3.4. THE CHAIN RULE
3.4
241
The Chain Rule
You will recall from Calculus I that we apply the chain rule when we have the
d
f (g (x)). The
composition of two functions, for example when computing dx
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 (x; y) where both x and y are functions of one or
more other 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
dz
@z dx @z dy
=
+
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. Each path is the partial derivative of the variable on top with respect to
the variable at the bottom.
2. Multiply all the partials which appear on a path.
3. Add all these products collected on each path.
242
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
z
x
y
t
t
Figure 3.6: z = f (x (t) ; y (t))
In our case, we obtain the following: There are two paths from z to t. The
@z dx
. The second path is z ! y ! t which
…rst one is z ! x ! t which gives
@x dt
@z dy
@z dx @z dy
dz
gives
. Adding these gives
+
which is
.
@y dt
@x dt
@y dt
dt
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 3.4.1 Find
du
dt
du
if u = x2
dt
y 2 and x = t2
=
@u dx @u dy
+
@x dt
@y dt
2x2t 2y3 cos t
=
4xt
=
=
=
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
3.4. THE CHAIN RULE
243
w
x
y
t
t
z
t
Figure 3.7: w = f (x (t) ; y (t) ; z (t))
3.4.2
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
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 3.4.2 Let z = sin (x + y) where x = 2st and y = s2 + t2 . Find
@s
@z
and
.
@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)
244
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
z
x
y
s
t
s
t
Figure 3.8: z = f (x (s; t) ; y (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 3.4.3 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 3.4.4 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
245
@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
246
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
247
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
=
@F
@x
@F
@y
=
Fx
Fy
dy
if 2xy y 3 + 1 x 2y = 0.
dx
Using the notation above, F (x; y) = 2xy y 3 + 1 x 2y
Example 3.4.5 Find
dy
dx
=
=
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
248
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
get
@F dx @F @z
+
=0
@x dx
@z @x
Now,
@x
= 1 so, we get
@x
@F @z
@F
+
=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 3.4.6 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 3.3.10 on page 229.
3.4.5
Problems
For the problems below, do the following:
Find
dw
using the chain rule.
dt
Find
dw
by …rst expressing w as a function of t.
dt
Evaluate
dw
at the given value of t.
dt
1. w = x2 + y 2 , x = cos t, y = sin t, t = .
2. w =
x y
1
+ , x = cos2 t, y = sin2 t, z = , t = 3.
z
z
t
3.4. THE CHAIN RULE
3. w = 2yex
249
ln z, x = ln t2 + 1 , y = tan
1
t, z = et , t = 1.
For the problems below, do the following:
Find
@z
@z
and
using the chain rule.
@u
@v
Find
@z
@z
and
by …rst expressing z as a function of u and v.
@u
@v
Evaluate
@z
@z
and
at the given point (u; v).
@u
@v
4. z = 4ex ln y, x = ln (u cos v), y = u sin v, (u; v) = 2;
4
.
For the problems below, do the following:
Find
@w
@w
and
using the chain rule.
@u
@v
Find
@w
@w
and
by …rst expressing w as a function of u and v.
@u
@v
Evaluate
@w
@w
and
at the given point (u; v).
@u
@v
5. w = xy + yz + xz, x = u + v, y = u
v, z = uv, (u; v) =
1
;1 .
2
For the problems below, do the following:
Find
@u @u
@u
,
and
using the chain rule.
@x @y
@z
Find
@u
@u @u
,
and
by …rst expressing u as a function of u and v.
@x @y
@z
Evaluate
6. u =
p
q
@u @u
@u
,
and
at the given point (x; y; z).
@x @y
@z
q
, p = x + y + z, q = x
r
y + z, r = x + y
z, (x; y; z) =
p
3; 2; 1 .
Draw a tree diagram and write a formula for the each derivative below.
7.
dz
for z = f (x; y), x = g (t), y = h (t).
dt
8.
