Simplify Derivative with Substitution
up vote
3
down vote
favorite
I try to evaluate:
$$ frac{partial}{partial x} log{u(x, y, z)}$$
Mathematica gives:
$$ frac{1}{x+y+z}$$
I want to simplify the expression with my function:
$$ frac{1}{u(x, y, z)}$$
How to do that?
Thanks.
u[x_, y_, z_] = x + y + z
Simplify[D[Log[u[x, y, z]], x]]
calculus-and-analysis simplifying-expressions
New contributor
add a comment |
up vote
3
down vote
favorite
I try to evaluate:
$$ frac{partial}{partial x} log{u(x, y, z)}$$
Mathematica gives:
$$ frac{1}{x+y+z}$$
I want to simplify the expression with my function:
$$ frac{1}{u(x, y, z)}$$
How to do that?
Thanks.
u[x_, y_, z_] = x + y + z
Simplify[D[Log[u[x, y, z]], x]]
calculus-and-analysis simplifying-expressions
New contributor
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I try to evaluate:
$$ frac{partial}{partial x} log{u(x, y, z)}$$
Mathematica gives:
$$ frac{1}{x+y+z}$$
I want to simplify the expression with my function:
$$ frac{1}{u(x, y, z)}$$
How to do that?
Thanks.
u[x_, y_, z_] = x + y + z
Simplify[D[Log[u[x, y, z]], x]]
calculus-and-analysis simplifying-expressions
New contributor
I try to evaluate:
$$ frac{partial}{partial x} log{u(x, y, z)}$$
Mathematica gives:
$$ frac{1}{x+y+z}$$
I want to simplify the expression with my function:
$$ frac{1}{u(x, y, z)}$$
How to do that?
Thanks.
u[x_, y_, z_] = x + y + z
Simplify[D[Log[u[x, y, z]], x]]
calculus-and-analysis simplifying-expressions
calculus-and-analysis simplifying-expressions
New contributor
New contributor
New contributor
asked Nov 22 at 17:33
R zu
1797
1797
New contributor
New contributor
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]
1/u[x, y, z]
A more general substitution:/. u[x_,y_,z_] -> Defer[u[x,y,z]]
– R zu
Nov 22 at 18:04
@Rzu, good point.
– kglr
Nov 22 at 18:05
add a comment |
up vote
4
down vote
An alternative is to define UpValues
instead of DownValues
of u
:
Derivative[1, 0, 0][u] ^:= 1&
Derivative[0, 1, 0][u] ^:= 1&
Derivative[0, 0, 1][u] ^:= 1&
D[Log[u[x, y, z]], x]
1/u[x, y, z]
What are UpValues and DownValues? The definition in the doc seems recursive:UpValue
"gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
– R zu
Nov 22 at 19:29
@Rzu Maybe you can check out the documentation forUpSetDelayed
andTagSetDelayed
.
– Carl Woll
Nov 22 at 19:39
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]
1/u[x, y, z]
A more general substitution:/. u[x_,y_,z_] -> Defer[u[x,y,z]]
– R zu
Nov 22 at 18:04
@Rzu, good point.
– kglr
Nov 22 at 18:05
add a comment |
up vote
7
down vote
accepted
D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]
1/u[x, y, z]
A more general substitution:/. u[x_,y_,z_] -> Defer[u[x,y,z]]
– R zu
Nov 22 at 18:04
@Rzu, good point.
– kglr
Nov 22 at 18:05
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]
1/u[x, y, z]
D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]
1/u[x, y, z]
edited Nov 22 at 18:05
answered Nov 22 at 17:36
kglr
173k8195400
173k8195400
A more general substitution:/. u[x_,y_,z_] -> Defer[u[x,y,z]]
– R zu
Nov 22 at 18:04
@Rzu, good point.
– kglr
Nov 22 at 18:05
add a comment |
A more general substitution:/. u[x_,y_,z_] -> Defer[u[x,y,z]]
– R zu
Nov 22 at 18:04
@Rzu, good point.
