guide:3c6a2362c1

[math] \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow} \newcommand{\ddxof}[1]{\frac{d #1}{d x}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\dydx}{\ddxof y} \newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}} \newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}} \newcommand{\dist}{\mathrm{distance}} \newcommand{\arccot}{\mathrm{arccot\:}} \newcommand{\arccsc}{\mathrm{arccsc\:}} \newcommand{\arcsec}{\mathrm{arcsec\:}} \newcommand{\arctanh}{\mathrm{arctanh\:}} \newcommand{\arcsinh}{\mathrm{arcsinh\:}} \newcommand{\arccosh}{\mathrm{arccosh\:}} \newcommand{\sech}{\mathrm{sech\:}} \newcommand{\csch}{\mathrm{csch\:}} \newcommand{\conj}[1]{\overline{#1}} \newcommand{\mathds}{\mathbb} [/math]

The Differential.

If [math]y = f(x)[/math], we have denoted the derivative of [math]f[/math] by [math]f'[/math], or [math]\frac{df}{dx}[/math], or [math]\frac{dy}{dx}[/math]. The value of the derivative at a number a is written [math]f'(a)[/math], or [math]\frac{df}{dx} (a)[/math], or [math]\frac{dy}{dx} (a)[/math]. Up to this point, the expressions [math]df[/math], [math]dy[/math], and [math]dx[/math] by themselves have had no meaning other than as parts of notations for the derivative. However, the cancellation suggested by the Chain Rule

[[math]] \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} [[/math]]

indicates that the derivative behaves like a ratio and suggests that it may be possible to sensibly regard it as such. In this section we shall define a mathematical object called the differential of a function, examples of which are [math]df[/math], [math]dy[/math], and [math]dx[/math]. The ratio of [math]df[/math], or [math]dy[/math], to [math]dx[/math] will be equal to the derivative.

File:Guide c5467 scanfig2 20.png

If [math]f[/math] is a function having a derivative at [math]a[/math], we define its differential at [math]a[/math], denoted by [math]d_{a}f[/math], to be the linear function whose value for any number [math]t[/math] is

[[math]] (d_{a}f)(t) = f'(a)t. [[/math]]

For example, if [math]f(x) = x^2 - 2x[/math], then the differential [math]d_{a}f[/math] is the function of [math]t[/math] defined by [math]f'(a)t = (2a - 2)t[/math]. In particular,

[[math]] (d_{3}f)(t)= [2 \cdot 3 - 2]t = 4t. [[/math]]

The value of the differential for a typical function [math]f[/math] is illustrated in Figure. By simply [math]df[/math] we mean the rule (or function) that assigns the linear function [math]d_{x}f[/math] to each number [math]x[/math] in the domain of [math]f'[/math].

Theorem

If [math]f[/math] and [math]u[/math] are differentiablefunctions, then

[[math]] \begin{equation} df(u) = f'(u)du. \end{equation} [[/math]]

Show Proof

This formula is an abbreviation of the equation

[[math]] \begin{equation} d_{x} f(u) = f'(u(x)) d_{x}u. \label{eq2.6.2} \end{equation} [[/math]]

The proof is an application of the Chain Rule. We first write down the two linear functions [math]d_{x}f(u)[/math] and [math]d_{x}u[/math]. By the definition of the differential they are

[[math]] \begin{eqnarray*} (d_{x}u)(t) &=& u'(x)t,\\ (d_{x}f(u))(t) &=& [[f(u)]'(x)]t. \end{eqnarray*} [[/math]]

The Chain Rule says that [math][f(u)]'(x) = f'(u(x))u'(x)[/math]. Hence

[[math]] \begin{eqnarray*} (d_{x} f(u) )(t) &=& [f'(u(x) )u'(x)]t \\ &=& f'(u(x))(d_{x}u)(t). \end{eqnarray*} [[/math]]

Thus (2) appears as an equality between linear functions, and the proof is complete.

■

If [math]u[/math] is the independent variable [math]x[/math], then (6.1) reduces to the formula

[[math]] \begin{equation} df(x) = f'(x)dx. \label{eq2.6.3} \end{equation} [[/math]]

Example

Evaluate the following differentials:

\item[a] [math]d(x^2 + 2)[/math], \item[b] [math]d\sqrt{x^2 + 3}[/math], \item[c] [math]d(2x^2-x)^7[/math].]

Using formula (3), we get immediately

\item[(a') [math]d(x^2 + 2) = 2xdx[/math] \item[(b') [math]d\sqrt{x^2 + 3} = x(x^2 + 3)^{-1/2} dx[/math]] \item[(c') [math]d(2x^2-x)^7 = 7(2x^2-x)^6(4x-1)dx[/math].]

