Page 19 - (1-1)eng

(1 - 1) The First Derivative of a Function
Example (10)

z 2 dy 1
If y = and z = 2x 1 , prove that = .
1  z 2 dx  2 x 1  2
Solution:
dy dy dz
(b) = ×
dx dz dx

 2 z 1 z 2   2 z  z 2  2
= 2 ×
 1  z 2  2 2 x 1
2 z 1
= ×
 1 z 2  2 2 x 1
2 2 x 1 1
= ×
 1  2 x 1  2 2 x 1
2
=
 2 x  2  2
2
=
4  x  1 2
1
=
2  x 1  2
Another method:
z 2
 y = and z = 2 x 1
1  z 2
2 x 1
 y =
1  2 x 1 

2 x 1
=
2 x  2


dy  2 2 x 2  2 2 x 1 
 =

dx 2 x 2  2
2
=
4  x 1  2

1
= 2
2  x 1 

Calculus 18 Unit (1)