) 4
y  2 cosu                                                cos x 3
To differentiate u we must use the chain rule
v  3x  7         u  v4
dv              du
3                4v 3                                   1.  3 cos2 x sin x
dx              dv
du                                                           2. 12 tan 2 x sec2 2 x
 4v 3  3  12v 3  12(3 x  7) 3                              2 cos(2 x  3)
dx                                                           3.
dy                                                                 sin 2 (2 x  3)
 2 sin u
du                                                           4. 12(1  e13 x )(e13 x  3x) 3
dy
 2 sin u  12(3 x  7) 3                                5. 10 sin(t  7) cos(t  7)  12 sin 2t cos 2t
dx                                                                   40
dy                                                           6.
 24(3 x  7) 3 sin(3 x  7) 4                                (3  4q )
dx
2t cost 2
7.        2
 2t cot t 2
sin t
3
3. x                                                           2 cost
tan 5t                                                   8.           2 cot t
sin t
Now x  3(tan 5t ) 1                                                       5
9.
u  tan 5t               x  3u 1                              2(5 x  3) ln(5 x  3)
du                       dx                    3
 5 sec2 5t               1  3u  2   2                   6 x 2 sin x 3
dt                       du                   u              10.         3
cos 2 x 3
dx       3
  2  5 sec2 5t
dt      u
dx         3                    cos2 5t      15
       2
 5 sec2 5t            
dt      tan 5t                     2
sin 5t cos2 5t
dx
 15 cos ec 2 5t
dt

Mathematics Support Centre,Coventry University, 2001

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