It is worthwhile learning to use the stronger formula (1). In problem (b), let [math]f[/math] be the function [math]f(u) = \sqrt{u}[/math]. If we set [math]u = x^2 + 3[/math], then [math]du = 2x dx[/math] and

[[math]] \begin{eqnarray*} d\sqrt{x^2 + 3} &=& df(u) = f'(u) du\\ &=& \frac{1}{2}u^{-1/2}du\\ &=& \frac{1}{2}(x^2 + 3)^{-1/2}2x dx. \end{eqnarray*} [[/math]]

Let us also do problem (c) using (1), but without explicitly making the substitution [math]u = 2x^2 - x[/math]. We get

[[math]] \begin{eqnarray*} d(2x^2 - x)^7 &=& 7(2x^2 - x)^{6} d(2x^2 - x)\\ &=& 7(2x^2 - x)^{6} (4x - 1) dx. \end{eqnarray*} [[/math]]

Formula (3) establishes the fact that the ratio of [math]df[/math] to [math]dx[/math] is equal to the derivative [math]f'[/math]. We can see this in greater detail by going back to the definitions:

[[math]] \begin{eqnarray*} (d_{a}f)(t) &=& f'(a)t, \\ (d_{a}x)(t) &=& x'(a)t. \end{eqnarray*} [[/math]]

Since [math]x[/math] is the identity function, its derivative is the constant function 1. Hence [math](d_{a}x)(t) = t.[/math] The ratio of the two linear functions [math]d_{a}f[/math] and [math]d_{a}x[/math] is thus the constant function

[[math]] \frac{d_{a}f}{d_{a}x} = \frac{f'(a)t}{t} = f'(a). [[/math]]

Having proved this formula for every a in the domain of [math]f'[/math], we can write it simply as

[[math]] \frac{df}{dx} (a) = f'(a) \;\;\; \mbox{or} \;\;\; \frac{df}{dx} = f'. [[/math]]

If [math]f[/math] and [math]g[/math] are differentiable functions, then it is easy to show that [math]d_{a}(f + g) = d_{a}f + d_{a}g[/math]. The proof involves only the definition of the differential plus the fact that the derivative of a sum is the sum of the derivatives. In detail:

[[math]] \begin{eqnarray*} [d_{a}(f + g)](t) &=& [(f + g)'(a)] t = [f' (a) + g'(a)] t\\ &=& f'(a)t + g'(a)t = (d_{a}f)(t) + (d_{a}g)(t)\\ &=& [d_{a}f + d_{a}g](t) . \end{eqnarray*} [[/math]]

The result is simply the equation

[[math]] d(f + g) = df + dg. [[/math]]

An analogous argument using the Product Rule for differentiation shows that [math]d_{a}(fg) = f(a)d_{a}g + g(a)d_{a}f[/math], or, more simply,

[[math]] d(fg) = f dg + g df. [[/math]]

For each one of the six differentiation rules proved in Section \secref{1.7} of Chapter \ref{chp 1}, there is an analogous rule in terms of differentials: Let [math]u[/math] and [math]v[/math] be differentiable functions, and [math]c[/math] a constant. Then

Proposition

\item [(i)] [math]d(u + v) = du + dv[/math], \item [(ii)] [math]d(cu)= cdu[/math], \item [(iii)] [math]dc= 0[/math], \item [(iv)] [math]d(uv) = u dv + v du[/math], \item [(v)] [math]du^r = ru^{r-1}du, \provx{where [/math]r[math] is any rational number,}[/math] \item [(vi)] [math]d \biggl(\frac{u}{v} \biggr) = \frac{vdu - udv}{v^2}[/math].

Note that we have replaced the analogue of (v) in the list in Section \secref{1.7} of Chapter \ref{chp 1} by the formula corresponding to the more powerful theorem of Chapter \ref{chp 1}.

Example

Find the differential [math]d(x^3 + \sqrt{x^2 + 2x})^7[/math]. Applying the above formulas successively, we get

[[math]] \begin{eqnarray*} d(x^3 + \sqrt{x^2 + 2x})^7 &=& 7(x^3 + \sqrt{x^2 + 2x})^6 d(x^3 + \sqrt{x^2 + 2x}) \;\;\;\mbox{by (v)}\\ &=& 7(x^3 + \sqrt{x^2 + 2x})^6 (dx^3 + d\sqrt{x^2 + 2x}) \;\;\;\mbox{by (i)}\\ &=& 7(x^3 + \sqrt{x^2 + 2x})^6 [3x^2 dx + \frac{1}{2}(x^2 + 2x)^{-1/2} d(x^2 + 2x)] \\ & & \mbox{by (v)}\\ &=& 7(x^3 + \sqrt{x^2 + 2x})^6 [3x^2 dx + \frac{1}{2}(x^2 + 2x)^{-1/2} (2x dx + 2 dx)] \\ & & \mbox{by (i) and (ii)} \\ &=& 7(x^3 + \sqrt{x^2 + 2x})^6 \biggl( 3x^2 +\frac{x+1}{\sqrt{x^2 + 2x}}\biggr) dx. \end{eqnarray*} [[/math]]

The derivative is therefore given by

[[math]] \frac{d(x^3 + \sqrt{x^2 + 2x})^7}{dx} = 7(x^3 + \sqrt{x^2 + 2x})^6 \biggl(3x^2 + \frac{x + 1}{\sqrt{x^2 + 2x}}\biggr). [[/math]]

