Page 15 - (1-1)eng

(1 - 1) The First Derivative of a Function

7) Chain rule y= ƒ ( z) and z = g ( x ) dy dy dz
 
dx dz dx
( ƒ  g ) ( x ) y = ƒ   y  = ƒ'   × g' ( x )
 g x
 g x

Example (5)
2
If y = ( ƒ  g ) ( x ) where ƒ ( x ) = x , g ( x ) = 5 x + 1 , find dy .
dx
Solution: Another method:

 y = ( ƒ  g ) ( x )  y = ( ƒ  g ) ( x )

= ƒ ( g ( x ) ) = ƒ ( g ( x ) )

2
= ( 5 x + 1 )  y = ƒ ( g ( x ) )  g ( x )

2
= 25 x + 10 x + 1 = 2 ( 5 x + 1 )  5

dy
 = 50 x + 10 = 50 x + 10
dx

Example (6)
2
2
4
If ƒ ( x ) = x + 2 x + 1 , x > 0 , find ƒ ( 9 ).
Solution: Another method:
2
2
2
2
2
4
 ƒ ( x ) = x + 2 x + 1  ƒ ( x ) = ( x + 1 )

2
3
2
 ƒ ( x )  2 x = 4 x + 4 x Put z = x
2
2
 ƒ ( x ) = 4x 3  4x  2x 2  2  ƒ ( z ) = ( z + 1 )
2x
2
 ƒ ( 9 ) = 2 ( 3 ) + 2  ƒ ( z ) = 2 ( z + 1 )

= 18 + 2 = 20  ƒ ( 9 ) = 2 ( 10 ) = 20

Calculus 14 Unit (1)