The derivative of a functionf at a pointa can be interpreted asthe slope of the tangent lineto the graph off at(a, f(a)). The tangent line is the graph of a linear function that is the best linear approximation tof near ain a sense that will be explained in this section.
Local Linear Approximations
Given a functionf that is dierentiable at the pointa, the tangent line to the graph of f at (a, f(a))is the graph of the equation
y=f(a) +f0(a) (xa). We will give a name to the underlying linear function:
Denition 1 The linear approximation to f based atais Ld(x) =f(a) +f0(a)(xa).
We refer toaasthe basepoint.
x y
a La
a, fa
Figure 1: The graph ofLa is a tangent line
Example 1 Letf(x) =x22x+ 4, as in Example 1 of Section 2.1. We showed thatf0(3) = 4 and the tangent line to to graph off at(3, f(3))is the graph of the equation
y=f(3) +f0(3) (x3) = 7 + 4 (x3). Thus, the linear approximation tof based at 3 is
L3(x) = 7 + 4 (x3).
Figure 2 illustrates the eect of zooming in towards the point(3, f(3)) = (3,7). Note that we can hardly distinguish between the graphs off and L3in the third frame. This indicates that L3(x)approximatesf(x)very well ifxis close to the basepoint 3. On the other hand, we do not expectL3(x)to approximatef(x)whenxis far from 3. The linear function L3 is a "local approximation" tof. ¤
3 6x
7 10 20 y
3, f3
2.6 3.4
5 9
3, f3
2.8 3.2
6.5 7.5
3, f3
Figure 2
Let’s assess the error in the approximation of f(x) by L3(x) algebraically. Since L3(x) is expected to be a good approximation tof whenx is near 3, it is convenient to setx= 3 +h, so thath(=x3) represents the deviation ofxfrom the basepoint 3. We have
L3(3 +h) = 7 + 4 (x3)|x=3+h= 7 + 4h.
Therefore,
f(3 +h)L3(3 +h) = (3 +h)22 (3 +h) + 4(7 + 4h)
= 9 + 6h+h262h+ 474h
=h2
140 CHAPTER 2. THE DERIVATIVE Thus, the absolute error is
|f(3 +h)L3(3 +h)|=h2. Note thath2is much smaller than|h|if|h|is small. For example,
¡102¢2
= 104and ¡ 103¢2
= 106.
Thus, the absolute error in the approximation of f(x) by L3(x) is much smaller than the distance ofx from the basepoint 3 ifxis close to3. This numerical fact is consistent with our graphical observation. ¤
Example 2 Let
f(x) = 1 x. a) DetermineL2, the linear approximation tof based at 2.
b) Calculatef(2 +h)andL2(2 +h)forh=10n,n= 1,2,3. Compare|f(2 +h)L2(2 +h)| with|h|.
Solution
a) By the power rule,
f0(x) = d dx
1 x
ả
= d dx
¡x1¢=x2= 1 x2. Therefore,f0(2) =1/4and
L2(x) =f(2) +f0(2) (x2) = 1 21
4(x2). Thus,
L2(2 +h) =1 2 1
4h.
Figure 3 shows the graphs of f and L2 (the dashed line) in a small viewing window that is centered at(2, f(2)) = (2,0.5).
1.8 1.9 2.1 2.2
0.46 0.48 0.52 0.54
Figure 3
b) Table 1 displays the required data. We see that |f(2 +h)L2(2 +h)| is much smaller than |h| for the values ofh that are considered. Indeed, f¡
2103¢
and L2¡
2103¢ are represented by the same decimal, rounded to 6 signicant digits. Therefore, the numbers support our analysis of the error in linear approximations. ¤
h f(2 +h) L2(2 +h) |f(2 +h)L2(2 +h)| 101 0.526 316 0.525 1.3×103
102 0.502 513 0.502 5 1.3×105 103 0.500 25 0.500 25 1.3×107
Table 1
Remark We have identied the rate of change of a functionf at a point awith f0(a), and f0(a)is the rate of the linear function La. The fact that La(x)approximates f(x)very well ifxis close toajusties this identication. After all, there is no question that the rate of change of the linear function
La(x) =f(a) +f0(a) (xa) isf0(a)at any point.
