Are all functions that have a primitive differentiable?












3














Are all functions that have a primitive differentiable ?



For some background, I know that not all functions that are integrable are differentiable. For example:



$$
f =
begin{cases}
0 & x neq 0 \
1 & x = 0
end{cases}
$$



Is integrable over $mathbb{R}$, and $int_{a}^{b} f(x) dx = 0$. However, a function $F(x)$ such that $F'(x) = f(x) space forall x in mathbb{R}$ does not exist.



But I can't find a counterexample to the statement: "All functions that have a primitive are differentiable". These functions are guaranteed to be continuous due to the fundamental theorem of calculus, but not every continuous function is differentiable, hence the question.



Thanks !






calculus







share|cite|improve this question
























  • Any continuous function on a closed bounded interval has a primitive, but not all continuous functions are differentiable.
    – Andrés E. Caicedo
    53 mins ago










  • Take a function that is $C^1$ but not $C^2$.
    – dmtri
    52 mins ago










  • @dmtri That is not enough. Your conditions allow for the existence of a discontinuous second derivative.
    – Andrés E. Caicedo
    51 mins ago










  • @Andres, you are right! I was focusing more on differentiability
    – dmtri
    48 mins ago


















3














Are all functions that have a primitive differentiable ?



For some background, I know that not all functions that are integrable are differentiable. For example:



$$
f =
begin{cases}
0 & x neq 0 \
1 & x = 0
end{cases}
$$



Is integrable over $mathbb{R}$, and $int_{a}^{b} f(x) dx = 0$. However, a function $F(x)$ such that $F'(x) = f(x) space forall x in mathbb{R}$ does not exist.



But I can't find a counterexample to the statement: "All functions that have a primitive are differentiable". These functions are guaranteed to be continuous due to the fundamental theorem of calculus, but not every continuous function is differentiable, hence the question.



Thanks !






calculus







share|cite|improve this question
























  • Any continuous function on a closed bounded interval has a primitive, but not all continuous functions are differentiable.
    – Andrés E. Caicedo
    53 mins ago










  • Take a function that is $C^1$ but not $C^2$.
    – dmtri
    52 mins ago










  • @dmtri That is not enough. Your conditions allow for the existence of a discontinuous second derivative.
    – Andrés E. Caicedo
    51 mins ago










  • @Andres, you are right! I was focusing more on differentiability
    – dmtri
    48 mins ago
















3












3








3







Are all functions that have a primitive differentiable ?



For some background, I know that not all functions that are integrable are differentiable. For example:



$$
f =
begin{cases}
0 & x neq 0 \
1 & x = 0
end{cases}
$$



Is integrable over $mathbb{R}$, and $int_{a}^{b} f(x) dx = 0$. However, a function $F(x)$ such that $F'(x) = f(x) space forall x in mathbb{R}$ does not exist.



But I can't find a counterexample to the statement: "All functions that have a primitive are differentiable". These functions are guaranteed to be continuous due to the fundamental theorem of calculus, but not every continuous function is differentiable, hence the question.



Thanks !






calculus







share|cite|improve this question















Are all functions that have a primitive differentiable ?



For some background, I know that not all functions that are integrable are differentiable. For example:



$$
f =
begin{cases}
0 & x neq 0 \
1 & x = 0
end{cases}
$$



Is integrable over $mathbb{R}$, and $int_{a}^{b} f(x) dx = 0$. However, a function $F(x)$ such that $F'(x) = f(x) space forall x in mathbb{R}$ does not exist.



But I can't find a counterexample to the statement: "All functions that have a primitive are differentiable". These functions are guaranteed to be continuous due to the fundamental theorem of calculus, but not every continuous function is differentiable, hence the question.



Thanks !






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 37 mins ago









dmtri

1,3881521




1,3881521










asked 1 hour ago









ninivert

233




233












  • Any continuous function on a closed bounded interval has a primitive, but not all continuous functions are differentiable.
    – Andrés E. Caicedo
    53 mins ago










  • Take a function that is $C^1$ but not $C^2$.
    – dmtri
    52 mins ago










  • @dmtri That is not enough. Your conditions allow for the existence of a discontinuous second derivative.
    – Andrés E. Caicedo
    51 mins ago










  • @Andres, you are right! I was focusing more on differentiability
    – dmtri
    48 mins ago




















  • Any continuous function on a closed bounded interval has a primitive, but not all continuous functions are differentiable.
    – Andrés E. Caicedo
    53 mins ago










  • Take a function that is $C^1$ but not $C^2$.
    – dmtri
    52 mins ago










  • @dmtri That is not enough. Your conditions allow for the existence of a discontinuous second derivative.
    – Andrés E. Caicedo
    51 mins ago










  • @Andres, you are right! I was focusing more on differentiability
    – dmtri
    48 mins ago


















