This concept can be understood by calculating the the
value of Δy
and dy when the value of x changes from 2.02 to 2.03 for the function
of y = x3 - 2x2 + 4.
Substitute a = 2.02 into y = y = x3 - 2x2 + 4,
f(a) = f(2.02) = 2.023 - 2(2.02)2 + 4 = 4.081608
so point A(2.02, 4.081608)
f(a) = f(2.03) = 2.033 - 2(2.03)2 + 4 = 4.123627
This gives point B(2.03, 4.123627)
Δy = f(2.03) - f(2.02)
= 4.123627 - 4.081608
= 0.042019 The derivative of y is
y ' = (x3 - 2x2 + 4) ' =
3x2 - 6
According to differential definition,
dy = y 'dx = (3x2 - 6)dx
In this example, dx = Δx = 2.03 -2.02 = 0.01.
Substituting dx = 0.01 and x = 2.02 into equation dy gives
dy = (3x2 - 6)dx = (3(2.02)2-6)(0.01) = 0.062412
Notice that Δy and dy are very close in this
case.The formula
f(a + Δx) ≈ f(a)
+ dy
or
f(a + Δx) ≈ f(a)
+ f '(x-a)
is called the linear approximation of the function at point a. |