Calculus I 03.09 Differentials

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3.9 Differentials

• Tangent line approximation.
• Compare \(dy\) with the actual change in \(y,\: \Delta y\).
• Estimate a propagated error using a differential.
• Use differentiation formulas to find the differential for a function.

Tangent Line Approximations

Newton’s Method uses a tangent line to approximate the graph for a function. Differentials use a straight line to approximate the graph for a function

Consider a function \(f\) that is differentiable at \(c\). The equation for the tangent line at the point \((c,f(c))\) is

\(y-f(c)\)	\(={f}'(c)(x-c)\)
\(y\)	\(=f(c)+{f}'(c)(x-c)\)

and is called the tangent line approximation, or linear approximation, for \(f\) at \(c\). Because \(c\) is a constant, \(y\) is a linear function for \(x\)-values. As \(x\) approaches \(c\), the limit for \(y\) is \(f(c)\). By choosing \(x\)-values close to \(c\), \(y\) values can be approximated.

Example 3.9.1 Using a Tangent Line Approximation

Tangent line approximation for \(f\) at the point \((0,1)\). Figure 3.9.1

Find the tangent line approximation for \(f(x)=1 + \sin x\) at the point \((0,1)\). Then compare the \(y\)-values for the linear function with those for \(f(x)\) on an open interval containing \(x=0\). Solution The derivative for \(f\) is \({f}'(x)=\cos x.\)    First derivative The equation for the tangent line to the graph for \(f\) at the point \((0,1)\) is \(y\) \(=f(0)+{f}'(0)(x-0)\) \(=1+(1)(x-0)\) \(=1+x\)     Tangent line approximation Table 3.9.1 compares the \(y\)-values produced by the linear approximation with \(f(x)\)-values near \(x=0\). Notice the close \(x\) is to 0, the better the approximation. This conclusion is shown in Figure 3.9.1. Table 3.9.1 x -0.5 -0.1 -0.01 0 0.01 0.1 0.5 \(f(x)=1 + \sin x\) 0.521 0.9002 0.9900002 1 1.0099998 1.0998 1.479 \(y=1+x\) 0.5 0.9 0.99 1 1.01 1.1 1.5 The linear approximation for \(f(x)=1 + \sin x\) depends on the tangency point. A different point on the graph for \(f\) would yield a different tangent line approximation

Differentials

When \(\Delta x\) is small, \(\Delta y=f(c+\Delta x)-f(c)\) is approximated by \({f}'(c)\Delta x\). Figure 3.9.2

When the tangent line to the graph for \(f\) at the point \((c,f(c))\) \(y=f(c)+{f}'(c)(x-c)\)    Tangent line at \((c,f(c))\) is used as an approximation for \(f\)'s graph, the quantity \(x-c\) is called the change in \(x\), and is denoted by \(\Delta x\), as shown in Figure 3.9.2. When \(\Delta x\) is small, the change in \(y\), denoted by \(\Delta y\), is approximated by \(\Delta y\) \(=f(c+\Delta x)-f(c)\)     Actual change in \(y\) \(\approx {f}'(c) \Delta x\)     Approximate change in \(y\) The quantity \(\Delta x\) is traditionally denoted by \(dx\), and is called the differential for \(x\). The expression \({f}'(x) dx\) is denoted by \(dy\), and is called the differential for \(y\). Definition 3.9.1 Differentials Let \(y=f(x)\) represent a function that is differentiable on an open interval containing \(x\). The differential for \(x\), denoted by \(dx\), is any nonzero real number. The differential for \(y\), denoted by \(dy\) is \(dy={f}'(x) dx\). In many applications the differential for \(y\) is used as an approximation for the change in \(y\). That is \(\Delta y \approx \text{ or } \Delta y \approx {f}'(x) dx\).

Example 3.9.2 Comparing \(\Delta y\) and \(dy\)

The change in \(y\), \(\Delta y\), is approximated by the differential for \(y\), \(dy\). Figure 3.9.3

Let \(y=x^2\). Find \(dy\) when \(x=1\) and \(dy=0.01\). Compare this value with \(\Delta y\) for \(x=1\) and \(\Delta x=0.01\). Solution Because \(y=f(x)=x^2\), then \({f}'(x)=2x\), and the differential \(dy\) is \(dy={f}'(x) dx={f}'(1)(0.01)=2(0.01)=0.02\).    Differential for \(y\) Using \(\Delta x=0.01\) the change in \(y\) is \(\Delta y=f(x+\Delta x)-f(x)=f(1.01)-f(1)=(1.01)^2-1^2=0.0201\). The geometric comparison between \(dy\) and \(\Delta y\) is shown in Figure 3.9.2. UThe values for \(dy\) and \(\Delta y\) converge as \(dx\) or \(\Delta x\) approaches 0.

In Example 3.9.2, the tangent line to the graph for \(f(x)=x^2\) at \(x=1\) is

\(y=2x-1\).    Tangent line to the graph for \(f\) at \(x=1\).

For \(x\)-values near 1, this line is close to the graph for \(f\), as shown in Figure 3.9.2 and in Table 3.9.2 below.

