Differentials and Antiderivatives

New notation: dy = f?(x)dx (y = f(x))
Both dy and f?(x)dx are called differentials. You can think of
dy
= f?(x)
dx
as a quotient of differentials. One way this is used is for linear approximations.
?y dy
?x ? dx
Example 1. Approximate 651/3
Method 1 (review of linear approximation method)
f(x)= x1/3 1
f?(x)= x?2/3
3 f(x) ? f(a)+ f?(a)(x ? a) 1
x1/3 ? a1/3 +3a?2/3(x ? a)
A good base point is a = 64, because 641/3 = 4. Let x = 65.
1 111
651/3 = 641/3 + 64?2/3(65 ? 64)=4+ (1)=4+ 48 ? 4.02
3 316
Similarly, 1
(64.1)1/3 ? 4+
480
Method 2 (review)
??1/3
1 11
651/3 = (64 + 1)1/3 = [64(1 + )]1/3 = 641/3[1+ ]1/3 =4 1+
64 6464 11
Next, use the approximation (1 + x)r ? 1+ rx with r = 3 and x = 64.
11 1
651/3 ? 4(1+ ( ))=4+
3 64 48 This is the same result that we got from Method 1.
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Method 3 (with differential notation)
y = x1/3|x=64 =4 ??1 111
dy =3x?2/3dx|x=64 =3 16 dx = 48dx
1
We want dx = 1, since (x + dx) = 65. dy = when dx = 1.
48 1
(65)1/3 =4+
48
What underlies all three of these methods is
y = x1/3 dy 1x?2/3dx =3|x=64
Anti-derivatives
F(x)= f(x)dx means that F is the antiderivative of f.
Other ways of saying this are:
F?(x)= f(x) or, dF = f(x)dx
Examples:
1. sin xdx = ?cos x + c where c is any constant.
n+1
x
2. xndx = n +1 + c for n ?= ?1.
dx
3.
x = ln |x| + c (This takes care of the exceptional case n = ?1 in 2.)
4.
sec2 xdx = tan x + c
dx 1
5.
?1 ? x2 = sin?1 x + c (where sin?1 x denotes “inverse sin” or arcsin, and not sin x)
6.
dx = tan?1(x)+ c
1+ x2
Proof of Property 2: The absolute value |x| gives the correct answer for both positive and negative
x. We will double check this now for the case x< 0:
ln |x|d dx ln(?x)
==
ln(?x)? d du ln(u)? du dx
where u = ?x.
d dx ln(?x)
=
1u(?1) = 1 ?x(?1) = 1x
2
Lecture 15 18.01 Fall 2006
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