In particular, iff0(a) = 0we declare that the rate of change off atais 0. This does not mean that we havef(x) =f(a)for eachxin some interval centered ata. On the other hand,
La(x) =f(a) +f0(a) (xa) =f(a), and the rate of change of the constant functionLa is 0. Since
f(x)=La(x) =f(a),
and the magnitude of the error can be expected to be much smaller than |xa| if |xa| is small, the restriction of f to a small interval centered at a is almost a constant function.
Therefore, it is reasonable to declare that the rate of change off atais 0.
Example 3 As in Example 2 of Section 2.3, where we determined the tangent line to the graph of cosine at(0,0), the linear approximation to cosine based at 0 is
L0(x) = cos (0) + d
dxcos (x)¯¯
¯¯x=0ả x= 1.
Thus, L0 is a constant function and its graph, i.e., the tangent line to the graph of cosine at (1,0), is a horizontal line, as shown in Figure 4.
Π2 Π2 x
1 1 y
Figure 4
Obviously, the rate of change of L0 is 0. We declare that the rate of change of cosine at 0 is also 0, even thoughcos (x)6= 0ifxdeviates from 0 slightly. This is justied in view of the fact thatcos (x)= 1ifx= 0, and the absolute error in the approximation is much smaller than|x| is|x|is small. For example,cos (0.01)= 0.999 95,|cos (0.01)1| = 5.0×105, and5.0×105 is much smaller than102. ¤
142 CHAPTER 2. THE DERIVATIVE
The Dierential
It is useful to consider all the local linear approximations to a given function at once by consid- ering the basepoint to be a variable. In this case it is convenient to work with dierences and a change in the notation seems to be in order. We will denote an increment along thex-axis by x. Thus,
f0(x) = lim
h0
f(x+h)f(x)
h = lim
x0
f(x+ x)f(x)
x .
Therefore,
f(x+ x)f(x)
x =f0(x) if|x|is small, so that
f(x+ x)f(x)=f0(x) x.
We will give the expressionf0(x) xa special name:
Denition 2 The dierential of the functionf is d f =f0(x)x.
Thus,df is a function of two independent variables, the basepoint xand the increment x. We can indicate this explicitly by writing
d f(x,x) =f0(x)x.
We have
f(x+ x)f(x)=d f(x,x)
if |x| is small. The idea behind the dierential is the same as the idea of local linear ap- proximations. The dierential merely keeps track of local linear approximations to a function as the basepoint varies. Note that d f(x,x)is the change corresponding to the incrementxalong the tangent line to the graph off at(x, f(x)), as illustrated in Figure 5.
x y
x x, fx
xx, fxx
x x
fx xfx x
dfx,x
Figure 5
Example 4 Letf(x) =
x. Approximate
4.1via the dierential off. Solution
Since
f0(x) = d dx
x= 1 2
x,
The dierential off is
df(x,x) =f0(x) x= 1 2
x(x) = x 2
x. It is natural to setx= 4 andx = 0.1 for the approximation of
4.1 = f(4.1)sincef(4) = 4 = 2. Thus,
4.12 =f(4.1)f(4)=df(4,0.1) = 0.1 24 = 0.1
4 = 0.025.
Therefore,
4.1 = 2 +³
4.12´
= 2 + 0.025 = 2.025.
We have
4.1= 2.024 85, rounded to 6 signicant digits, and
¯¯¯
4.12.025¯¯¯= 1.5×104. Note that the absolute error in the approximation of
4.1 via the dierential is much smaller thanx= 0.1.¤
Remark 1 As we saw in Section 2.4, the rate of change of the position f(t)of an object in one-dimensional motion at time t is the instantaneous velocityv(t). If the time increment is t >0is small then
f(t+ t)f(t)=df(t,t) =f0(t) t=v(t) t.
Thus, the displacement over the time time interval[t, t+ t]is approximatelyv(t) tiftis small.
For example, iff(t) = cos (t)thenv(t) =sin (t)so that f(t+ t)f(t)=sin (t) t.
