CHAIN RULE (SEVERAL VARIABLES)
DR KUENZER
Typed by: Yusuf Saber
Example:
 x 2  x22   f1  x1 , x2  
0
f ( x1 , x2 )   1
 , x0   
  

0
 x1  x1 x2   f 2  x1 , x2  
Solution
The derivative of f  x1, x2  is given by
 f1, x  x1 , x2 
f   x1 , x2    1
 f 2, x  x1 , x2 
 1
2x2 
 2x
 1

x1 
1  x2
0
f   0,0   
1
f1, x2  x1, x2  

f 2, x2  x1, x2  
0

0
0
We need to linearly approximate the function at x0    , we thus use the following
0
general formula
f  x   f  x0   f   x0  x  x0   R2  x 
with lim x  x0
x  x0
1
 R2 ( x) .
Here:
0   x1  0 
  R2  x 

0  x2  0 
 x12  x22   0   0

  
 x1  x1 x2   0  1
f  x
f  x0 
f   x0 
x  x0
Hence
 x 2  x22   0   x12  x22 
R2  x    1
   

 x1  x1 x2   x1   x1 x2 
error term
 x x 
 x2  x2 
2
 2

 1

1
2
 x1   x12  x22 
 x1  x2 
1

  0 ,
lim   
 lim 
 lim 


  
0
0
x  0   x2 
 x1 x2  x 0  x1 x2  x 0  1 1   0 
0
 x2  x2 
 x2  x2 
1 
1
2
 2


as to be expected.
2
1
2
Theorem (Chain Rule):
Let f : D 

m
be a function that is differentiable in x0  D , g : E 

n
that is differentiable in f  x0   E .
k
be a function
m
(Assume that there is an   0 such that B  x0   D , and B  f  x0   E  , and assume that
f ( D)  E .)
Recall:
The composition of f followed by g is given by
g
f  x   g  f  x  
def
Then the derivative of such a function is given by
 g f   x   g   f  x
0
k n



f   x0 
k m

mn
inner derivative
outer derivative
Note:
The number of variables determine the number of columns in the resulting matrix and the
number of functions determines the number of rows.
Proof:
Consider the linear approximation of  g f  in k0
g
f  x   g  f  x    g  f  x0    g   f  x0    f  x   f  x0    R2  f  x  
 g  f  x0    g   f  x0    f   x0    x  x0   R2  x    R2  f  x  


  g f  x0   g   f  x0    f   x0    x  x0   g   f  x0    R2  x   R2  f  x   ,
 g f   x0 
and x  x0
1
 g   f  x  R  x   0 , and
0
2
x  x0
1
R2  f  x    0 for x  x0
Since we have constructed a linear approximation with matrix g   f  x2    f   x0  , we
know that  g f   x0   g   f  x0   f   x0 
Example:
x
Let g    g  x, y   x  y
 y
r 
 r sin  
Let f    f  r ,   

 
 r cos  
We want to calculate  g f   r ,   .
g:
2

1
f:
2

2
By direct calculation:
g
g
f  r,   g  f  r ,    r sin   r cos
f   r ,     sin   cos 
r cos   r sin  
Now we apply the chain rule to recalculate  g f  (r ,  ) .
r cos  
 sin 
f   r,   

 r sin  
 cos 
g  x, y   1 1
We can therefore find
g
f   r ,   g   f  r ,    f   r , 
r cos  
 sin 
1  

 r sin  
 cos 
  sin   cos  r cos   r sin 
 1

which is the same.
Example:
 u  x 
Let f  x   
, where u  x  and v  x  are arbitrary differentiable functions in
 v  x  


 y
let g    g  y, z   y  z .
z 
We want to rederive the product rule from one-dimensional calculus.
From the question we see that
Now:
g
1

2
g:
2

1
 u  x 
f  x   g  f  x    g 
 u  x  v x ,
 v  x  


hence
g f  x  u x v x 

On the other hand,
f:
       
g   y, z    z
 u  x  
f  x  
 v  x  


y
,
Hence
g
f   x   g   f  x    f   x 
 u  x  
 v  x  u  x   
 v  x  


 v  x   u  x   u  x  v  x 
Hence
u  x   v  x   v  x   v  x   u  x  v  x  ,
as to be expected.
Example:
Calculate the derivative
d x2 1
t sin  xt 2  dt

x
dx
def
 f ( x)
We use the following trick. Define
h  y, z, w   t 1 sin  wt 2  dt ,
def
z
y
simply by replacing all occurrences of x by different variables, and let
x 
def 

g  x    x2 
x 
 
record the replacements. Then
 h g  x   h  g  x    h  x, x 2 , x 
  t 1 sin  t 2 x  dt
x2
x
 f  x
The derivative of g is given by
1 
 
g  x    2 x 
1 
 
The derivative of h is given by
h  y, z , w    hy  y, z , w  hz  y, z , w  hw  y , z , w  
Now we must find each of those terms separately.
By the Fundamental Theorem, we get, for a constant a
d x
u  t  dt  u  x 
dx a
So
hy  y, z, w  


y
d
  t 1 sin  t 2 w  dt   y 1 sin  y 2 w 
z
dy
And
hz  y, z , w  
d
dz
 t
z
1
y
Finally

sin  t 2 w  dt  z 1 sin  z 2 w 

z
d
  t 1 sin  t 2 w  dt
y
dw
z d

t 1 sin  t 2 w  dt
y dw
hw  y, z , w  

  t 1 cos  t 2 w  t 2 dt
z
y
1 z
2tw cos  t 2 w  dt

y
2w
z
1
sin  t 2 w  

y
2w 
1
sin  z 2 w   sin  y 2 w  


2w 

Therefore
h  y, z , w    hy  y, z , w  hz  y, z , w  hw  y , z , w  

   y 1 sin  y 2 w 

z 1 sin  z 2 w 
Hence
f '( x)
 h  g ( x)  g ( x)
1
 
 hy  x, x , x  hz  x, x , x  hw  x, x , x   2 x 
1
 
1
  x 1 sin( x 3 ) 1  x 2 sin( x 5 )  2 x  x 1 sin( x 5 )  sin( x 3 ) 
2
3
5
  x 1 sin( x 3 )  x 1 sin( x 5 ).
2
2

2
2
2

1
sin  z 2 w   sin  y 2 w   

2w 