The derivative of a function can often be used to approximate certain function values with a surprising degree of accuracy. To do this, the concept of the differential of the independent variable and the dependent variable must be introduced.
The definition of the derivative of a function y = f(x) as you recall is
![](https://s3.amazonaws.com/dev-hmhco-vmg-craftcms-public/_cliffsnotes/assets/39398.gif)
which represents the slope of the tangent line to the curve at some point ( x, f(x)). If Δ x is very small (Δ x ≠ 0), then the slope of the tangent is approximately the same as the slope of the secant line through ( x, f(x)). That is,
![](https://s3.amazonaws.com/dev-hmhco-vmg-craftcms-public/_cliffsnotes/assets/39400.gif)
The differential of the independent variable x is written dx and is the same as the change in x, Δ x. That is,
![](https://s3.amazonaws.com/dev-hmhco-vmg-craftcms-public/_cliffsnotes/assets/39402.gif)
The differential of the dependent variable y, written dy, is defined to be
![](https://s3.amazonaws.com/dev-hmhco-vmg-craftcms-public/_cliffsnotes/assets/39404.gif)
The conclusion to be drawn from the preceding discussion is that the differential of y(dy) is approximately equal to the exact change in y(Δ y), provided that the change in x (Δ x = dx) is relatively small. The smaller the change in x, the closer dy will be to Δ y, enabling you to approximate function values close to f(x) (Figure ) .
![](http://s3.amazonaws.com/prod-hmhco-vmg-craftcms-public/_cliffsnotes/assets/39422.jpg)
Figure 1 Approximating a function with differentials.
Example 1: Find dy for y = x 3 + 5 x −1.
![](https://s3.amazonaws.com/dev-hmhco-vmg-craftcms-public/_cliffsnotes/assets/39406.gif)
Example 2: Use differentials to approximate the change in the area of a square if the length of its side increases from 6 cm to 6.23 cm.
Let x = length of the side of the square. The area may be expressed as a function of x, where y = x 2. The differential dy is
![](https://s3.amazonaws.com/dev-hmhco-vmg-craftcms-public/_cliffsnotes/assets/39408.gif)
Because x is increasing from 6 to 6.23, you find that Δ x = dx = .23 cm; hence,
![](https://s3.amazonaws.com/dev-hmhco-vmg-craftcms-public/_cliffsnotes/assets/39410.gif)
The area of the square will increase by approximately 2.76 cm 2 as its side length increases from 6 to 6.23. Note that the exact increase in area (Δ y) is 2.8129 cm 2.
Example 3: Use differentials to approximate the value of
to the nearest thousandth.
Because the function you are applying is
, choose a convenient value of x that is a perfect cube and is relatively close to 26.55, namely x = 27. The differential dy is
![](https://s3.amazonaws.com/dev-hmhco-vmg-craftcms-public/_cliffsnotes/assets/39412.gif)
Because x is decreasing from 27 to 26.55, you find that Δ x = dx = −.45; hence,
![](https://s3.amazonaws.com/dev-hmhco-vmg-craftcms-public/_cliffsnotes/assets/39414.gif)
which implies that
will be approximately 1/60 less that
; hence,
to the nearest thousandth.
Note that the calculator value of
is 2.983239874, which rounds to the same answer to the nearest thousandth!