dh df df dg
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
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— = — .dt dt y
x(x , y)
(x+x , y+y)
fchange in f in x direction — x
x
fchange in f in y direction — y
y
y
x(x , y)
(x+x , y+y)
fchange in f in x direction — x
x
fchange in f in y direction — y
y
dxchange in x — t
dt
dychange in y — t
dt
x =
y =
f dx — — t x dt
f dy — — t y dt
f = total change in f =(change in f in x direction) + (change in f in y direction) f dy
+ — — t y dt
f dx— — tx dt
f h — = —
t t
f dy+ — — y dt
f dx— —x dt
Taking the limit of both sides as t 0, it can be shown that
dh f dx f dy— = — — + — — =dt x dt y dt
f h— = —=t t
f dy+ — — y dt
f dx— —x dt
Taking the limit of both sides as t 0, it can be shown that
dh f dx f dy f dz— = — — + — — + — — =dt x dt y dt z 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.
For instance if h(t) = f(x(t),y(t),z(t)) , then
dxf f —— — dtx y
dy—dt
dx—dt
dyf f f —— — — dtx y z
dz—dt
= Df(x,y) c (t)
= Df(x,y,z) c (t)
f(u,v) = u2ev – uv3 x = cos t , y = sin t
Find dh/dt where h(t) = f(x(t),y(t)) .
Using the chain rule, we havedh f dx f dy— = — — + — — =dt x dt 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 .
Using the chain rule, we havedu u dx u dy u dz— = — — + — — + — — =dt x dt y dt z dt
u = xeyz x = et , y = t , z = sin t
Find du/dt .
Using the chain rule, we havedu u dx u dy u dz— = — — + — — + — — =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)) = fg(x,y) is
_ _| h h || — — ||_ x y _|
_ _| f u f v f u f
v | = | — — + — — — — + — — |
|_ u x v x u yv y _|
D(f g) =
= Df Dgf f
= — —u v
u u— —x y
v v— —x 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)) = [f1g(x,y), f2g(x,y)] = fg(x,y) is_ _
| h1 h1 || — — || x y || || h2 h2 || — — ||_ x y _|
_ _ _ _| f1 f1 | | u u || — — | | — — |
= | u v | | x y || | | || | | || f2 f2 | | v v || — — | | — — ||_ u v _| |_ x y _|
= Df Dg
D(f g) =
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)) = fg(x,y).
_ _ _ _ _ _| h h | | f f | | u u || — — | = | — — | | — — | |_ x y _| | u v | | x y |
|_ _| | || || v v || — — ||_ x y _|
_ _ _ _| | | || v u | | 2x –2y |
= | | | ||_ _| | | =
| || 2x 2y || ||_ _|
Using the chain rule, we havef : R R g : R R h : R R2 1 2 2 2 1
_ _| |
| v (2x) + u (2x) v (–2y) + u (2y) ||_ _|
_ _| |
= | (x2 + y2)(2x) + (x2 – y2)(2x) (x2 + y2)(–2y) + (x2 – y2)(2y) ||_ _|
_ _| |
= | 4x3 –4y3 | .|_ _|
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 | | | | — — | = | 4x3 –4y3 | .|_ x y _| |_ _|
Is it possible to define gf ? 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)) = fg(x,y,z).
_ _| u u u || — — — |
_ _ _ _ | 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 _|
Using the chain rule, we have
_ _ _ _| | | || 2u 2v –1 | | 2xy x2 0 |
= | | | ||_ _| | 0 2y 0 | =
| || –ze–xz 0 –xe–xz ||_ _|
f : R R g : R R h : R R3 1 3 3 3 1
_ _| |
| 2u (2xy) + ze–xz 2u (x2) + 2v (2y) xe–xz ||_ _|
_ _| |
= | 2x2y (2xy) + ze–xz 2x2y (x2) + 2y2 (2y) xe–xz ||_ _|_ _| |
= | 4x3y2 + ze–xz 2x4y + 4y3 xe–xz ||_ _|
Since f(u,v,w) = u2 + v2 – w and g(x,y,z) = (x2y , y2 , e–xz) , it is easy to see that h(x,y,z) = x4y2 + y4 – e–xz , after which it is easy to see that_ _ _ _
| h h h | | | | — — — | = | 4x3y2 + ze–xz 2x4y + 4y3 xe–xz | .|_ x y z _| |_ _|
Is it possible to define gf ? 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)) = fg(x,y) at the point (x0 , y0) = (2 , –3) .
Df(u,v) = Dg(x,y) =
g(2 , –3) =
2 –8
2u 0
0 4v3
3x2 0
5 – 2y
(u0 , v0) = g(x0 , y0) = (12 , 1)
Dfg(2 , –3) = Df(12 , 1) Dg(2 , –3) =
f : R R g : R R h : R R2 3 2 2 2 3
2 –8
24 0
0 4
12 0=
5 6
–16 –48
288 0
20 24
Is it possible to define gf ? No