Any continuous function on a closed bounded interval has a primitive, but not all continuous functions are differentiable.
– Andrés E. Caicedo
53 mins ago




Any continuous function on a closed bounded interval has a primitive, but not all continuous functions are differentiable.
– Andrés E. Caicedo
53 mins ago












Take a function that is $C^1$ but not $C^2$.
– dmtri
52 mins ago




Take a function that is $C^1$ but not $C^2$.
– dmtri
52 mins ago












@dmtri That is not enough. Your conditions allow for the existence of a discontinuous second derivative.
– Andrés E. Caicedo
51 mins ago




@dmtri That is not enough. Your conditions allow for the existence of a discontinuous second derivative.
– Andrés E. Caicedo
51 mins ago












@Andres, you are right! I was focusing more on differentiability
– dmtri
48 mins ago






@Andres, you are right! I was focusing more on differentiability
– dmtri
48 mins ago












2 Answers
2






active

oldest

votes


















3














A primitive for $f(x)=x^{1/3}$ is $F(x)=frac{3}{4}x^{4/3}$, but $f'(0)$ doesn't exist, because $$limlimits_{xto0}dfrac{f(x)-f(0)}{x-0}=limlimits_{xto0}dfrac{x^{1/3}}{x}=limlimits_{xto0}dfrac{1}{x^{2/3}}=infty$$






share|cite|improve this answer























  • Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
    – Andrés E. Caicedo
    48 mins ago






  • 1




    @AndrésE.Caicedo I edited to be more clarifying
    – Martín Vacas Vignolo
    40 mins ago



















1














Any bounded function that has discontinuity at a single point is integrable but of course that function will be non differentiable.






share|cite|improve this answer








New contributor




Akash Roy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • @Martin has justified my statement.
    – Akash Roy
    49 mins ago











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055852%2fare-all-functions-that-have-a-primitive-differentiable%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














A primitive for $f(x)=x^{1/3}$ is $F(x)=frac{3}{4}x^{4/3}$, but $f'(0)$ doesn't exist, because $$limlimits_{xto0}dfrac{f(x)-f(0)}{x-0}=limlimits_{xto0}dfrac{x^{1/3}}{x}=limlimits_{xto0}dfrac{1}{x^{2/3}}=infty$$






share|cite|improve this answer























  • Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
    – Andrés E. Caicedo
    48 mins ago






  • 1




    @AndrésE.Caicedo I edited to be more clarifying
    – Martín Vacas Vignolo
    40 mins ago
















3














A primitive for $f(x)=x^{1/3}$ is $F(x)=frac{3}{4}x^{4/3}$, but $f'(0)$ doesn't exist, because $$limlimits_{xto0}dfrac{f(x)-f(0)}{x-0}=limlimits_{xto0}dfrac{x^{1/3}}{x}=limlimits_{xto0}dfrac{1}{x^{2/3}}=infty$$






share|cite|improve this answer























  • Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
    – Andrés E. Caicedo
    48 mins ago






  • 1




    @AndrésE.Caicedo I edited to be more clarifying
    – Martín Vacas Vignolo
    40 mins ago














3












3








3






A primitive for $f(x)=x^{1/3}$ is $F(x)=frac{3}{4}x^{4/3}$, but $f'(0)$ doesn't exist, because $$limlimits_{xto0}dfrac{f(x)-f(0)}{x-0}=limlimits_{xto0}dfrac{x^{1/3}}{x}=limlimits_{xto0}dfrac{1}{x^{2/3}}=infty$$






share|cite|improve this answer














A primitive for $f(x)=x^{1/3}$ is $F(x)=frac{3}{4}x^{4/3}$, but $f'(0)$ doesn't exist, because $$limlimits_{xto0}dfrac{f(x)-f(0)}{x-0}=limlimits_{xto0}dfrac{x^{1/3}}{x}=limlimits_{xto0}dfrac{1}{x^{2/3}}=infty$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 42 mins ago

























answered 51 mins ago









Martín Vacas Vignolo

3,549522




3,549522












  • Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
    – Andrés E. Caicedo
    48 mins ago






  • 1




    @AndrésE.Caicedo I edited to be more clarifying
    – Martín Vacas Vignolo
    40 mins ago


















  • Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
    – Andrés E. Caicedo
    48 mins ago






  • 1




    @AndrésE.Caicedo I edited to be more clarifying
    – Martín Vacas Vignolo
    40 mins ago
















Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
– Andrés E. Caicedo
48 mins ago




Almost. What you wrote is $f'(x)$ for $xne0$; you need an additional argument to ensure that $f'(0)$ does not exist.
– Andrés E. Caicedo
48 mins ago




1




1




@AndrésE.Caicedo I edited to be more clarifying
– Martín Vacas Vignolo
40 mins ago




@AndrésE.Caicedo I edited to be more clarifying
– Martín Vacas Vignolo
40 mins ago











