Let y=x^3, find the differential

Question

Let y=x^3, find the differential dy of y=x^3 at x=2

Answer 1

y=x^3

y' = 3x^2

y'(2) = 3(2)^2 = 3(4) = 12

y' = dy/dx = 12 dy= 12dx

(1.9)^3 = 6.859

dx= 2-1.9 = 0.1

dy= 12(0.1) = 1.2

y(2) = (2)3 = 8

8/1.2 = 6.667 which is close to 6.859

Answer 2

The differential of the function f, denoted df, is defined as df=(f'(x))dx.

Note that there are three variables here:

df, dx, and x.

In this case, since y=x^3, y'=3x^2. So the general differential of y is

dy=(3x^2)dx.

At the specific point x=2,

dy = 12dx.

Recall that dy gives the approximate change in y (approximated by using a linear approximation), given a change dx, relative to the point x.

So, in this specific case, at x=2, if x changes by dx, y will change by approximately 12*dx.

A quick example: x = 2. If we change x to 2.1, that means dx=0.1.

dy = 12dx = 12(.1)=1.2.

So, as x changes from 2 to 2.1, y will change by about 1.2 (i.e. from 8 to about 9.2).

Note that (2.1)^3=9.261 (rounded), so it's a fairly good approximation.

As you've learned, however, a linear approximation is typically less and less accurate the further you move from the reference point, i.e. the larger dx is