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
    35 mins ago










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










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










  • @Andres, you are right! I was focusing more on differentiability
    – dmtri
    30 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
    35 mins ago










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










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










  • @Andres, you are right! I was focusing more on differentiability
    – dmtri
    30 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 19 mins ago









dmtri

1,3881521




1,3881521










asked 44 mins 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
    35 mins ago










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










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










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




















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










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










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










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


















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




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












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




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












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




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












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






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












2 Answers
2






active

oldest

votes


















2














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
    30 mins ago






  • 1




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



















0














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
    31 mins ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














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
    30 mins ago






  • 1




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
















2














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
    30 mins ago






  • 1




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














2












2








2






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 24 mins ago

























answered 34 mins ago









Martín Vacas Vignolo

3,539522




3,539522












  • 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
    30 mins ago






  • 1




    @AndrésE.Caicedo I edited to be more clarifying
    – Martín Vacas Vignolo
    23 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
    30 mins ago






  • 1




    @AndrésE.Caicedo I edited to be more clarifying
    – Martín Vacas Vignolo
    23 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
30 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
30 mins ago




1




1




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




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











0














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
    31 mins ago
















0














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
    31 mins ago














0












0








0






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 31 mins ago









Akash Roy

93




93




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
    31 mins ago


















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
















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




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


















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