– kglr
Nov 22 at 18:05
A more general substitution:
/. u[x_,y_,z_] -> Defer[u[x,y,z]]
– R zu
Nov 22 at 18:04
A more general substitution:
/. u[x_,y_,z_] -> Defer[u[x,y,z]]
– R zu
Nov 22 at 18:04
@Rzu, good point.
– kglr
Nov 22 at 18:05
@Rzu, good point.
– kglr
Nov 22 at 18:05
add a comment |
up vote
4
down vote
An alternative is to define UpValues
instead of DownValues
of u
:
Derivative[1, 0, 0][u] ^:= 1&
Derivative[0, 1, 0][u] ^:= 1&
Derivative[0, 0, 1][u] ^:= 1&
D[Log[u[x, y, z]], x]
1/u[x, y, z]
What are UpValues and DownValues? The definition in the doc seems recursive:UpValue
"gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
– R zu
Nov 22 at 19:29
@Rzu Maybe you can check out the documentation forUpSetDelayed
andTagSetDelayed
.
– Carl Woll
Nov 22 at 19:39
add a comment |
up vote
4
down vote
An alternative is to define UpValues
instead of DownValues
of u
:
Derivative[1, 0, 0][u] ^:= 1&
Derivative[0, 1, 0][u] ^:= 1&
Derivative[0, 0, 1][u] ^:= 1&
D[Log[u[x, y, z]], x]
1/u[x, y, z]
What are UpValues and DownValues? The definition in the doc seems recursive:UpValue
"gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
– R zu
Nov 22 at 19:29
@Rzu Maybe you can check out the documentation forUpSetDelayed
andTagSetDelayed
.
– Carl Woll
Nov 22 at 19:39
add a comment |
up vote
4
down vote
up vote
4
down vote
An alternative is to define UpValues
instead of DownValues
of u
:
Derivative[1, 0, 0][u] ^:= 1&
Derivative[0, 1, 0][u] ^:= 1&
Derivative[0, 0, 1][u] ^:= 1&
D[Log[u[x, y, z]], x]
1/u[x, y, z]
An alternative is to define UpValues
instead of DownValues
of u
:
Derivative[1, 0, 0][u] ^:= 1&
Derivative[0, 1, 0][u] ^:= 1&
Derivative[0, 0, 1][u] ^:= 1&
D[Log[u[x, y, z]], x]
1/u[x, y, z]
answered Nov 22 at 19:26
Carl Woll
65.8k285171
65.8k285171
What are UpValues and DownValues? The definition in the doc seems recursive:UpValue
"gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
– R zu
Nov 22 at 19:29
@Rzu Maybe you can check out the documentation forUpSetDelayed
andTagSetDelayed
.
– Carl Woll
Nov 22 at 19:39
add a comment |
What are UpValues and DownValues? The definition in the doc seems recursive:UpValue
"gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
– R zu
Nov 22 at 19:29
@Rzu Maybe you can check out the documentation forUpSetDelayed
andTagSetDelayed
.
– Carl Woll
Nov 22 at 19:39
What are UpValues and DownValues? The definition in the doc seems recursive:
UpValue
"gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "– R zu
Nov 22 at 19:29
What are UpValues and DownValues? The definition in the doc seems recursive:
UpValue
"gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "– R zu
Nov 22 at 19:29
@Rzu Maybe you can check out the documentation for
UpSetDelayed
and TagSetDelayed
.– Carl Woll
Nov 22 at 19:39
@Rzu Maybe you can check out the documentation for
UpSetDelayed
and TagSetDelayed
.– Carl Woll
Nov 22 at 19:39
add a comment |
R zu is a new contributor. Be nice, and check out our Code of Conduct.
R zu is a new contributor. Be nice, and check out our Code of Conduct.
R zu is a new contributor. Be nice, and check out our Code of Conduct.
R zu is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f186514%2fsimplify-derivative-with-substitution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown