The subscript s indicates that s is being held constant. Then dividing by dt, we have du/dt as before. We could also use derivatives instead of differentials. By an equation like (5.1), we have du dz əz ət ди ди дх ди ду (7.2) Əx dt dy dt where we have written all thet derivatives as partials since u, x, y, and z depend on both s and t. Using (7.2), we get du = (2x + 2y)(2t)+ (2x – In z)(-2t) + (-2) (2) = 4yt + 2t ln z – 2y It is sometimes useful to write chain rule formulas in matrix form (for matrix multiplication, see Chapter 3, Section 6). Given, as above, u = f(x,y, z), x(s, t), y(s, t), z(s,t), we can write equations like (7.2) in the following matrix form: as ди ди ds dt ди ди ди dx dy dz, ду ду (7.3) ds dz dz ds dt Sometimes (7.3) is written in the abbreviated form d(u) a(s,t) a(u) a(x, y, z) d(x, y, z) d(s,t) which is reminiscent of dy dy dx %3D dt dx dt but be careful of this for two reasons: (a) It may be helpful in remembering the formula but to use it you must understand that it means the matrix product (7.3). (b) The symbol d(u, v)/a(x, y) usually means a determinant rather than a matrix of partial derivatives see Chapter 5, Section 4]. Again in these problems, you may say, why not just substitute? Look at the following problem. Example 2. Find du/ðs, du/dt, given u = x² + 2xy – y ln z and x = s+t2, y= s – t2, z = 2t. We find du = 2x dx + 2x dy + 2y dx – 2dz – In z dy = (2x + 2y)(ds + 2t dt) + (2x – In 2)(ds – 2t dt) – 2 (2 dt) 2y = (4x + 2y – In z) ds + ( 4yt + 2t ln z – dt. Then ди = 4x + 2y – In z, ди = 4yt + 2t I z 2y Əs If we want just one derivative, say du/at, we can save some work by letting ds = 0 to start with. To make it clear that we have done this, we write (2 dt) du, = (2x + 2y)(2t dt) + (2x – In z)(-2t dt) – 2y 4yt + 2t ln z dt.
The subscript s indicates that s is being held constant. Then dividing by dt, we have du/dt as before. We could also use derivatives instead of differentials. By an equation like (5.1), we have du dz əz ət ди ди дх ди ду (7.2) Əx dt dy dt where we have written all thet derivatives as partials since u, x, y, and z depend on both s and t. Using (7.2), we get du = (2x + 2y)(2t)+ (2x – In z)(-2t) + (-2) (2) = 4yt + 2t ln z – 2y It is sometimes useful to write chain rule formulas in matrix form (for matrix multiplication, see Chapter 3, Section 6). Given, as above, u = f(x,y, z), x(s, t), y(s, t), z(s,t), we can write equations like (7.2) in the following matrix form: as ди ди ds dt ди ди ди dx dy dz, ду ду (7.3) ds dz dz ds dt Sometimes (7.3) is written in the abbreviated form d(u) a(s,t) a(u) a(x, y, z) d(x, y, z) d(s,t) which is reminiscent of dy dy dx %3D dt dx dt but be careful of this for two reasons: (a) It may be helpful in remembering the formula but to use it you must understand that it means the matrix product (7.3). (b) The symbol d(u, v)/a(x, y) usually means a determinant rather than a matrix of partial derivatives see Chapter 5, Section 4]. Again in these problems, you may say, why not just substitute? Look at the following problem. Example 2. Find du/ðs, du/dt, given u = x² + 2xy – y ln z and x = s+t2, y= s – t2, z = 2t. We find du = 2x dx + 2x dy + 2y dx – 2dz – In z dy = (2x + 2y)(ds + 2t dt) + (2x – In 2)(ds – 2t dt) – 2 (2 dt) 2y = (4x + 2y – In z) ds + ( 4yt + 2t ln z – dt. Then ди = 4x + 2y – In z, ди = 4yt + 2t I z 2y Əs If we want just one derivative, say du/at, we can save some work by letting ds = 0 to start with. To make it clear that we have done this, we write (2 dt) du, = (2x + 2y)(2t dt) + (2x – In z)(-2t dt) – 2y 4yt + 2t ln z dt.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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![The subscript s indicates that s is being held constant. Then dividing by dt, we