In particular, f
³ 6 + 0.1´
f
³ 6
´=sin³ 6
´(0.1) =0.1
2 =0.05.
The()sign indicates that the motion is in the negative direction.
The Traditional Notation for the Dierential
We wrote
df(x,x) =f0(x)x.
Traditionally, the incrementxis denoted bydxwithin the context of dierentials. Thus, df(x, dx) =f0(x)dx.
If we use the Leibniz notation forf0(x), we have df(x, dx) = df
dx(x)dx.
144 CHAPTER 2. THE DERIVATIVE
x y
x x, fx
xdx, fxdx
xdx
fxdxfx
dx df
Figure 6: The geometric meaning of the dierential
We usually do not bother to indicate that the dierential depends onxanddx, and write df = df
dxdx.
This is convenient and traditional notation, but you should keep in mind that the “fraction”
df dx
is a symbolic fraction, and that the symboldxthat appears as the denominator does not have the same meaning asdxthat stands for the increment in the value of the independent variable.
The expression
df = df dxdx is analogous to the expression
f= f
xx, wherex6= 0andf =f(x+ x)f(x).
If we refer to the function asy=y(x), we can write dy= dy
dxdx The above expression is analogous to the expression
y= y
xx, wherex6= 0andy=y(x+ x)y(x).
Example 5 Letf(x) =x1/3 a) Determine the dierentialdf.
b) Make use of the dierentialdf to approximate(8.01)1/3. Determine the absolute error in the approximation by treating the value that is obtained from your calculator as the exact value.
Compare with the deviation from the basepoint that you have chosen.
Solution a)
df = df dxdx=
d dx
³ x1/3
´ả
dx= 1
3x2/3
ả
dx= 1 3x2/3dx.
b) Since8.01is close to 8, andf(8) = 81/3= 2, the natural choice for the basepoint is 8. Thus, dx= 8.018 = 0.01. The value of the dierential corresponding tox= 8anddx= 0.01is
à 1 3¡82/3¢
!
(0.01) = 0.01 3 (4) = 0.01
12 . Therefore,
(8.01)1/32 =f(8.01)f(8)= 0.01 12 so that
(8.01)1/3= 2 +0.01
12 = 2.000 83
A calculator tells us that (8.01)1/3 = 2.000 83, rounded to 6 signicant digits. Thus, the approximation via the dierential gave us the same decimal, rounded to 6 signicant digits.
There is a nonzero of course. Indeed,
¯¯¯¯
2 +0.01 12
ả
(8.01)1/3¯¯
¯¯= 3.5×107.
Thus, the absolute error in the approximation is much smaller than0.01, the deviation of8.01 from the basepoint 8. ¤
Example 6 The volume of a spherical ball of radiusris V = 4
3r3. a) Determine the dierentialdV.
b) Use the dierential to approximate the change in the volume of the ball if the ball is inated and its radius increases from20centimeters to20.1centimeters.
Solution a) We have
dV dr = d
dr 4
3r3
ả
= 4 3 d
dr
¡r3¢
= 4 3¡
3r2¢
= 4r2. Therefore,
dV = dV
drdr= 4r2dr.
Note that4r2is the surface area of sphere of radiusr. Therefore, the change in the volume of a spherical ball that corresponds to a small change in the radius can be approximated by the product of the area of its boundary and the increment of the radius.
b) In particular,
V(20.1)V(20)= 4¡ 202¢
(0.1)= 502.655 (cm3). The actual change in the volume is
V(20.1)V (20) =4
3(20.1)3 4
3(20)3= 505.172
(cm3). Therefore, the error in the approximation of the change in the volume via the dierential is approximately2.517¡
cm3¢
. This may not be considered to be a small number. On the other hand, the relative error is usually more appropriate in assessing error. Thus,
(V(20.1)V(20))4¡ 202¢
(0.1)
V(20) = 2.517
33510.3 = 7.5×105,
146 CHAPTER 2. THE DERIVATIVE and this numberis small.
We can also approximate the relative change in the volume, i.e., V(20.1)V (20)
V(20) , via the dierential by calculating
dV(20,0.1) V(20) = 4¡
202¢ (0.1)
V (20) = 1.5×102. This approximates
V(20.1)V(20)
V(20) = 1.507 51×102 with an error that is approximately7×105. ¤
The Accuracy of Local Linear Approximations
Theorem 1 Assume thatf is dierentiable at a,and thatLd is the linear approxi- mation tof based at a. We have
f(a+h) =Ld(a+h) +hq(h), where
k0limq(h) = 0.
