The direct proof of the chain rule
Advanced Calculus II, Math 411
Summer Session 2010
1. Intro. In this note we’ll go over the proof of the chain rule as a corollary of the
Mean Value Lemma and Mean Value Proposition given in the book.
2. Theorem. Let g : Rn →
− R, and f1 , . . . , fn : Rm →
− R have continuous 1st order
partials. Consider the composition,
call
k(x) = g(f1 (x), . . . , fn (x)) = (g ◦ F)(x)
Let x ∈ Rm . To simplify notation write y = (f1 (x), . . . , fn (x)), so that k(x) = g(y).
Picture
Then k has first order partials, Di k, and
Di k(x) =
m
X
Dj g(y) · Di fj (x)
j=1
3. Remark. Notice the use of the notation Dj . This is to avoid confusion about
what we are taking the derivative with respect to. You need to treat g as a function
of f1 , . . . , fn , so Dj g means the partial of g “with respect to” fj .
4. Example. One common example is the case when the fi are change of coordinates. For example, write
x = r cos θ
y = r sin θ.
Here, x and y are taking the role of the fi . Then
∂x
= cos θ
∂r
∂x
= −r sin θ
∂θ
∂y
= sin θ
∂r
∂y
= r cos θ.
∂θ
This allows us to compute the radial rate of change of a function given to us in
terms of x and y.
5. Example. The charge of a particle in R2 is given by the equation
C(x, y) = yex
How fast is the charge changing as the particle moves away from the origin?
First note that
D1 C(x, y) = yex
D2 C(x, y) = ex
1
b θ) = C(f1 (r, θ), f2 (r, θ)), f1 (r, θ) = r cos θ and f2 (r, θ) = r sin θ. Fix
Define C(r,
a specific point (r, θ) →
7− (f1 (r, θ), f2 (r, θ)). By the chain rule,
b θ) = D1 C(f1 (r, θ), f2 (r, θ)) · D1 f1 (r, θ) + D2 C(f1 (r, θ), f2 (r, θ)) · D1 f2 (r, θ)
D1 C(r,
= f2 (r, θ)ef1 (r,θ) · cos θ + ef1 (r,θ) · sin θ
= r sin θer cos θ · cos θ + er cos θ · sin θ
6. Remark. In practice, the above notation is unwieldy and unnecessary. Instead,
we generally write
∂C
∂C ∂x ∂C ∂y
=
+
∂r
∂x ∂r
∂y ∂r
This notation can sometimes cause confusion because when you write ∂C/∂x and
∂C/∂y, you generally mean that they are functions of x and y. Here however, they
are thought of as functions of r and θ. Be careful.
Another way this notation is useful is that you can remember the chain rule
easily by using diagrams like the following
+3 x @
r..
G @@@
....
@@@@@
.. .
@@@@
...
@ $
....
....
;C C
....
....
/ y
θ
The bold arrows show all the paths we must take to get from r to C, from which
you can read of which partials you need to take.
7. Proof of Theorem. The goal is to examine the quotient
k(x + tei ) − k(x)
t
whose limit (as t →
− 0) is Di k(x).
For the sake of argument, fix t > 0. By the (usual) Mean Value Theorem applied
to each fj ,
∃ θj , 0 < θj < t
such that fj (x + tei ) − fj (x) = t · Di fj (x + θj ei )
Let h = F(x + tei ) − F(x). By The Mean Value Proposition applied to g,
∃ z1 , . . . , zn , kzj − yk < khk
such that g(y + h) − g(y) =
n
X
hj · Dj g(zj ).
j=1
But k(x + tei ) − k(x) = g(y + h) − g(y) and hj = fj (x + tei ) − fj (x), so
k(x + tei ) − k(x) =
n
X
j=1
hj · Dj g(zj ) =
n
X
t · Di fj (x + θj ei ) · Dj g(zj )
j=1
Dividing through by t. By construction, as t →
− 0, θj →
− 0 and zj →
− y. Therefore
n
n
k(x + tei ) − k(x) X
t→
−0 X
=
Di gj (x + θj ei ) · Dj f (zj ) −−−→
Di gj (x) · Dj f (y)
t
j=1
j=1
where on the last step we have used the fact that Di gj and Dj f are all continuous
functions (as zj →
− y, Dj f (zj ) →
− Dj f (y), etc.)
8. Corollary. Let x ∈ Rn , and f : O →
− R be a continuously differentiable function
on some open set O 3 x, and p ∈ Rn any vector. Then the partials form a basis
for the space of directional derivatives, specifically:
∂f
(x) = h∇f (x), pi
∂p
9. Proof. Recall that we can write the directional derivative as a one dimensional
derivative. Pick ε > 0 such that Bε (x) ⊂ O. Define g : (−ε, ε) →
− R by
g(t) = f (x + tp).
Then basically by definition, g 0 (0) =
d
g(t) =
dt
n
X
∂f
∂p (x).
By the chain rule,
n
Di f (x + tp) ·
i=1
Now just set t = 0, get g 0 (0) =
Pn
i=1
X
d
Di f (x + tp) · pi
(x + tp) =
dt
i=1
Di f (x) · pi = h∇f (x), pi.