082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

10.5 Rate of change at a general point 365
y
dy = y(x + dx) – y(x)
y(x + dx)
y(x)
x
x + dx
dx
x
Figure10.9
Therate ofchange ofyat apoint is foundby letting δx → 0.
Example10.6 Findtherateofchangeofy(x) = 2x 2 +3x.Calculatetherateofchangeofywhenx = 2
and whenx = −3.
Solution Giveny(x) = 2x 2 +3x
then
Hence
So,
y(x + δx) = 2(x + δx) 2 +3(x + δx)
= 2x 2 +4xδx +2(δx) 2 +3x +3δx
y(x + δx) −y(x) =2(δx) 2 +4xδx +3δx
( ) y(x + δx) −y(x)
rate ofchangeofy = lim
δx→0 δx
= lim
δx→0
( 2(δx) 2 +4xδx +3δx
δx
)
= lim
δx→0
(2δx+4x+3)=4x+3
Whenx = 2,therateofchangeofyis4(2) +3 = 11.Whenx = −3,therateofchange
ofyis4(−3) +3 = −9.Apositiverateofchangeshowsthatthefunctionisincreasing
at that particular point. A negative rate of change shows that the function is decreasing
atthatparticular point.
( ) δy
Therateofchangeofyiscalledthederivativeofy.Wedenote lim by dy
δx→0 δx dx .This
ispronounced ‘deeyby deex’.
rate ofchangeofy(x) = dy ( ) y(x + δx) −y(x)
dx = lim
δx→0 δx
Note thatthe notation dy δy
means lim
dx δx→0 δx .