Page 20 - (1-1)eng

(1 - 1) The First Derivative of a Function

9) Parametric function First derivative

If x , y are each expressed in terms of
a third variable t say, called the parameter, (a) Find dx and dy separately.
then x = ƒ( t ) , y = g ( t ) give the dt dt
parametric form of the equation relating x
and y. dy dy dt
(b) Use  
dx dt dx

Example (11)

3
2
(a) If y = t and x = t , then find dy .
dx


x 1 x 1 dy
(b) If y = and z = , find .


x 1 x 1 dz
Solution:
3
2
(a) y = t x = t
2
dy = 3 t dx = 2 t
dt dt
dy dy dt 1 3
2
=  = 3 t  = t
dx dt dx 2t 2


x 1 x 1
(b) y = z =


x 1 x 1
dy 1  x 1  1 x    1 dz 1  x 1  1 x 1   
= =
dx  x 1  2 dx  x 1  2
 2 2
= , =
 x 1  2  x 1  2

dy dy dx  2  x 1  2  x 1  2
=  =  = –  

dz dx dz  x 1  2 2  x 1 
Another method:
y z = 1
1
y = = z – 1
z


dy  x 1   2  x 1  2
– 2
= – z = –   = –  


dz  x 1   x 1 
Example (12)
3
2
Find the rate of change of x + 16 with respect to x – 4
Solution:
2
3
Let z = x + 16 and y = x – 4
dz dy
2
= 2 x = 3 x
dx dx
dz dz dx 1 2
 =  = 2 x  =
dy dx dy 3 x 2 3 x
Calculus 19 Unit (1)