CHAPTER 1. 1 — FORMS §1. Differentials: Basic Rules Differentials ...

§1. Differentials: Basic Rules
CHAPTER 1. 1 — FORMS
Differentials, usually regarded as fanciful objects, but barely mentioned in elementary
calculus textbooks, are now receiving our full attention as they deserve. We will see that
these fanciful objects have fantastic powers.
Since they are fanciful objects, it is rather difficult to tell exactly what they are.
However, from our experiences in calculus, spiced by our vivid imagination, we have some
good ideas about how they should behave. Let us start with putting down some “rules of
thumb” governing their behavior:
Rule DF1 (Linearity). d(αu + βv) = αdu + βdv where α, β ∈ R.
Rule DF2 (Leibniz’s Rule or Product Rule). d(uv) = udv + vdu.
Rule DF3 (Chain Rule). du = du
dv if u is a (nice) function of v.
dv
These rules will soon be applied to concrete examples in the routine manner. Before doing
so, we describe an idea leading to such rules swiftly, ignoring mathematical rigor in the
process. (This idea is a bit misleading; however it serves our purpose well at present.)
The expression du here is interpreted as a kind of “polished” change (or increment)
in u, which differs from the real change ∆u by an amount (designated by o(∆u) in the
following identity) so small that practically it may be ignored:
∆u = du + o(∆u).
Here, both ∆u and du are small amounts of u, but their difference o(∆u), comparing to
both ∆u and du, is much smaller. Suppose that we have two quantities u and v, and
due to a slight disturbance or other good reasons, u is changed by an amount ∆u to
u + ∆u and v is changed by ∆v to v + ∆v. Then their product w ≡ uv is changed into
(u + ∆u)(v + ∆v) = uv + u∆v + v∆u + (∆u)(∆v). Thus, the increment in w is given by
∆w = (u + ∆u)(v + ∆v) − uv = u∆v + v∆u + (∆u)(∆v).
Now we “polish” this identity: ∆u and ∆v become du and dv respectively, and ∆w becomes
dw = d(uv). What happens to the product (∆u)(∆v)? Because both ∆u and ∆v are small,
their product is much smaller comparing to each of them and hence can be “polished away”.
1

The final outcome is d(uv) = udv +vdu, namely the Leibniz rule given above. The identity
in Rule DF1 can be obtained in the same fashion, except a lot easier. I leave Rule DF3
for you to derive in this fashion.
You would say, this looks like something familiar in calculus: d(uv) = udv + vdu
d
dv
seems to be nothing new but the usual product rule dxuv = u dx
+ v du
dx
written in a
different way. So what is such a big deal? Well, the usual product rule only applies to
functions of a single variable x. The rates of change du
dx
and dv
dx
of u and v are measured
against the change in x. For differentials, what causes du and dv to emerge is not specified
and immaterial. It may due to the change of one variable, say x, as you learned in the
first year calculus. It may due to the changes of several variables. When I write down the
identity d(uv) = udv + vdu, I do not have to specify what variables are, or even how many
variables are involved. The cause for the “changes” du, dv in u, v may well be a mild
earthquake — I don’t care. So the product rule for differentials here has a much wider
appeal. In particular, it applies to the several variable case.
Let us derive some simple consequences of these rules of thumb. From Rule DF2 it
follows immediately that d(u 2 ) = 2udu. By mathematical induction, we can show that
d(u n ) = nu n−1 du
holds for all positive integers n. (This is a good exercise for reinforcing your skill in
induction. Do this.) On the other hand, we have d1 = d(1 2 ) = 2d1. Hence d1 = 0. So, if
c is a constant, by Rule DF1, we have dc = cd1 = 0. This is not surprising: we cannot see
any change in a constant c and hence its differential should be zero.
