How is the notation $frac{d}{dx} (f^3)(1)$ interpreted?
How is the following notation interpreted?
$$frac{d}{dx} (f^3)(1)$$
Does this evaluate to $3cdot f(1)^2cdot f'(1) $, or is it simply the derivative of a constant and equal to 0?
calculus derivatives
edited Nov 12 '18 at 16:12
user587192
1,747215
asked Nov 12 '18 at 1:21
Is12Prime
12119
|
show 2 more comments
How is the following notation interpreted?
$$frac{d}{dx} (f^3)(1)$$
Does this evaluate to $3cdot f(1)^2cdot f'(1) $, or is it simply the derivative of a constant and equal to 0?
calculus derivatives
edited Nov 12 '18 at 16:12
user587192
1,747215
asked Nov 12 '18 at 1:21
Is12Prime
12119
1
Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 '18 at 1:27
It was in my textbook
– Is12Prime
Nov 12 '18 at 1:27
Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Don Thousand
Nov 12 '18 at 1:30
2
In that case, I'd cautiously vouch for your evaluation.
– Don Thousand
Nov 12 '18 at 1:31
1
I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 '18 at 8:19
|
show 2 more comments
How is the following notation interpreted?
$$frac{d}{dx} (f^3)(1)$$
Does this evaluate to $3cdot f(1)^2cdot f'(1) $, or is it simply the derivative of a constant and equal to 0?
calculus derivatives
edited Nov 12 '18 at 16:12
user587192
1,747215
asked Nov 12 '18 at 1:21
Is12Prime
12119
How is the following notation interpreted?
$$frac{d}{dx} (f^3)(1)$$
Does this evaluate to $3cdot f(1)^2cdot f'(1) $, or is it simply the derivative of a constant and equal to 0?
calculus derivatives
calculus derivatives
edited Nov 12 '18 at 16:12
user587192
1,747215
asked Nov 12 '18 at 1:21
Is12Prime
12119
edited Nov 12 '18 at 16:12
user587192
1,747215
asked Nov 12 '18 at 1:21
Is12Prime
12119
edited Nov 12 '18 at 16:12
user587192
1,747215
edited Nov 12 '18 at 16:12
user587192
1,747215
edited Nov 12 '18 at 16:12
user587192
1,747215
1,747215
asked Nov 12 '18 at 1:21
Is12Prime
12119
asked Nov 12 '18 at 1:21
Is12Prime
12119
asked Nov 12 '18 at 1:21
Is12Prime
12119
12119
1
Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 '18 at 1:27
It was in my textbook
– Is12Prime
Nov 12 '18 at 1:27
Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Don Thousand
Nov 12 '18 at 1:30
2
In that case, I'd cautiously vouch for your evaluation.
– Don Thousand
Nov 12 '18 at 1:31
1
I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 '18 at 8:19
|
show 2 more comments
1
Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 '18 at 1:27
It was in my textbook
– Is12Prime
Nov 12 '18 at 1:27
Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Don Thousand
Nov 12 '18 at 1:30
2
In that case, I'd cautiously vouch for your evaluation.
– Don Thousand
Nov 12 '18 at 1:31
1
I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 '18 at 8:19
1
1
Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 '18 at 1:27
Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 '18 at 1:27
It was in my textbook
– Is12Prime
Nov 12 '18 at 1:27
It was in my textbook
– Is12Prime
Nov 12 '18 at 1:27
Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Don Thousand
Nov 12 '18 at 1:30
Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Don Thousand
Nov 12 '18 at 1:30
2
2
In that case, I'd cautiously vouch for your evaluation.
– Don Thousand
Nov 12 '18 at 1:31
In that case, I'd cautiously vouch for your evaluation.
– Don Thousand
Nov 12 '18 at 1:31
1
1
I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 '18 at 8:19
I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 '18 at 8:19
|
show 2 more comments
1 Answer
1
active
oldest
votes
I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:
$$
frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
$$
answered Nov 12 '18 at 1:32
Mike R.
1,339212
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: "69"
};
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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2fmath.stackexchange.com%2fquestions%2f2994700%2fhow-is-the-notation-fracddx-f31-interpreted%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:
$$
frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
$$
answered Nov 12 '18 at 1:32
Mike R.
1,339212
add a comment |
I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:
$$
frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
$$
answered Nov 12 '18 at 1:32
Mike R.
1,339212
add a comment |
I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:
$$
frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
$$
answered Nov 12 '18 at 1:32
Mike R.
1,339212
I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:
$$
frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
$$
answered Nov 12 '18 at 1:32
Mike R.
1,339212
edited Nov 12 '18 at 1:39
answered Nov 12 '18 at 1:32
Mike R.
1,339212
answered Nov 12 '18 at 1:32
Mike R.
1,339212
answered Nov 12 '18 at 1:32
Mike R.
1,339212
1,339212
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics 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.
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%2fmath.stackexchange.com%2fquestions%2f2994700%2fhow-is-the-notation-fracddx-f31-interpreted%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
1
Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 '18 at 1:27
It was in my textbook
– Is12Prime
Nov 12 '18 at 1:27
Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Don Thousand
Nov 12 '18 at 1:30
2
In that case, I'd cautiously vouch for your evaluation.
– Don Thousand
Nov 12 '18 at 1:31
1
I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 '18 at 8:19