@w
@w
and
for w = h (x; y; z), x = f (u; v), y = g (u; v) and z = k (u; v).
@u
@v
9.
@w
@w
and
for w = g (x; y), x = g (u; v) and y = k (u; v).
@u
@v
250
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
10.
@z
@z
and
for z = f (x; y), x = g (t; s) and y = h (t; s).
@t
@s
11.
@w
@w
and
for w = g (u), u = h (s; t).
@s
@t
12.
@w
@w
and
for w = f (x; y), x = g (r) and y = h (s).
@r
@s
dy
Assuming the relations below de…ne y as an implicit function of x, …nd
dx
at the given point.
13. x3
2y 2 + xy = 0, (1; 1)
14. x2 + xy + y 2
7 = 0, (1; 2)
Assuming the relations below de…ne z as an implicit function of x and y,
@z
@z
…nd
and
at the given point.
@x
@y
15. z 3
xy + yz + y 3
2 = 0, (1; 1; 1).
16. sin (x + y) + sin (y + z) + sin (x + z) = 0, ( ; ; ).
@w
when r = 1, s =
@r
z = sin (r + s).
17. Find
2
1, if w = (x + y + z) , x = r s, y = cos (r + s),
Additional problems.
18. If z = f (x; y) where f is di¤erentiable, x = g (t), y = h (t), g (3) = 2,
dz
g 0 (3) = 5, h (3) = 7, h0 (3) = 4, fx (2; 7) = 6, fy (2; 7) = 8, …nd
dt
when t = 3.
19. The pressure of 1 mole of perfect gas is increasing at a rate of 0.05 kPa/s
and the temperature is increasing at a rate of 0.15 K/s. Knowing that
dV
P V = 8:31T …nd
when P = 20kP a and T = 320K.
dt
20. If z = f (x; y) where x = r cos and y = r sin
@z
@z
and
.
@r
@
2
@z
2. Show that
+
@x
1. Find
@z
@y
2
=
@z
@r
2
+
1
r2
@z
@
@z
@z
+
= 0 (hint: let u = x
@x @y
chain rule to compute the partial derivatives).
21. If z = f (x
y), show that
2
y then use the
3.4. THE CHAIN RULE
3.4.6
251
Answers
For the problems below, do the following:
Find
dw
using the chain rule.
dt
Find
dw
by …rst expressing w as a function of t.
dt
Evaluate
dw
at the given value of t.
dt
1. w = x2 + y 2 , x = cos t, y = sin t, t = .
dw
dt
=
=
w = cos2 t + sin2 t = 1 hence
dw
dt
2. w =
@w dx @w dy
+
@x dt
@y dt
0
dw
= 0.
dt
=0
t=
x y
1
+ , x = cos2 t, y = sin2 t, z = , t = 3.
z
z
t
dw
dt
=
=
@w dx @w dy @w dz
+
+
@x dt
@y dt
@z dt
1
w=t
Hence
dw
dt
3. w = 2yex
dw
=1
dt
=1
t=3
ln z, x = ln t2 + 1 , y = tan
dw
dt
=
=
1
t, z = et , t = 1.
@w dx @w dy @w dz
+
+
@x dt
@y dt
@z dt
4t tan 1 t + 1
252
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
w = 2 tan
So
dw
dt
1
t t2 + 1
dw
= 4t tan
dt
1
= 4 tan
1+1=
1
t
t+1
+1
t=1
For the problems below, do the following:
Find
@z
@z
and
using the chain rule.
@u
@v
Find
@z
@z
and
by …rst expressing z as a function of u and v.
@u
@v
Evaluate
@z
@z
and
at the given point (u; v).
@u
@v
4. z = 4ex ln y, x = ln (u cos v), y = u sin v, (u; v) = 2;
@z
@u
=
=
4
.