By Rule DF3, we have: d(1/x) = −dx/x 2 , d √ x = −dx/2 √ x, d(1 + x 2 ) = 2xdx,
d sin x = cos x dx, d cos x = − sin x dx, d tan x = sec 2 x dx, d sec x = tan x sec x dx,
de x = e x dx,
d log x = dx
, d arcsin x =
x
dx
dx
√ , d arctan x = , etc.,
1 − x2 1 + x2 (log x here is the same as ln x). Formulae of derivatives you learned in the first year
calculus are very handy.
Example 1.1. Find d(e cos t ) and d(log sec x).
Solution: Let u = cos t. From d(e u ) = e u du we have
d(e cos t ) = d(e u ) = e u du = e cos t d(cos t) = e cos t (− sin t dt) = − sin t e cos t dt.
2

cos t
You should feel comfortable with the following way of taking the differential of e
without introducing a new variable u: d(e cos t ) = e cos t d(cos t) = − sin t e cos t dt. Similarly,
1
d(log sec x) = d log = d(− log cos x)
cos x
= −d(log cos x) = −
d cos x
cos x
= −− sin x dx
cos x
= tan x dx.
(Alternatively, d(log sec x) = (sec x) −1 d sec x = (sec x) −1 sec x tan x dx = tan x dx.)
Example 1.2. Find dr and d(1/r) for the “radius function” r = x 2 + y 2 + z 2 .
Solution: Applying “d” on both sides of the identity r 2 = x 2 + y 2 + z 2 , we obtain
d(r 2 ) = d(x 2 ) + d(y 2 ) + d(z 2 ), or 2rdr = 2xdx + 2ydy + 2zdz. Thus we arrive at
dr =
xdx + ydy + zdz
r
= xdx + ydy + zdz
x 2 + y 2 + z 2 .
Also, d(1/r) = −dr/r 2 = −(xdx + ydy + zdz)/r 3 = −(xdx + ydy + zdz)/(x 2 + y 2 + z 2 ) 3/2 .
This example illustrates how differentials work for functions of several variables.
u
Example 1.3. Derive the quotient rule d =
v
vdu − udv
v2 .
Solution: Let w ≡ u/v. Then u = vw and hence du = vdw + wdv, which gives a
relation between du, dv and dw. Use this relation to write dw in terms of du and dv:
dw =
which is the desired identity.
du − wdv
v
= du − u
v dv
v
= vdu − udv
v 2
Example 1.4. It is well-known that the connection between the polar coordinates
(r, θ) and the rectangular coordinates (x, y) is given by
Find dθ in rectangular coordinates.
x = r cos θ, y = r sin θ. (1.1)
Solution: We have dx = d(r cos θ) = cos θ dr + r d cos θ = cos θ dr + r(− sin θ) dθ,
and similarly dy = sin θ dr + r cos θ dθ. This gives − sin θ dx + cos θ dy = r dθ. Replacing
sin θ by y/r and cos θ by x/r, we have
dθ = 1
r
− y
r
x
dx +
r dy
= −ydx + xdy
r 2
3
= xdy − ydx
,
x2 , (1.2)
+ y2

which is the final answer. For an obvious reason, dθ is called an angular form. As we
will see at the end of the present section, putting the angular form as dθ is misleading.
Identity (1.2) will be needed in the future for defining winding number of a loop around
the origin.
Example 1.5. Find the equation of the tangent line to the ellipse
at a point (x0, y0) on this ellipse.
x 2
a
2 + y2
Solution: Applying “d” to both sides of (1.3), we obtain
= 1. (1.3)
b2 2x 2y
dx + dy = 0, (1.4)
a2 b2 which gives dy/dx = −b 2 x/a 2 y. The slope of the required tangent line is dy/dx at (x0, y0),
that is, −b 2 x0/a 2 y0. So the equation for the tangent line is y − y0 = (−b 2 x0/a 2 y0)(x − x0),
or (x0x − x 2 0)/a 2 + (y0y − y 2 0)/b 2 = 0, which can be rewritten as
due to the fact that (x0, y0) is on the ellipse.
