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]]

asked Nov 22 at 17:33

R zu

1797

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]]

asked Nov 22 at 17:33

R zu

1797

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]]

asked Nov 22 at 17:33

R zu

1797

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

asked Nov 22 at 17:33

R zu

1797

asked Nov 22 at 17:33

R zu

1797

asked Nov 22 at 17:33

R zu

1797

asked Nov 22 at 17:33

R zu

1797

asked Nov 22 at 17:33

R zu

1797

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]]

edited Nov 22 at 18:05

answered Nov 22 at 17:36

kglr

174k9196402

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]

answered Nov 22 at 19:26

Carl Woll

66k385171

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

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\$","\$"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

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

Name

Required, but never shown

2 Answers
2

active

oldest

votes

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]]

edited Nov 22 at 18:05

answered Nov 22 at 17:36

kglr

174k9196402

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]]

edited Nov 22 at 18:05

answered Nov 22 at 17:36

kglr

174k9196402

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]]

edited Nov 22 at 18:05

answered Nov 22 at 17:36

kglr

174k9196402

D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]

edited Nov 22 at 18:05

answered Nov 22 at 17:36

kglr

174k9196402

edited Nov 22 at 18:05

answered Nov 22 at 17:36

kglr

174k9196402

answered Nov 22 at 17:36

kglr

174k9196402

answered Nov 22 at 17:36

kglr

174k9196402

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

@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]

answered Nov 22 at 19:26

Carl Woll

66k385171

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

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]

answered Nov 22 at 19:26

Carl Woll

66k385171

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

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]

answered Nov 22 at 19:26

Carl Woll

66k385171

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]

answered Nov 22 at 19:26

Carl Woll

66k385171

answered Nov 22 at 19:26

Carl Woll

66k385171

answered Nov 22 at 19:26

Carl Woll

66k385171

answered Nov 22 at 19:26

Carl Woll

66k385171

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

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 for UpSetDelayed and TagSetDelayed.
– 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

@Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
– Carl Woll
Nov 22 at 19:39

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Mathematica Stack Exchange!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Cfrgtkky