One-variable chain rule
Example. You learned in your first calculus course that
d
1
dx ln(sin(x)) = sin(x) cos(x).
Math 311-102
The general rule for differentiating a composite function f ◦ g is
d
0
0
dx f (g(x)) = f (g(x))g (x).
Harold P. Boas
Alternate notation. If y = f (u) and u = g(x), then
boas@tamu.edu
Example. If you forgot the formula for
you work it out?
d
dx
dy
dx
=
dy du
du dx .
sin−1 (x), how could
Solution. Put y = sin−1 (x), which implies that sin(y) = x. We
want to find
1
dx/dy
dy
dx ,
and the chain rule implies that this equals
q
√
= 1 − x2 , it
1
= cos(y)
. Since cos(y) = 1 − sin2 (y)
d
follows that dx sin−1 (x) = √ 1 2 .
1−x
Math 311-102
June 20, 2005: slide #1
Math 311-102
Example: multi-variable chain rule
Linear functions
  
1
x
  
Example. If f  y  = 4
7
z
matrix (Jacobian matrix) of
 
x
2 3
 
5 6  y , what is the derivative
z
8 9
f?


à !
à ! Ã
!
à !
x+y
x
r
r cos(θ)
x


=g
=
(polar
If f
=  x2 + y and
y
θ
r sin(θ)
y
xy
coordinates), find the derivative of the composite function f ◦ g.
Solution. The derivative matrix of f (with respect to x and y)

 

1 1
1
1

 

equals 2x 1  = 2r cos(θ)
1 , and the derivative
y x
r sin(θ) r cos(θ)
Ã
!
cos(θ) −r sin(θ)
matrix of g equals
; the product equals
sin(θ)
r cos(θ)


cos(θ) + sin(θ)
−r sin(θ) + r cos(θ)


2r cos2 (θ) + sin(θ) −2r 2 cos(θ) sin(θ) + r cos(θ) .
2
2
2
2
2r sin(θ) cos(θ)
−r sin (θ) + r cos (θ)
Answer. The derivative matrix has rows equal to the gradients
of the component functions of f , so the Jacobian matrix is


1 2 3


4 5 6.
7 8 9
In other words, a linear function is its own linear approximation.
You compose linear functions by multiplying their matrices.
Therefore the derivative matrix of a (multi-variable) composite
function equals the product of the derivative matrices.
Math 311-102
June 20, 2005: slide #2
June 20, 2005: slide #3
Math 311-102
June 20, 2005: slide #4
Example continued
Coordinate transformations

  
x+y
u

  
Alternate notation. If  v  =  x2 + y and
xy
w
à ! Ã
!
x
r cos(θ)
=
, then
y
r sin(θ)




∂u ∂u
∂u ∂u


 ∂x
 ∂r
∂y 
∂θ 
 ∂x ∂x

 


∂v   ∂r ∂θ 
∂v   ∂v
 ∂v

=
.

∂y ∂y
 ∂r
  ∂x
∂y 
∂θ

  ∂w ∂w 

∂w ∂w
∂r ∂θ
∂x
∂y
∂r
∂θ
For example,
Math 311-102
Question.
coordinate transformation
à ! à Is the polar
!
x
r cos(θ)
=
invertible?
y
r sin(θ)
p
Answer. You can solve explicitly for r = x2 + y2 and
y
θ = tan−1 ( x ) = cot−1 yx , but there are two problems. There is a
global problem that θ is ambiguous by addition of multiples
of 2π. There is a local problem that θ is not defined when
x = y = 0. The local problem is addressed by the
Inverse function theorem. A transformation is locally
invertible at points where the Jacobian matrix is invertible.
³
´
cos(θ) −r sin(θ)
Example. Since det sin(θ) r cos(θ) = r, the theorem confirms
∂v
∂v ∂x ∂v ∂y
=
+
.
∂θ
∂x ∂θ
∂y ∂θ
that the polar coordinate transformation is locally invertible
when r 6= 0.
June 20, 2005: slide #5
Math 311-102
June 20, 2005: slide #6