The derivative rules for multivariable functions stated Theorem 10 on
page 151 are analogous to derivative rules from single variable calculus.
Example 1 (page 152) illustrates the quotient rule.
Recall from single-variable calculus that if f(x) and g(t) are each a
differentiable function from R1 to R1, and h = f(g(t)), then the chain rule
tells us that dh df
df dg
—= — = — —
or
h  (t) = f  (t) = f  (g(t)) g  (t) .
dt dt
dx dt
Now suppose f(x,y) is a differentiable function from R2 to R1 and c(t) =
(x(t),y(t)) is a differentiable function (path) from R1 to R2. Consider the
derivative of h(t) = f(c(t)) = foc(t) = f(x(t),y(t)) with respect to t, that is,
dh
df
f
— = — .
change in f in x direction  — x
dt
dt y
x
(x+x , y+y)
f
(x , y)
change in f in y direction  — y
x
y
y
(x+x , y+y)
(x , y)
x
f
f dx
change in f in x direction  — x  — — t
x
x dt
f dy
f
change in f in y direction  — y  — — t
y dt
y
dx
x = change in x  — t
dt
dy
y = change in y  — t
dt
f = total change in f =
(change in f in x direction) + (change in f in y direction) 
f dx
f dy
— — t + — — t
x dt
y dt
f h f dx f dy

— = —  — — + — — Taking the limit of both sides as
t t x dt y dt t  0, it can be shown that
f h f dx f dy
— = —= — — + — —
t t x dt y dt
Taking the limit of both sides as
t  0, it can be shown that
dx
dh
f dx
f dy
f
f
—
— = —— + —— = —
—
dt
= Df(x,y) c  (t)
dt
x dt
y dt
x
y
dy
—
dt
This chain rule can be extended in the natural way to a
situation where f is a differentiable function from Rn to R1
and c(t) is a differentiable function (path) from R1 to Rn.
dx
—
dt
For instance if h(t) = f(x(t),y(t),z(t)) , then
dh
f dx
f dy
f dz
— = —— + —— + —— =
dt
x dt
y dt
z dt
f
—
x
f
—
y
= Df(x,y,z) c  (t)
f
—
z
dy
—
dt
dz
—
dt
f(u,v) = u2ev – uv3
x = cos t , y = sin t
Find dh/dt where h(t) = f(x(t),y(t)) .
dh
f dx
Using the chain rule, we have — = — —
dt
x dt
+
f dy
—— =
y dt
(2xey – y3)(–sin t) + (x2ey – 3xy2)(cos t) =
–2(cos t)(sin t)esin t + sin4t + (cos3t)esin t – 3(cos2t)(sin2t) .
Note how the same result can be obtained by first expressing
h(t) = f(x(t),y(t)) in terms of t and then differentiating.
u = xeyz
x = et , y = t , z = sin t
Find du/dt .
du
u dx u dy
u dz
Using the chain rule, we have — = — — + — — + — — =
dt
x dt
y dt
z dt
u = xeyz
x = et , y = t , z = sin t
Find du/dt .
du
u dx u dy
u dz
Using the chain rule, we have — = — — + — — + — — =
dt
x dt
y dt
z dt
eyzet + xzeyz(1) + xyeyz(cos t) =
et(sin t)et + et(sin t)et(sin t) + et t et(sin t) cos t =
(1 + sin t + t cos t) et(1+sin t) .
Note how the same result can be obtained by first expressing
u in terms of t and then differentiating.
Suppose f(u,v) is a differentiable function from R2 to R1, and g(x,y) =
[u(x,y) , v(x,y)] is a differentiable function from R2 to R2. Then, the
derivative matrix for h(x,y) = f(u(x,y),v(x,y)) = fg(x,y) is
_
|
D(f g) = |
|_
=
=
h
—
x
_
| f u
| ——
|_ u x
f
—
u
_
h
—
y
+
f
—
v
|
|
_|
f v
——
v x
u
—
x
f u
——
u y
_
+
u
—
y
= Df Dg
v
—
x
v
—
y
f v
|
—— |
v y _|
Suppose f(u,v) = [f1(u,v) , f2(u,v)] is a differentiable function from R2 to
R2, and g(x,y) = [u(x,y) , v(x,y)] is a differentiable function from R2 to R2.