have du/dt as before. We could also use derivatives instead of differentials. By an
equation like (5.1), we have
du dz
əz ət
ди
ди дх
ди ду
(7.2)
Əx dt
dy dt
where we have written all thet derivatives as partials since u, x, y, and z depend
on both s and t. Using (7.2), we get
du
= (2x + 2y)(2t)+ (2x – In z)(-2t) +
(-2) (2) = 4yt + 2t ln z –
2y
It is sometimes useful to write chain rule formulas in matrix form (for matrix
multiplication, see Chapter 3, Section 6). Given, as above, u = f(x,y, z), x(s, t),
y(s, t), z(s,t), we can write equations like (7.2) in the following matrix form:
as
ди ди
ds dt
ди ди ди
dx dy dz,
ду
ду
(7.3)
ds
dz
dz
ds
dt
Sometimes (7.3) is written in the abbreviated form
d(u)
a(s,t)
a(u) a(x, y, z)
d(x, y, z) d(s,t)
which is reminiscent of
dy
dy dx
%3D
dt
dx dt
but be careful of this for two reasons: (a) It may be helpful in remembering the
formula but to use it you must understand that it means the matrix product (7.3).
(b) The symbol d(u, v)/a(x, y) usually means a determinant rather than a matrix
of partial derivatives see Chapter 5, Section 4].
Again in these problems, you may say, why not just substitute? Look at the
following problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fab776f79-d15e-4871-96c0-7ec59008ecc3%2F6a40ce13-1e11-48f8-a7cd-4f499dfac432%2F8hdpmz_processed.png&w=3840&q=75)
Transcribed Image Text:The subscript s indicates that s is being held constant. Then dividing by dt, we
have du/dt as before. We could also use derivatives instead of differentials. By an
equation like (5.1), we have
du dz
əz ət
ди
ди дх
ди ду
(7.2)
Əx dt
dy dt
where we have written all thet derivatives as partials since u, x, y, and z depend
on both s and t. Using (7.2), we get
du
= (2x + 2y)(2t)+ (2x – In z)(-2t) +
(-2) (2) = 4yt + 2t ln z –
2y
It is sometimes useful to write chain rule formulas in matrix form (for matrix
multiplication, see Chapter 3, Section 6). Given, as above, u = f(x,y, z), x(s, t),
y(s, t), z(s,t), we can write equations like (7.2) in the following matrix form:
as
ди ди
ds dt
ди ди ди
dx dy dz,
ду
ду
(7.3)
ds
dz
dz
ds
dt
Sometimes (7.3) is written in the abbreviated form
d(u)
a(s,t)
a(u) a(x, y, z)
d(x, y, z) d(s,t)
which is reminiscent of
dy
dy dx
%3D
dt
dx dt
but be careful of this for two reasons: (a) It may be helpful in remembering the
formula but to use it you must understand that it means the matrix product (7.3).
(b) The symbol d(u, v)/a(x, y) usually means a determinant rather than a matrix
of partial derivatives see Chapter 5, Section 4].
Again in these problems, you may say, why not just substitute? Look at the
following problem.
Transcribed Image Text:Example 2. Find du/ðs, du/dt, given u = x² + 2xy – y ln z and
x = s+t2, y= s – t2, z = 2t.
We find
du = 2x dx + 2x dy + 2y dx – 2dz – In z dy
= (2x + 2y)(ds + 2t dt) + (2x – In 2)(ds – 2t dt) – 2 (2 dt)
2y
= (4x + 2y – In z) ds + ( 4yt + 2t ln z –
dt.
Then
ди
= 4x + 2y – In z,
ди
= 4yt + 2t I z
2y
Əs
If we want just one derivative, say du/at, we can save some work by letting ds = 0
to start with. To make it clear that we have done this, we write
(2 dt)
du, = (2x + 2y)(2t dt) + (2x – In z)(-2t dt) –
2y
4yt + 2t ln z
dt.
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