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
    44 mins ago










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










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










  • @Andres, you are right! I was focusing more on differentiability
    – dmtri
    38 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
    44 mins ago










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










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










  • @Andres, you are right! I was focusing more on differentiability
    – dmtri
    38 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 27 mins ago









dmtri

1,3881521




1,3881521










asked 52 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
    44 mins ago










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










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










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




















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










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










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










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


















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




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












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




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












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




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












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






@Andres, you are right! I was focusing more on differentiability
– dmtri
38 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
    39 mins ago






  • 1




    @AndrésE.Caicedo I edited to be more clarifying
    – Martín Vacas Vignolo
    31 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
    39 mins ago











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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
    39 mins ago






  • 1




    @AndrésE.Caicedo I edited to be more clarifying
    – Martín Vacas Vignolo
    31 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
    39 mins ago






  • 1




    @AndrésE.Caicedo I edited to be more clarifying
    – Martín Vacas Vignolo
    31 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 32 mins ago

























answered 42 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
    39 mins ago






  • 1




    @AndrésE.Caicedo I edited to be more clarifying
    – Martín Vacas Vignolo
    31 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
    39 mins ago






  • 1




    @AndrésE.Caicedo I edited to be more clarifying
    – Martín Vacas Vignolo
    31 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
39 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
39 mins ago




1




1




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




@AndrésE.Caicedo I edited to be more clarifying
– Martín Vacas Vignolo
31 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
    39 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
    39 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 40 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
    39 mins ago


















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
















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




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


















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