Then, the derivative matrix for h(x,y) = [h1(x,y),h2(x,y)] = f(u(x,y),v(x,y))
= [f1g(x,y), f2g(x,y)] = fg(x,y) is
_
_
| h1
h1
|
| —
—
|
| x
y
|
|
D(f g) = |
| h2
h2
|
| —
—
|
|_ x
y
_|
_
_
_
_
| f1
f1 |
| u
u
|
| —
—
|
| —
—
|
= | u
v
|
| x
y
|
|
|
|
|
|
|
|
| = Df Dg
| f2
f2 |
| v
v
|
| —
—
|
| —
—
|
|_ u
v _|
|_ x
y _|
Look at the general chain rule stated in Theorem 11 on page 153.
f(u,v) = uv
g(x,y) = (x2 – y2 , x2 + y2)
Find the derivative matrix for h(x,y) = f(u(x,y),v(x,y)) = fg(x,y).
f : R 2 R 1
g : R 2 R 2 h : R 2 R 1
Using the chain rule, we have
_
_
_
_
_
| h
h
|
| f
f
|
| u
u
| —
—
| = | —
—
|
| —
—
|_ x
y
_|
| u
v
|
| x
y
|_
_|
|
|
| v
v
| —
—
|_ x
y
_
_
_
_
|
|
|
|
| v
u
|
| 2x
–2y |
= |
|
|
|
|_
_|
|
|
=
|
|
| 2x
2y |
|
|
|_
_|
_
|
|
|
|
|
|
|
_|
_
|
| v (2x) + u (2x)
|_
_
v (–2y) + u (2y)
|
|
_|
_
_
|
|
2
2
2
2
2
2
2
2
= | (x + y )(2x) + (x – y )(2x)
(x + y )(–2y) + (x – y )(2y) |
|_
_|
_
_
|
|
3
3
=
|
4x
–4y
| .
|_
_|
Since f(u,v) = uv and g(x,y) = (x2 – y2 , x2 + y2) , it is easy to see that
h(x,y) = (x2 – y2)(x2 + y2) = x4 – y4 , after which it is easy to see that
_
_
_
_
| h
h
|
|
|
3
3
| —
—
| = |
4x
–4y
| .
|_ x
y
_|
|_
_|
Is it possible to define gf ? No
f(u,v,w) = u2 + v2 – w
g(x,y,z) = (x2y , y2 , e–xz)
Find the derivative matrix for h(x,y,z) = f(u(x,y,z),v(x,y,z),w(x,y,z)) =
fg(x,y,z).
f : R 3 R 1
g : R 3 R 3 h : R 3 R 1
_
_
| u u u |
Using the chain rule, we have
| — — — |
_
_
_
_
| x y z |
| h
h
h |
| f f f |
|
|
| —
—
— | = | — — — |
| v v v |
|_ x
y
z _|
| u v w |
| — — — |
|_
_|
| x y z |
|
|
| w w w |
| — — — |
|_ x y
_
_
_
_ z _|
|
|
|
|
2
| 2u 2v –1 |
| 2xy
x
0
|
= |
|
|
|
|_
_|
| 0
2y
0
|
=
|
|
| –ze–xz 0
–xe–xz |
|_
_|
_
|
| 2u (2xy) + ze–xz
|_
_
|
= | 2x2y (2xy) + ze–xz
|_
_
|
= | 4x3y2 + ze–xz
|_
_
2u (x2) + 2v (2y)
2x2y
(x2)
+
2y2
2x4y + 4y3
(2y)
xe–xz
|
|
_|
xe–xz
xe–xz
_
|
|
_|
_
|
|
_|
Since f(u,v,w) = u2 + v2 – w and g(x,y,z) = (x2y , y2 , e–xz) , it is easy to
4y2 + y4 – e–xz , after which it is easy to see that
see
that
h(x,y,z)
=
x
_
_
_
_
| h
h
h
|
|
|
| —
—
—
| = | 4x3y2 + ze–xz 2x4y + 4y3 xe–xz
| .
|_ x
y
z _|
|_
_|
Is it possible to define gf ? No
f(u,v) = (2u – 8v , u2 , v4)
g(x,y) = (x3 + 4 , 5x – y2)
Find the derivative matrix for h(x,y) = f(u(x,y),v(x,y)) = fg(x,y) at the
point (x0 , y0) = (2 , –3) . f : R 2 R 3
g : R 2 R 2 h : R 2 R 3
Df(u,v) =
2
–8
2u
0
0
4v3
Dg(x,y) =
(u0 , v0) = g(x0 , y0) = g(2 , –3) = (12 , 1)
Dfg(2 , –3) = Df(12 , 1) Dg(2 , –3) =
3x2
0
5
– 2y
2
–8
24
0
0
4
12
5
Is it possible to define gf ? No
0
6
=
–16
–48
288
0
20
24