x0x y0y
+ = 1, (1.5)
a2 b2 In the rest of the present section we describe how to compute a line integral
γ ω,
where γ is a path and ω is a differential form. Suppose that the variables in ω are
x1, x2, . . . , xn, which can be put together as a vector variable x = (x1, x2, . . . , xn). We
may write ω as ω = n
k=1 Fkdxk where Fk = Fk(x) ≡ Fk(x1, x2, . . . , xn) are functions
in variables xj’s. The path γ can be described in parametric equations as xk = xk(t),
1 ≤ k ≤ n, or, in vector form, x = x(t), where the parameter t is running in some interval,
say I = [a, b]. First consider the special case n = 1 and write ω = F (x)dx. When γ is the
usual path in the real line moving from a to b, that is, x ≡ x(t) = t (a ≤ t ≤ b), the integral
γ ω clearly should be interpreted as the usual definite integral b
F (t)dt. In the general
a
case, we “pull back” the differential form ω in n variables xj’s via γ to get a differential
form in one variable t, denoted by γ∗ω, and then define
γ ω as b
a γ∗ω. The pullback
γ∗ω of ω here is obtained by the substitutions xj = xj(t) (1 ≤ j ≤ n) into ω. Thus, when
ω = n k=1Fkdxk as before, we have γ∗ω = n k=1 Fk(x(t))dxk(t) = n k=1 Fk(x(t))x ′ k (t)dt.
The line integral
γ ω is, by definition, the integral of γ∗ω over the interval [a, b] :
b
ω = γ ∗ ω, with γ ∗ ω = n
γ
a
k=1
dxk
n
Fk(x(t)) dt, where ω =
dt k=1 Fk(x)dxk.
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Using this definition, the actual computation of line integrals is quite straightforward.
Example 1.6. Compute the line integral
γ
xdy + ydz + zdx for three paths linking
the origin (0, 0, 0) to the point (1, 1, 1), the first path being γ = α : x = t, y = t, z = t
(0 ≤ t ≤ 1), the second being γ = β : x = t, y = t 2 , z = t 3 (0 ≤ t ≤ 1), and the third being
γ = η : x = t 2 , y = t 4 , z = t 6 (0 ≤ t ≤ 1).
Solution: Put ω = xdy + ydz + zdx. Do the pull-backs α ∗ ω = tdt + tdt + tdt = 3tdt,
β ∗ ω = td(t 2 ) + t 2 d(t 3 ) + t 3 dt = (2t 2 + 3t 4 + t 3 )dt and η ∗ ω = (4t 5 + 6t 9 + 2t 7 )dt. Hence
we get
1
α ω = 1
0
0 (4t5 + 6t 9 + 2t 7 )dt = 91
60
3 3tdt = 2 , β ω = 1
0 (2t2 + 3t4 + t3 )dt = 2
3
, the same as
β
+ 3
5
+ 1
4
91 = 60 , and ω = η
ω. Notice that γ is just a reparametrization of
β. This indicates that the fact that a line integral in general depends on the path but not
on its parametrization.
Example 1.7. Compute the line integral
ω ≡
γ
γ
xdy − ydx
x2 ,
+ y2 where γ is the path of going around the unit circle once in the anti-clockwise direction
described by the parametric equations x = cos t and y = sin t with 0 ≤ t ≤ 2π,
Solution: The differential form ω here is just the angular form dθ in Example 1.4.
So the line integral here is the change of the polar angle θ when a point moves around the
circle once. Thus the answer should be 2π. But let us just follow the definition described
above to compute this line integral. The pull-back γ ∗ ω is
γ ∗ ω =
x(t)dy(t) − y(t)dx(t)
x(t) 2 + y(t) 2
= cos t d sin t − sin t d cos t
cos 2 t + sin 2 t
= dt.
So
γ ω = 2π
dt = 2π. Notice that if γ goes around the unit circle n times instead of
0
once, then the resulting line integral is 2πn
dt = 2πn.