The task of computing the differential of a complicated function of [math]x[/math] amounts to successively working the differential operator d through the given expression from left to right. At each stage one uses the correct one of formulas (i) through (vi), or formula (1), until one finally reaches [math]dx[/math], and the process stops. The derivative can then be obtained by dividing the resulting equation by [math]dx[/math]. Note that an equation of the form [math]df(x) = ...[/math] will always contain the symbol [math]d[/math] on the right side. Equations such as [math]dx^5 = 5x^4[/math] are not only false; they are nonsense. (Correct version: [math]dx^5 = 5x^4 dx[/math].) Example

Consider the functions

\item [(a) [math]y = (4x^3 + 3x^2 + 1)^2,[/math]] \item [(b) [math]y = \frac{x^2 -1}{x^2 + 1}[/math]] \item [(c) [math]z = 3y^{5/3}.[/math]]

Find the differential of each:

[[math]] \begin{eqnarray*} \mbox{(a')}\; dy &=& 2(4x^3 + 3x^2 + 1) d(4x^3 + 3x^2 + 1) \\ &=& 2(4x^3 + 3x^2 + 1)(12x^2 dx + 6x dx) \\ &=& 12x(4x^3 + 3x^2 + 1)(2x + 1) dx, \end{eqnarray*} [[/math]]

[[math]] \begin{eqnarray*} \mbox{(b')}\; dy &=& \frac{(x^2 + 1 ) d(x^2 - 1 ) - (x^2 - 1 ) d(x^2 + 1 )}{(x^2 + 1 )^2}\\ &=& \frac{(x^2 + 1)2xdx - (x^2 - 1)2xdx}{(x^2 + 1)^2} = \frac{4x dx}{(x^2 + 1)^2}, \end{eqnarray*} [[/math]]

[[math]] \mbox{(c')}\; dz = (3)(\frac{5}{3})y^{2/3}dy= 5y^{2/3}dy. [[/math]]

If we consider the composition of the function [math]y[/math] in (b) with the function [math]z[/math] in (c), we get for the differential of the composition

[[math]] dz = 5y^{2/3} \frac{4x dx}{(x^2 + 1)^2} = 5 \biggl(\frac{x^2 - 1}{x^2 + 1} \biggr)^{2/3} \frac{4x}{(x^2 + 1)^2} dx. [[/math]]

One traditional interpretation of the differential, which is especially useful in physics, is that of an “infinitesimal.” If [math]y = f(x)[/math], we know that [math]dy = f'(x) dx[/math]. Now [math]dx[/math] is the function that assigns to every real number a the linear function defined by [math](d_{a}x)(t) = x'(a)t = t[/math]; i.e., it assigns the identity function. Hence we can interpret [math]dx[/math] as simply another independent variable. Then [math]dy[/math] is the variable whose value for a given [math]x[/math] and [math]dx[/math] is shown in Figure. (Compare this illustration with Figure.) The difference between the value of [math]f[/math] at [math]x[/math] and at [math]x + dx[/math] is denoted by [math]\Delta y[/math] in the figure. If [math]dx[/math] is chosen [math]y[/math]-axis very small, then the difference between [math]dy[/math] and [math]\Delta y[/math] is relatively negligible. Hence [math]dy[/math] measures the resulting change in the value of [math]y = f(x)[/math] corresponding to an infinitesimal change [math]dx[/math] in the variable [math]x[/math].

File:Guide c5467 scanfig2 21.png

Example

The height [math]h[/math] of a square pyramid is found to be 100 feet, and the length [math]x[/math] of one edge of its base is measured to be 160 feet. The volume [math]V[/math] of the pyramid is given by the formula [math]V = \frac{1}{3}hx^2[/math]. What error in the computed volume will result from an error of 4 inches in the measurement of [math]x[/math]? If we consider [math]h[/math] as fixed, and [math]V[/math] as a function of [math]x[/math], then

[[math]] dV = \frac{1}{3}h dx^2 = \frac{2}{3}hx dx. [[/math]]

Since 4 inches is small compared with 160 feet, we set [math]dx[/math] = 4 inches = [math]\frac{1}{3}[/math] foot. The resulting change in volume is then approximately

[[math]] dV = \frac{2}{3}(100)(160)\frac{1}{3} = \frac{32}{9} 1000 = 3555 \;\mbox{feet}^3. [[/math]]

The percentage error in volume is

[[math]] \frac{dV}{V} = \frac{ {\frac{2}{3}} hx dx}{{\frac{1}{3}}hx^2} = 2 \frac{dx}{x}. [[/math]]

We compute [math]\frac{dx}{x} = \frac{\frac{1}{3}}{160} = 0.0021 = 0.21\%[/math], and so the percentage error in volume is only [math]0.42\%[/math].

\end{exercise}

General references

Doyle, Peter G. (2008). "Crowell and Slesnick's Calculus with Analytic Geometry" (PDF). Retrieved Oct 29, 2024.