Thus,hq(h)represents the error in the approximation off byLaatx=a+h. Since the error is the product of h and q(h), and q(h) 0 as h 0, its magnitude is much smaller than
|h|=|xa|if xis close to the basepointa.
With reference to Example 1,q(h) =h2. Proof
As in Example 1, we will setx=a+h, so that h=xarepresents the deviation of xfroma and has small magnitude ifxis neara. We have
La(a+h) = f(a) +f0(a) (xa)|xa=h=f(a) +f0(a)h.
Therefore,
f(a+h)La(a+h) = f(a+h)(f(a) +f0(a)h)
= (f(a+h)f(a))f0(a)h
= h
f(a+h)f(a)
h f0(a)
ả
Let’s set
q(h) = f(a+h)f(a)
h f0(a),
so thatq(h)is the dierence between the dierence quotient and the derivative. We have
h0limq(h) = lim
h0
f(a+h)f(a)
h f0(a)
ả
= 0, since the dierence quotient approaches the derivative ash0.
Thus,
f(a+h)La(a+h) =hq(h), so that
f(a+h) =La(a+h) +hq(h), wherelimh0q(h) = 0.¥
The analysis of the error in the approximation of dierences via the dierential is along similar lines:
Theorem 2 Assume thatf is dierentiable at x. Then,
f(x+ x)f(x) =d f(x,x) + x q(x), where
{0lim q(x) =0.
Proof We have
f(x+ x)f(x)df(x,x) = f(x+ x)f(x)f0(x) x
= x
f(x+ x)f(x)
x f0(x)
ả . If we set
q(x) = f(x+ x)f(x)
x f0(x), then
f(x+ x)f(x)df(x,x) = xq(x). We have
x0lim q(x) = lim
x0
f(x+ x)f(x)
x f0(x)
ả
= 0, since
x0lim
f(x+ x)f(x)
x .f0(x). Thus,
f(x+ x)f(x)df(x,x) = xq(x), wherelimx0q(x) = 0.
Problems
In problems 1 - 6,
a) DetermineLa, the linear approximation tof based ata,
b) Make use ofLa(b)to approximatef(b)if such a pointbis indicated,
c) [C] Calculate the absolute error in the approximation off(b) by La(b) and compare with
|ba|.
d)[C]Plot the graphs off andLa in a suciently small viewing window centered at(a, f(a)) that demonstrates the accuracy of the linear approximation neara.
148 CHAPTER 2. THE DERIVATIVE 1.
f(x) =x2+x, a= 3, b= 3.01 2.
f(x) =x4, a= 1, b= 0.98 3.
f(x) = 1
x2, a= 0.5, b= 0.502
4.
f(x) =x2/3, a= 8, b= 7.8 5.
f(x) =x1/4, a= 16, b= 16.2 6.
f(x) = sin (x), a= 3, b=
3 0.1 In problems 7 and 8,
a) Determine the dierential off,
b) Approximatef(b)via the the dierential off’ 7.
f(x) =
x, b= 24.9.
8.
f(x) = 1
x2, b= 2.2.
In problems 9 - 12 approximate the given number via the dierential (You need to choose an appropriate function and basepoint).
9. (26.5)1/3 10.. 1
3.9
11. sin 3
4 + 0.1
ả
12. cos³
6 0.2´ 13. LetA(r)be the area of a disk of radiusr.
a) Approximate the change in the area corresponding to a change in the radius from 10 meters to 10.1 meters via the dierential.
b) Calculate the relative error in the approximation of part a).
14. LetV(r)be the volume of a right circular cone of height 10 meter and base radiusrmeters.
a) Approximate the change in the volume corresponding to a change in the radius from 4 meters to 4.2 meters via the dierential.
b) Calculate the relative error in the approximation of part a).