@z @x @z @y
+
@x @u @y @u
4 cos v (ln (u sin v) + 1)
and
@z
@v
=
=
@z @x @z @y
+
@x @v
@y @v
4u sin v ln (u sin v) +
4u cos2 v
sin v
z = 4u cos v ln (u sin v)
So
@z
= 4 cos v (ln (u sin v) + 1)
@u
and
@z
=
@v
4u sin v ln (u sin v) +
4u cos2 v
sin v
p
@z
2;
= 2 (ln 2 + 2)
@u
4
p
@z
2;
= 2 2 (ln 2 2)
@v
4
3.4. THE CHAIN RULE
253
For the problems below, do the following:
Find
@w
@w
and
using the chain rule.
@u
@v
Find
@w
@w
and
by …rst expressing w as a function of u and v.
@u
@v
Evaluate
@w
@w
and
at the given point (u; v).
@u
@v
5. w = xy + yz + xz, x = u + v, y = u
@w
@u
=
=
v, z = uv, (u; v) =
1
;1 .
2
@w @x @w @y
@w @z
+
+
@x @u
@y @u
@z @u
4uv + 2u
and
@w
@v
=
=
@w @x @w @y @w @z
+
+
@x @v
@y @v
@z @v
2
2v + 2u
w = u2
So
and
v 2 + 2u2 v
@w
= 2u + 4uv
@u
@w
=
@v
@w
and
@w
2v + 2u2
1
;1
2
@u
1
;1
2
@v
=3
=
3
2
For the problems below, do the following:
Find
@u @u
@u
,
and
using the chain rule.
@x @y
@z
Find
@u @u
@u
,
and
by …rst expressing u as a function of u and v.
@x @y
@z
254
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
Evaluate
6. u =
p
q
@u @u
@u
,
and
at the given point (x; y; z).
@x @y
@z
q
, p = x + y + z, q = x
r
@u
@x
=
=
y + z, r = x + y
z, (x; y; z) =
@u @p @u @q
@u @r
+
+
@p @x
@q @x
@r @x
0
and
@u
@y
=
=
@u @p @u @q
@u @r
+
+
@p @y
@q @y
@r @y
z
2
(z
y)
and
@u
@z
=
=
@u @p @u @q
@u @r
+
+
@p @z
@q @z
@r @z
y
2
(z
y)
u=
So
and
and
y
z
y
@u
=0
@x
@u
z
=
2
@y
(z y)
@u
=
@z
(z
@u
p
@u
p
and
and
@u
y
2
y)
3; 2; 1
=0
@x
3; 2; 1
=1
@y
p
3; 2; 1
=
@z
2
p
3; 2; 1 .
3.4. THE CHAIN RULE
255
Draw a tree diagram and write a formula for the each derivative below.
7.
8.
dz
for z = f (x; y), x = g (t), y = h (t).
dt
dz
@z dx @z dy
=
+
dt
@x dt
@y dt
@w
@w
and
for w = h (x; y; z), x = f (u; v), y = g (u; v) and z = k (u; v).
@u
@v
@w @x @w @y
@w @z
@w
=
+
+
@u
@x @u
@y @u
@z @u
and
9.
@w
@w
and
for w = g (x; y), x = g (u; v) and y = k (u; v).
@u
@v
@w
@w @x @w @y
=
+
@u
@x @u
@y @u
and
10.
12.
@w @x @w @y
@w
=
+
@v
@x @v
@y @v
@z
@z
and
for z = f (x; y), x = g (t; s) and y = h (t; s).
@t
@s
@z
@z @x @z @y
=
+
@t
@x @t
@y @t
and
11.
@w
@w @x @w @y @w @z
=
+
+
@v
@x @v
@y @v
@z @v
@z @x @z @y
@z
=
+
@s
@x @s
@y @s
@w
@w
and
for w = g (u), u = h (s; t).
@s
@t
dw @u
@w
=
@s
du @s
and
@w
dw @u
=
@t
du @t
@w
@w
and
for w = f (x; y), x = g (r) and y = h (s).