0
A path γ in the xy–plane described by parametric equations x = x(t) and y = y(t)
with a ≤ t ≤ b is called a closed path or a loop if its end points coincide, that is,
−1 (x(a), x(b)) = (y(a), y(b)). Given a loop γ not passing the origin, the integral (2π) γ ω
(where ω is the angular form) is called the winding number of γ about the origin and is
denoted by W(γ, 0):
W(γ, 0) = 1
ω ≡
2π γ
1
2π γ
5
xdy − ydx
x 2 + y 2
.

The above example tells is that the winding number of the loop going around the unit
circles n times is exactly n. In §3.1 we will see that W(γ, 0) is always an integer.
Forming pullbacks is a nice and easy operation. For example, we have
Rule PB1. (g ◦ f) ∗ ω = f ∗ (g ∗ ω).
Here, g ◦ f is the composite of g and f: if f sends x = (x1, . . . , xn) to y = (y1, . . . , ym)
and g sends y to z = (z1, . . . , zℓ), then g ◦ f sends x to z: (g ◦ f)(x) = z = g(y) = g(f(x)).
For a differential form ω in the z-space, g ∗ ω is a differential form in y-space with every
occurrence of zj (1 ≤ j ≤ ℓ) in ω replaced by zj = gj(y1, . . . , yn). Similarly, the pull
back f ∗ (g ∗ ω) of g ∗ ω is a differential form in x-space obtained by another substitution.
Certainly the result of consecutive substitutions by g followed by f is the same as the
single substitution by their composite g ◦ f. So Rule PB1 is clear. From this rule we can
deduce the identity
Indeed,
g◦γ
ω =
γ
g ∗ ω.
g◦γ ω = b
a (g ◦ γ)∗ ω = b
a γ∗ (g ∗ ω) =
γ g∗ ω.
A differential form ω is said to be exact if it is the differential of some function f,
that is ω = df. In that case, the pull back of ω = df = n k=1 (∂f/∂xk)dxk is given by
γ ∗ ω = n
k=1
∂f
∂xk
dxk(t) = d(f(x(t))) ≡
x(t)
d
f(x(t)) dt
dt
(1.7)
(the subscript x(t) indicates the point at which ∂f/∂xk is evaluated) and hence we have
γ ω = b
a γ∗ω = b d
a dtf(x(t)) dt = f(x(b)) − f(x(a)). We conclude:
γ
df = f(the terminal point of γ) − f(the initial point of γ). (1.8)
In particular, if γ is a loop, that is, x(b) = x(a), then
ω = 0. Thus the line integral γ γ ω
depends only on the end points but not on the path γ linking them. This may be called
the path-independence property of line integrals for exact forms. The differential form
xdy + ydz + zdx in Example 1.6 does not have this property, because its integrals along
two paths α and β with same end points are different. Notice that (1.7) actually shows
that γ ∗ df = dγ ∗ f, where γ ∗ f, as a function of t is f(x(t)). This is not surprising because
forming pull backs (or substitution) and taking differentials are independent actions like
“kicking and punching”. In general, we have
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Rule PB2. g ∗ df = d(g ∗ f) ≡ d(f ◦ g).
Now we explain why putting the angular form as dθ is problematic. In rectangular
coordinates, the angular form is ω = (−ydx + xdy)/(x 2 + y 2 ), which is defined everywhere
except the origin. The expression dθ wrongfully suggests the exactness of ω. If ω were
exact, then
ω would be zero for any closed path γ. But, as shown in Example 1.7, when
γ
γ goes around the unit circle once in anticlockwise direction, the line integral
ω turns
out to be 2π, not 0. The trouble here is caused by the fact that, although ω is defined on
the punctured plane R 2 \{(0, 0)} (the Euclidean plane with the origin removed), θ cannot
be properly defined on the punctured plane. We may let the domain D for θ be the plane
with the negative part of the x-axis removed: D = {(x, y) : y = 0 or x > 0}. Then θ
becomes a genuine function defined on D satisfying −π < θ < π, called the principal value
of “arg”.
