3.4. THE CHAIN RULE
3.4
139
The Chain Rule
You will recall from Calculus I that we apply the chain rule when we have the
d
composition of two functions, for example when computing dx
f (g (x)). The
chain rule applies in similar situations when dealing with functions of several
variables. For example, f (x; y) is a function of two variables. However, we may
be computing the partials of f (u; v) where both u and v are functions of one or
more variables. We look at di¤erent cases.
3.4.1
Partials of f (x; y) where x and y are functions of one
other variable t
Before we start, let us remind the reader that if a variable depends on several
variables then the derivatives are partial derivatives and we will use the partial
@z
but if a variable depends only on one other variable,
derivative notation as in
@x
dx
we will use the notation from calculus of functions of one variable as in
.
dt
Now, suppose we have z = f (x; y), x = g (t), y = h (t). Then, in this case
we can also think of z as a function of t. So, there is only one derivative to
dz
compute,
. It can be computed as follows:
dt
@z dx @z dy
dz
=
+
dt
@x dt
@y dt
(3.1)
To remember this formula, we can use a tree structure shown in …gure 3.6
as follows.
1. Draw the tree from top to bottom. Assuming z = f (x; y), x = g (t),
y = h (t), we start with z.
2. From z, we draw a branch for each variable z depends on, x and y in this
case, so we draw two branches.
3. From each of these variables, we repeat the procedure, that is draw a
branch for each variable it depends on. Both x and y only depend on t,
so we draw one branch from each.
The tree is interpreted as follows. Since z is ultimately a function of t, look
at all the paths from z to t.
1. Multiply all the partials which appear on a path.
2. Add all these products collected on each path.
In our case, we obtain the following: There are two paths from z to t. The
@z dx
…rst one is z ! x ! t which gives
. The second path is z ! y ! t which
@x dt
@z dy
@z dx @z dy
dz
gives
. Adding these gives
+
which is
.
@y dt
@x dt
@y dt
dt
140
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
z
x
y
t
t
Figure 3.6: z = f (x (t) ; y (t))
Similarly, if w = f (x; y; z) and x; y; z are functions of t, then the corresponding tree structure is shown in …gure 3.7.
Again, w is ultimately a function of t. So, there is only one derivative to
dw
compute,
. Using the interpretation outlines above, we obtain the following
dt
formula:
@w dx @w dy @w dz
dw
=
+
+
dt
@x dt
@y dt
@z dt
and so on.
Example 244 Find
du
if u = x2
dt
du
dt
=
@u dx @u dy
+
@x dt
@y dt
2x2t 2y3 cos t
=
4xt
=
=
=
3.4.2
y 2 and x = t2
6 y cos t
2
4 t
4t
3
1, y = 3 sin t.
1 t
4t
6 (3 sin t) cos t
18 sin t cos t
Partials of f (x; y) where x and y are functions of two
variables s and t
We can use the same tree structure as above to compute partial derivatives in
this case. Suppose we have z = f (x; y), x = g (s; t), y = h (s; t). So, in this
case, z is a function of s and t. So, there are two partial derivatives to compute:
@z
@z
and
. The corresponding tree structure is shown in …gure 3.8.
@s
@t
3.4. THE CHAIN RULE
141
w
x
y
t
z
t
t
Figure 3.7: w = f (x (t) ; y (t) ; z (t))
z
x
s
y
t
s
Figure 3.8: z = f (x (s; t) ; y (s; t))
t
142
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
The …rst partials of z with respect to x or t are computed as follows:
@z
@s
@z
@t
@z @x @z @y
+
@x @s
@y @s
@z @x @z @y
+
@x @t
@y @t
=
=
(3.2)
(3.3)
@z
Example 245 Let z = sin (x + y) where x = 2st and y = s2 + t2 . Find
and
@s
@z
.
@t
@z
@s
@z @x @z @y
+
@x @s
@y @s
= cos (x + y) (2t) + cos (x + y) (2s)
=
=
2t cos 2st + s2 + t2 + 2s cos st + s2 + t2
=
2 cos (s + t)
2
(s + t)
and
@z
@t
=
@z @x @z @y
+
@x @t
@y @t
cos (x + y) (2s) + cos (x + y) (2t)
=
2 cos (s + t)
=
Example 246 Let u = x2
@u
@u
and
.
@s
@t
@u
@s
=
=
@u
@t
2
2xy + 2y 3 where x = s2 ln t and y = 2st3 . Find
@u @x @u @y
+
@x @s
@y @s
(2x 2y) (2s ln t) +
=
2s2 ln t
=
@u @x @u @y
+
@x @t
@y @t
=
2s2 ln t
(s + t)
2x + 6y 2
4st3 (2s ln t) +
4st3
s2
t
+
2t3
2s2 ln t + 24s2 t6
2s2 ln t + 24s2 t6
Example 247 Given z = f (x; y), x = r2 + s2 and y = 2rs …nd
2t3
6st2
@z
@2z
and
@r
@r2
3.4. THE CHAIN RULE
Computation of
143
@z
.