1














Any bounded function that has discontinuity at a single point is integrable but of course that function will be non differentiable.






share|cite|improve this answer








New contributor




Akash Roy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • @Martin has justified my statement.
    – Akash Roy
    49 mins ago
















1














Any bounded function that has discontinuity at a single point is integrable but of course that function will be non differentiable.






share|cite|improve this answer








New contributor




Akash Roy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • @Martin has justified my statement.
    – Akash Roy
    49 mins ago














1












1








1






Any bounded function that has discontinuity at a single point is integrable but of course that function will be non differentiable.






share|cite|improve this answer








New contributor




Akash Roy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









Any bounded function that has discontinuity at a single point is integrable but of course that function will be non differentiable.







share|cite|improve this answer








New contributor




Akash Roy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer






New contributor




Akash Roy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 49 mins ago









Akash Roy

193




193




New contributor




Akash Roy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Akash Roy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Akash Roy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • @Martin has justified my statement.
    – Akash Roy
    49 mins ago


















  • @Martin has justified my statement.
    – Akash Roy
    49 mins ago
















@Martin has justified my statement.
– Akash Roy
49 mins ago




@Martin has justified my statement.
– Akash Roy
49 mins ago


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055852%2fare-all-functions-that-have-a-primitive-differentiable%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Morgemoulin

Image
Morgemoulin — miejscowość i gmina — Państwo   Francja Region Grand Est Departament Moza Okręg Verdun Kod INSEE 55357 Powierzchnia 6,82 km² Populacja  (1990) • liczba ludności 75 • gęstość 11 os./km² Kod pocztowy 55400 Położenie na mapie Francji Morgemoulin 49°14′N   5°35′E / 49,233333   5,583333 Portal Francja Morgemoulin – miejscowość i gmina we Francji, w regionie Grand Est, w departamencie Moza. Według danych na rok 1990 gminę zamieszkiwało 75 osób, a gęstość zaludnienia wynosiła 11 osób/km² (wśród 2335 gmin Lotaryngii Morgemoulin plasuje się na 970. miejscu pod względem liczby ludności, natomiast pod względem powierzchni na miejscu 848.). Bibliografia | Francuski urząd statystyczny ( fr. ) . p   •   d   •   e Gminy okręgu Verdun Abaucourt-Hautecourt • Aincreville • Ambly-sur-Meuse • Amel-sur-l'Étang • Ancemont • Arrancy-sur-Crusne • Aubréville • Autré...
Read more

Scott Moir

Image
Scott Moir Tessa Virtue i Scott Moir w 2016 r. Reprezentacja   Kanada Data i miejsce urodzenia 2 września 1987 London Wzrost 180 cm Konkurencja Pary taneczne Partner sportowy Tessa Virtue Trener Marie-France Dubreuil, Patrice Lauzon, Romain Haguenauer Poprzednio: Marina Zujewa, Oleg Epstein, Igor Szpilband, Johnny Johns, Carol Moir, Paul MacIntosh, Suzanne Killing Klub Montreal International School of Skating Poprzednio: Arctic Edge FSC Ilderton Skating Club Lokalizacja treningowa Montreal Poprzednio: Canton Kitchener Rekordy życiowe ISU przed sezonem 2018/2019 Nota łączna 206,07 Igrzyska Olimpijskie 2018 Taniec krótki 83,67 Igrzyska Olimpijskie 2018 Taniec dowolny 122,40 Igrzyska Olimpijskie 2018 Dorobek medalowy Reprezentacja   Kanada Igrzyska olimpijskie Złoto Vancouver 2010 pary taneczne Złoto Pjongczang 2018 pary taneczne Złoto Pjongczang 2018 drużynowo Srebro S...
Read more

Souastre

Image
Souastre — miejscowość i gmina — Herb Państwo   Francja Region Hauts-de-France Departament Pas-de-Calais Okręg Arras Kod INSEE 62800 Powierzchnia 7,18 km² Populacja  (1990) • liczba ludności 352 • gęstość 49 os./km² Kod pocztowy 62111 Położenie na mapie Francji Souastre 50°09′N   2°34′E / 50,150000   2,566667 Portal Francja Souastre – miejscowość i gmina we Francji, w regionie Hauts-de-France, w departamencie Pas-de-Calais. Według danych na rok 1990 gminę zamieszkiwały 352 osoby, a gęstość zaludnienia wynosiła 49 osób/km² (wśród 1549 gmin regionu Nord-Pas-de-Calais Souastre plasuje się na 887. miejscu pod względem liczby ludności, natomiast pod względem powierzchni na miejscu 507.). Bibliografia | Francuski urząd statystyczny ( fr. ) . p   •   d   •   e Gminy okręgu Arras Ablain-Saint-Nazaire Ablainzevelle Acheville Achicourt Achiet-le-Grand Achiet-le-Pe...
Read more