The definition of the derivative of a function *y* = *f(x)* as you recall is

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,

The differential of the independent variable *x* is written *dx* and is the same as the change in *x*, Δ *x*. That is,

The differential of the dependent variable *y*, written *dy*, is defined to be

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 ) .

**Figure 1 **Approximating a function with differentials.

**Example 1:** Find *dy* for *y* = *x* ^{3} + 5 *x* −1.

**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

Because *x* is increasing from 6 to 6.23, you find that Δ *x* = *dx* = .23 cm; hence,

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

Because the function you are applying is *x* that is a perfect cube and is relatively close to 26.55, namely *x* = 27. The differential *dy* is

Because *x* is decreasing from 27 to 26.55, you find that Δ *x* = *dx* = −.45; hence,

which implies that

to the nearest thousandth.

Note that the calculator value of