@r
@z
@r
=
=
Computation of
@2z
@r2
@z @x @z @y
+
@x @r
@y @r
@z
@z
2r
+ 2s
@x
@y
@2z
@r2
@
@r
@
@r
@z
@r
@z
@z
2r
+ 2s
=
@x
@y
@r
@z
@ @z
= 2
+ 2r
+2
@r
@x
@r @x
@z
@ @z
@ @z
= 2
+ 2r
+ 2s
@x
@r @x
@r @y
=
@s
@r
@z
@y
+ 2s
@
@r
@z
@y
(3.4)
We compute separately
@
@r
@z
@x
=
=
@
@z @x
@
+
@x @x @r
@y
@2z
@2z
2r 2 + 2s
@x
@y@x
@z
@x
@
@x
@z
@y
@y
@r
(3.5)
and
@
@r
@z
@y
@z @x
@
+
@y @r
@y
@2z
@2z
= 2r
+ 2s 2
@x@y
@y
=
@y
@r
(3.6)
If f has continuous second partial derivatives, then combining Equations
3.4, 3.5 and 3.6 gives
2
2
@2z
@z
@2z
2@ z
2@ z
=
2
+
4r
+
8rs
+
4s
@r2
@x
@x2
@y@x
@y 2
If f does not have continuous partial derivatives, then we cannot combine
the mixed partials, and we get an expression slightly more complicated:
@2z
@z
@2z
@2z
@2z
@2z
=2
+ 4r2 2 + 4rs
+ 4rs
+ 4s2 2
2
@r
@x
@x
@y@x
@x@y
@y
144
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
F
x
y
x
x
Figure 3.9: F (x; f (x))
3.4.3
General Case
Suppose f is a function of n variables x1 , x2 , :::, xn and each xi is in turn a
function of m variables t1 , t2 , :::, tm . f is then a function of m variables. The
m …rst partials of f are:
@f
@f @x1
@f @x2
@f @xn
=
+
+ ::: +
@ti
@x1 @ti
@x2 @ti
@xn @ti
for each i = 1; 2; :::; m.
3.4.4
Implicit Di¤erentiation
The chain rule can be used to derive a simpler method for …nding the derivative
of an implicitly de…ned function.
y de…ned implicitly as a function of x in a relation of the form F (x; y) =
0
Suppose that F (x; y) = 0 de…nes y as an implicit function of x we will call
dy
. We do so by di¤erentiating both sides of
y = f (x). We wish to …nd
dx
F (x; y) = 0 with respect to x. The right side is easy to di¤erentiate and we will
skip the details here! To di¤erentiate the left side with respect to x, F (x; y),
we will use the chain rule, remembering that F (x; y) = F (x; f (x)). So, F is
ultimately a function of x. The corresponding tree structure is shown in …gure
3.9.We obtain:
3.4. THE CHAIN RULE
145
F
x
y
x
z
y x
y
Figure 3.10: F (x; y; f (x; y))
@F dx @F
+
@x dx
@y
@F
@F
+
@x
@y
dy
dx
dy
dx
dy
dx
=
0
=
0
dy
if 2xy y 3 + 1
dx
Using the notation above, F (x; y) = 2xy
Example 248 Find
dy
dx
=
=
@F
@x
@F
@y
=
x
Fx
Fy
=
2y = 0.
y3 + 1
x
2y
Fx
Fy
2x
2y 1
3y 2 2
z de…ned implicitly as a function of x and y in a relation of the form
F (x; y; z) = 0
Suppose that F (x; y; z) = 0 de…nes z as an implicit function of x and y that
is F (x; y; z) = F (x; y; f (x; y)). We can use a tree structure to …nd the partial
derivatives. It is shown in …gure 3.10. The dotted lines indicate that if x were
a function of y, it is where we would have drawn a branch. Since x and y are
the independent variables, they do not depend on each other. So, x is not a
function of y and y is not a function of x.
If we di¤erentiate each side with respect to x, using the tree structure, we
146
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
get
@F @x @F @z
+
=0
@x @x
@z @x
Now,
@x
= 1 so, we get
@x
@F
@F @z
+
=0
@x
@z @x
That is
@z
=
@x
@F
@x
@F
@z
=
Fx
Fz
@z
=
@y
@F
@y
@F
@z
=
Fy
Fz
Similarly, we can show that
Example 249 Find
@z
if z is de…ned implicitly by
@x
x3 + y 3 + z 3 + 6xyz = 1
Using the formula above with F (x; y; z) = x3 + y 3 + z 3 + 6xyz
@z
@x
=
=
=
1, we see that
Fx
Fz
3x2 + 6yz
3z 2 + 6xy
x2 + 2yz
z 2 + 2xy
Compare this solution with the solution of example 234 on page 134.
3.4.5
Assignment
Do odd# 1 - 33 at the end of 11.4 in your book.