@r
@s
@w
@w dx
=
@r
@x dr
and
@w
@w dy
=
@s
@y ds
256
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
dy
Assuming the relations below de…ne y as an implicit function of x, …nd
dx
at the given point.
13. x3
2y 2 + xy = 0, (1; 1)
dy
=
dx
3x2 + y
4y + x
Hence
4
dy (1; 1)
=
dx
3
14. x2 + xy + y 2
7 = 0, (1; 2)
dy
=
dx
2x + y
x + 2y
Hence
dy (1; 2)
=
dx
4
5
Assuming the relations below de…ne z as an implicit function of x and y,
@z
@z
and
at the given point.
…nd
@x
@y
15. z 3
xy + yz + y 3
2 = 0, (1; 1; 1).
@z
=
@x
3z 2
y
+y
So
@z (1; 1; 1)
1
=
@x
4
and
@z
=
@y
x + z + 3y 2
3z 2 + y
So
@z (1; 1; 1)
=
@y
3
4
16. sin (x + y) + sin (y + z) + sin (x + z) = 0, ( ; ; ).
@z
Fx
=
=
@x
Fz
1
@z
Fy
=
=
@y
Fz
1
and
3.4. THE CHAIN RULE
257
@w
when r = 1, s =
@r
z = sin (r + s).
2
17. Find
@w
@r
=
=
1, if w = (x + y + z) , x = r s, y = cos (r + s),
@w @x @w @y @w @z
+
+
@x @r
@y @r
@z @r
2 (r s + cos (r + s) + sin (r + s)) (1
So
sin (r + s) + cos (r + s))
@w (1; 1)
= 12
@r
Additional Problems
18. If z = f (x; y) where f is di¤erentiable, x = g (t), y = h (t), g (3) = 2,
dz
g 0 (3) = 5, h (3) = 7, h0 (3) = 4, fx (2; 7) = 6, fy (2; 7) = 8, …nd
dt
when t = 3.
dz
@f (x; y) dx (t) @f (x; y) dy (t)
=
+
dt
@x
dt
@y
dt
When t = 3, we have
dz
dt
=
(6) (5) + ( 8) ( 4)
=
62
19. The pressure of 1 mole of perfect gas is increasing at a rate of 0.05 kPa/s
and the temperature is increasing at a rate of 0.15 K/s. Knowing that
dV
P V = 8:31T …nd
when P = 20kP a and T = 320K.
dt
8:31T
We have V =
so
P
dV
@V dT
@V dP
=
+
dt
@T dt
@P dt
=
0:270 08L=s
20. If z = f (x; y) where x = r cos and y = r sin
1. (a) Find
@z
@z
and
.
@r
@
@z
@r
=
=
@z @x @z @y
+
@x @r
@y @r
@z
@z
cos
+ sin
@x
@y
and
@z
@
=
=
@z @x @z @y
+
@x @
@y @
@z
@z
r sin
+ r cos
@x
@y
258
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
2
@z
@z
+
@x
@y
The right hand side is
2
(b) Show that
@z
@r
2
+
1
r2
@z
@
=
2
=
cos
=
@z
@x
@z
@r
2
+
1
r2
@z
@z
+ sin
@x
@y
2
+
@z
@y
2
@z
@
2
+
1
r2
r sin
@z
@z
+ r cos
@x
@y
2
@z
@z
+
= 0 (hint: let u = x y then use the
@x @y
chain rule to compute the partial derivatives).
@z
dz @u
dz
@z
dz @u
dz
Let u = x y, then
=
=
and
=
=
.
@x
du @x
du
@y
du @y
du
Therefore,
21. If z = f (x
y), show that
@z
@z
+
@x @y
=
=
fx = cos (x + ct)
dz
du
0
dz
du
2 sin (2x + 2ct)
so
fxx
Hence the result.
=
sin (x + ct)
4 cos (2x + 2ct)
=
(sin (x + ct) + 4 cos (2x + 2ct))
2