Table 3.9.2
x	0.5	0.9	0.99	1	1.01	1.1	1.5
\(f(x)=x^2\)	0.25	0.81	0.9801	1	1.0201	1.21	2.25
\(y=2x-1\)	0	0.8	0.98	1	1.02	1.2	2

Error Propagation

Physicists and engineers liberally use approximations for \(\Delta y\) by \(dy\). One way is estimating propagated errors by physical measuring devices. For example, let \(x\) represent the measured value for a variable and let \(x +\Delta x\) represent the exact value, then \(\Delta x\) is the error in measurement. If the measured value \(x\) is used to compute another value \(f(x)\), then the difference between \(f(x+\Delta x)\) and \(f(x)\) is the propagated error.

Example 3.9.3 Error Estimation

Ball bearing with measured radius that is correct to within 0.01 inch. Figure 3.9.4

A ball bearing's measured radius is 0.7 inch, as shown in the Figure 3.9.4. The measurement is correct to within 0.01 inch. Estimate the propagated error in the ball bearing as the variable volume \(V\). Solution The formula for a spherical volume is $$V=\frac{4}{3} \pi r^3$$ where \(r\) is the sphere's radius. Write \(r=0.7\)    Measured radius and \(-0.01 \leq \Delta r \leq 0.01.\)    Possible error To approximate the propagated error in the volume, differentiate \(V\) to obtain $$\frac{dV}{dr}=4 \pi r^2$$ and write \(V\) \(\approx dV\)     Approximate \(\Delta V\) by \(dV\) \(=4 \pi r^2 dr\) \(=4\pi ( \color{red}{0.7})^2(\color{red}{\pm 0.01})\)     Substitue for \(r\) and \(dr\) \(\approx \pm 0.06158\) cubic inch. The propagated error is about 0.06 cubic inch.

Calculating Differentials

The differentiation rules in Chapter 2 can be written in differential form. For example, let \(u\) and \(v\) be differentiable functions for \(x\) and by Definition 3.9.1 Differentials we have

\du={u}'dx\)

and

\dv={v}'dx.\)

The Product Rule's differential form is written as

\(d[uv]\)	$$=\frac{d}{dx}[uv] dx$$	Differential for \(uv\)
	\(=[u{v}'+v{u}'] dx\)	Product Rule
	\(=u{v}' dx + v{u}' dx\)
	\(=u dv + v du\)

Differential Formulas

Let \(u\) and \(v\) be differentiable functions for \(x\).

Constant Multiple Rule:	\(d[cu]=c du\)
Sum or difference:	\(d[u \pm v]=du \pm dv\)
Product:	\( d[uv]=u dv + v du\)
Quotient	$$d \left [ \frac{u}{v} \right ] = \frac{v du - u dv}{v^2}$$

Example 3.9.4 Finding Differentials

The notation in Example 3.9.4 is called the Leibniz notation[1] for derivatives and differentials, named after the German mathematician Gottfried Wilhelm Leibniz[2]. It makes remembering several important calculus formulas easy by using an algebra like notation. For example, in Leibniz notation, the Chain Rule

$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$

would appear true because the \(du\)'s divide out. Even thought this reasoning is incorrect, the notation does help remember the Chain Rule.

Example 3.9.5 Finding the Differential for a Composite Function

Example 3.9.6 Finding the Differential for a Composite Function

Differentials are used to approximate function values. For the function given by \(y=f(x)\), use the formula

\(f(x+\Delta x) \approx f(x)+dy=f(x)+{f}'(x) dx\)

which is derived from the approximation

\(\Delta y=f(x+ \Delta x) - f(x) \approx dy\)

which was derived from the tangent line approximation discussed earlier. The key to using this formula is to choose a value for \(x\) that makes the calculations easier, as shown in Example 3.9.7.

Example 3.9.7 Approximating Function Values

Use differentials to approximate \(sqrt{16.5}\).
Solution Using \(f(x)=\sqrt{x}\), the differential is

$$f(x+\Delta x) \approx f(x)+{f}'(x) dx = \sqrt{x} + \frac{1}{2\sqrt{x}} dx.$$

Choosing \(x=16\) and \(dx= 0.5\), obtains the approximation.

$$f(x+\Delta x)=\sqrt{16.5} \approx \sqrt{16}+\frac{1}{2\sqrt{x}}(0.5) = 4 + \left ( \frac{1}{8} \right ) \left ( \frac{1}{2} \right ) = 4.0625$$

Figure 3.9.5

The tangent line approximation to \(f(x)=\sqrt{x}\) at \(x=16\) is the line \(g(x)=\frac{1}{8}x+2\). For \(x\)-values near 16, the graphs for \(f\) and \(g\) are close together, as shown in Figure 3.9.5. For example, \(f(16.5)=\sqrt{16.5} \approx 4.0620\) and $$g(16.5)=\frac{1}{8}(16.5) +2 = 4.0625.$$ Notice in Figure 3.9.5 the two graphs appear to coincide near the tangency point \((16,4)\). The farther from the tangency point the linear approximations is less accurate.

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Parent Article: Calculus I 03 Differentiation Applications