Page 16 - (1-1)eng

(1 - 1) The First Derivative of a Function

Example (7)
2 z dy
2
If y = and z = x – 3 , find
z  2 dx
Solution:
2z
2
 y = and z = x – 3
z  2
dy dy dz
 = 
dx dz dx

2  z 2    2 z 
 1

=   2 x
 z 2  2
4
2 z   2 z

=   2 x
 z 2  2
4

= 2   2 x
  x 2 3   2 
8x
=
 x 2 1  2

Another method:
2z
2
 y = and z = x – 3
z  2
2  x 2 3 
=
 x 2 3   2

2 x 2 6
=
x 2 1


dy 4x x 2 1  2x 2x 2 6 
 =
dx  x 2 1  2
3
3
4x 4x  4x 12x
=
 x 2 1  2
8x
= 2
 x 2 1 

Calculus 15 Unit (1)