{"id":299,"date":"2023-09-20T22:48:55","date_gmt":"2023-09-20T22:48:55","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/differentials\/"},"modified":"2024-08-05T02:11:41","modified_gmt":"2024-08-05T02:11:41","slug":"linear-approximations-and-differentials-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/linear-approximations-and-differentials-learn-it-2\/","title":{"raw":"Linear Approximations and Differentials: Learn It 2","rendered":"Linear Approximations and Differentials: Learn It 2"},"content":{"raw":"

Differentials and Amount of Error<\/h2>\r\n

Computing Differentials<\/h3>\r\n

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.<\/p>\r\n

When we first looked at derivatives, we used the Leibniz notation [latex]dy\/dx[\/latex] to represent the derivative of [latex]y[\/latex] with respect to [latex]x[\/latex]. Although we used the expressions [latex]dy[\/latex] and [latex]dx[\/latex] in this notation, they did not have meaning on their own.<\/p>\r\n

Here we see a meaning to the expressions [latex]dy[\/latex] and [latex]dx[\/latex]. Suppose [latex]y=f(x)[\/latex] is a differentiable function. Let [latex]dx[\/latex] be an independent variable that can be assigned any nonzero real number, and define the dependent variable [latex]dy[\/latex] by<\/p>\r\n

[latex]dy=f^{\\prime}(x) \\, dx[\/latex]\/<\/div>\r\n

It is important to notice that [latex]dy[\/latex] is a function of both [latex]x[\/latex] and [latex]dx[\/latex]. The expressions [latex]dy[\/latex] and [latex]dx[\/latex] are called differentials<\/strong>.<\/p>\r\n

We can divide both sides of the equation by [latex]dx[\/latex], which yields<\/p>\r\n

[latex]\\frac{dy}{dx}=f^{\\prime}(x)[\/latex]<\/div>\r\n

This is the familiar expression we have used to denote a derivative. The first equation is known as the differential form<\/strong> of the second one.<\/p>\r\n

\r\n

differentials<\/h3>\r\n

Differentials, denoted as [latex]dy[\/latex] and [latex]dx[\/latex], provide a method to estimate the rate of change of a function [latex]y=f(x) [\/latex] due to a small change in [latex]x[\/latex].<\/p>\r\n

 <\/p>\r\n

By representing the derivative [latex]\\frac{dy}{dx}=f^{\\prime}(x)[\/latex] in terms of differentials, [latex]dy[\/latex] can be understood as the change in [latex]y[\/latex] resulting from an infinitesimal increment [latex]dx[\/latex] in [latex]x[\/latex].<\/p>\r\n<\/section>\r\n

\r\n

For each of the following functions, find [latex]dy[\/latex] and evaluate when [latex]x=3[\/latex] and [latex]dx=0.1[\/latex].<\/p>\r\n

    \r\n\t
  1. [latex]y=x^2+2x[\/latex]<\/li>\r\n\t
  2. [latex]y= \\cos x[\/latex]<\/li>\r\n<\/ol>\r\n

    [reveal-answer q=\"fs-id1165042926541\"]Show Solution[\/reveal-answer]
    \r\n[hidden-answer a=\"fs-id1165042926541\"]<\/p>\r\n

    The key step is calculating the derivative. When we have that, we can obtain [latex]dy[\/latex] directly.<\/p>\r\n

      \r\n\t
    1. Since [latex]f(x)=x^2+2x[\/latex], we know [latex]f^{\\prime}(x)=2x+2[\/latex], and therefore
      \r\n
      [latex]dy=(2x+2) \\, dx[\/latex].<\/div>\r\n

      When [latex]x=3[\/latex] and [latex]dx=0.1[\/latex],<\/p>\r\n

      [latex]dy=(2 \\cdot 3+2)(0.1)=0.8[\/latex].<\/div>\r\n<\/li>\r\n\t
    2. Since [latex]f(x)= \\cos x[\/latex], [latex]f^{\\prime}(x)=\u2212\\sin (x)[\/latex]. This gives us
      \r\n
      [latex]dy=\u2212\\sin x \\, dx[\/latex].<\/div>\r\n

      When [latex]x=3[\/latex] and [latex]dx=0.1[\/latex],<\/p>\r\n

      [latex]dy=\u2212\\sin (3)(0.1)=-0.1 \\sin (3)[\/latex].<\/div>\r\n<\/li>\r\n<\/ol>\r\n

      Watch the following video to see the worked solution to this example.<\/p>\r\n