Linear Approximations and Differentials: Learn It 2

Differentials and Amount of Error

Computing Differentials

We have seen that linear approximations can be used to estimate function values. They can also be used to estimate the amount a function value changes as a result of a small change in the input. To discuss this more formally, we define a related concept: differentials. Differentials provide us with a way of estimating the amount a function changes as a result of a small change in input values.

When we first looked at derivatives, we used the Leibniz notation dy/dxdy/dx to represent the derivative of yy with respect to xx. Although we used the expressions dydy and dxdx in this notation, they did not have meaning on their own.

Here we see a meaning to the expressions dydy and dxdx. Suppose y=f(x)y=f(x) is a differentiable function. Let dxdx be an independent variable that can be assigned any nonzero real number, and define the dependent variable dydy by

dy=f(x)dxdy=f(x)dx/

It is important to notice that dydy is a function of both xx and dxdx. The expressions dydy and dxdx are called differentials.

We can divide both sides of the equation by dxdx, which yields

dydx=f(x)dydx=f(x)

This is the familiar expression we have used to denote a derivative. The first equation is known as the differential form of the second one.

differentials

Differentials, denoted as dydy and dxdx, provide a method to estimate the rate of change of a function y=f(x)y=f(x) due to a small change in xx.

 

By representing the derivative dydx=f(x)dydx=f(x) in terms of differentials, dydy can be understood as the change in yy resulting from an infinitesimal increment dxdx in xx.

For each of the following functions, find dydy and evaluate when x=3x=3 and dx=0.1dx=0.1.

  1. y=x2+2xy=x2+2x
  2. y=cosxy=cosx

We now connect differentials to linear approximations. Differentials can be used to estimate the change in the value of a function resulting from a small change in input values.

Consider a function ff that is differentiable at point aa. Suppose the input xx changes by a small amount. We are interested in how much the output yy changes. If xx changes from aa to a+dxa+dx, then the change in xx is dxdx (also denoted ΔxΔx), and the change in yy is given by

Δy=f(a+dx)f(a)Δy=f(a+dx)f(a)

Instead of calculating the exact change in yy, however, it is often easier to approximate the change in yy by using a linear approximation.

For xx near aa, f(x)f(x) can be approximated by the linear approximation

L(x)=f(a)+f(a)(xa)L(x)=f(a)+f(a)(xa)

Therefore, if dxdx is small,

f(a+dx)L(a+dx)=f(a)+f(a)(a+dxa)f(a+dx)L(a+dx)=f(a)+f(a)(a+dxa)

That is,

f(a+dx)f(a)L(a+dx)f(a)=f(a)dx.f(a+dx)f(a)L(a+dx)f(a)=f(a)dx.

In other words, the actual change in the function ff if xx increases from aa to a+dxa+dx is approximately the difference between L(a+dx)L(a+dx) and f(a)f(a), where L(x)L(x) is the linear approximation of ff at aa. By definition of L(x)L(x), this difference is equal to f(a)dxf(a)dx.

In summary,

Δy=f(a+dx)f(a)L(a+dx)f(a)=f(a)dx=dyΔy=f(a+dx)f(a)L(a+dx)f(a)=f(a)dx=dy

Therefore, we can use the differential dy=f(a)dxdy=f(a)dx to approximate the change in yy if xx increases from x=ax=a to x=a+dxx=a+dx. We can see this in the following graph.

A function y = f(x) is shown along with its tangent line at (a, f(a)). The tangent line is denoted L(x). The x axis is marked with a and a + dx, with a dashed line showing the distance between a and a + dx as dx. The points (a + dx, f(a + dx)) and (a + dx, L(a + dx)) are marked on the curves for y = f(x) and y = L(x), respectively. The distance between f(a) and L(a + dx) is marked as dy = f’(a) dx, and the distance between f(a) and f(a + dx) is marked as Δy = f(a + dx) – f(a).
Figure 5. The differential dy=f(a)dxdy=f(a)dx is used to approximate the actual change in yy if xx increases from aa to a+dxa+dx.

We now take a look at how to use differentials to approximate the change in the value of the function that results from a small change in the value of the input. Note the calculation with differentials is much simpler than calculating actual values of functions and the result is very close to what we would obtain with the more exact calculation.

Let y=x2+2xy=x2+2x.

Compute ΔyΔy and dydy at x=3x=3 if dx=0.1dx=0.1.