Exercises
1. Calculate the following differentials: (a) d(e −x2 /2 y
), (b) d(log log x), (c) d sin x ,
(d) d (xy ), (e) d(log r), where r = x2 + y2 + z2 .
2. Find the tangent plane to the sphere x 2 + y 2 + z 2 = 3 2 at the point P0 = (2, −2, −1)
by using the method described in the remark after Example 1.5.
3. Derive the product rule d(uv) = udv + vdu from d(u 2 ) = 2udu. (Hint: compute
d (u + v) 2 in two different ways.)
4. Derive the product rule for three functions: d(uvw) = uvdw + uwdv + vwdu. Write
an expression for d(u1u2 · · · un), giving the product rule for n functions.
5. The logarithmic differential of u, denoted by ℓu here (this is not a standard notation),
is defined to be du/u. Verify each of the following identities: (a) ℓ(uv) = ℓu + ℓv,
(b) ℓ(u/v) = ℓu − ℓv, (c) ℓ(u v ) = v((log u)ℓv + ℓu), (d) ℓ(e v ) = v ℓv and (e) ℓ log v =
ℓv/ log v. Part (a) can be generalized as ℓ(u1u2 · · · un) = ℓu1 + ℓu2 + · · · + ℓun. Does
it help you to write down the product rule for n functions, which is asked in the last
exercise?
6. A point P = (x, t) in the xt-plane is said to be inside the future cone C + if t > 0
and t 2 − x 2 > 0. The hyperbolic distance of such a point to the origin is given by
r = √ t 2 − x 2 . The hyperbolic coordinates (r, θ) of P is related to (x, t) by x = r sinh θ,
7
γ

t = r cosh θ, where sinh θ = (e θ − e −θ )/2, cosh θ = (e θ + e −θ )/2 are the hyperbolic
sine and cosine functions. Find dr and dθ in terms of dx and dt.
7. Compute the differentials drx, dry, dθx and dθy, where
rx = x
r , ry = y
r , θx = − y
r2 , θy = x
,
r2
r = x2 + y2
.
(This exercise will be useful for computing the Laplacian in polar coordinates.)
8. Consider the spherical coordinates (r, θ, φ) of a point P , where r is the distance to the
origin, θ is the longitude and φ is the latitude of P . Its relation to the rectangular
coordinates is given by x = r cos θ cos φ, y = r sin θ cos φ and z = r sin φ. (The conven-
tion of spherical coordinates here differs from some books.) Express the differentials
dr, dθ and dφ in rectangular coordinates. (Hint: The computation is rather tedious.
However, Example 1.2 helps.)
9. Compute the line integral
γ xdy − ydx, where γ is the path given by x = cos2 t,
y = sin 2 t (0 ≤ t ≤ π/2). Note that γ moves from (1, 0) to (0, 1) along the line
x + y = 1. What happens when you change the parametrization of this path to
x = 1 − t, y = t, or to x = 1 − t 2 , y = t 2 ?
10. For a loop γ, why is
xdy + ydx always zero, while xdy − ydx in general is not?
γ γ
For a loop γ, we have
1 xdy = γ 2 (xdy − ydx). Why?
γ
11. We have seen that if a differential form ω is exact, then any line integral
ω depends
only on the end points of γ. Prove that the converse of this statement is also true by
following the steps described below. Assume that ω = F1dx1 + F2dx2 + · · · + Fndxn
is a differential form having the property that any line integral
ω depends only
on the end points of γ. Fix any point p in the domain of ω. For any point x, let
f(x) =
ω, where γ is any path from p to x. This definition of f(x) makes sense
γ
because of the assumption on ω here. Complete the proof by checking ∂f/∂xk = Fk,
which uses
∂f
∂xk
x
= lim
h→0
f(x + hej) − f(x)
h
where ej is the jth vector of the natural basis of R n , namely ej = (δj1, δj2 . . . , δjn),
where δjk is Kronecker’s delta: it is 0 if j = k and is 1 if j = k.
8
γ
γ