Page 17 - (1-1)eng

(1 - 1) The First Derivative of a Function
Example (8)
z x dy
If y = , z = , prove that = 1.


z 1 1 x dx
Solution:
dy dy dz
(a) = ×
dx dz dx


x
  1 z 1       1 1 x   1  
z
 1
= ×
 z 1  2  1 x  2
1 1
= ×
 z  1 2  1 x  2
1 1
= 
 x  2  1 x  2
  
1
1
  x 
1
= 2
 x   
1
   1     1 x 

  x
1
= = 1
 x   x  2
1

Another method:
z x
 y = and z =
z 1 1  x

x

 y = 1  x   1 x 

x 1   1 x 
1  x

x
=
x   1 x 
x
=
1
= x

dy
 = 1
dx

Calculus 16 Unit (1)