Cancellation rules for partial derivatives











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Is it possible to do some kind of simplifications on an expression like



$$
f : x, y to mathbb{R} \
frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}} = frac{partial f}{partial y}
$$



So that $frac{partial f}{partial y}$ in the denominator reduces $frac{partial^2 f}{partial y^2}$ to $frac{partial f}{partial y}$ like $frac{a^2}{a}=a$, without knowing the function $f$.



I would say no, that's not allowed, but I'm just wondering.



Thank you.










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  • I think that the closest you have to it is the chain rule: $$ frac{partial f}{partial u} = frac{partial x}{partial u} frac{partial f}{partial x} $$
    – rafa11111
    Nov 26 at 22:59












  • No, a second derivative has nothing to do with the square of the first derivative. This quotient cannot be simplified.
    – Yves Daoust
    Nov 26 at 23:00















up vote
3
down vote

favorite












Is it possible to do some kind of simplifications on an expression like



$$
f : x, y to mathbb{R} \
frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}} = frac{partial f}{partial y}
$$



So that $frac{partial f}{partial y}$ in the denominator reduces $frac{partial^2 f}{partial y^2}$ to $frac{partial f}{partial y}$ like $frac{a^2}{a}=a$, without knowing the function $f$.



I would say no, that's not allowed, but I'm just wondering.



Thank you.










share|cite|improve this question






















  • I think that the closest you have to it is the chain rule: $$ frac{partial f}{partial u} = frac{partial x}{partial u} frac{partial f}{partial x} $$
    – rafa11111
    Nov 26 at 22:59












  • No, a second derivative has nothing to do with the square of the first derivative. This quotient cannot be simplified.
    – Yves Daoust
    Nov 26 at 23:00













up vote
3
down vote

favorite









up vote
3
down vote

favorite











Is it possible to do some kind of simplifications on an expression like



$$
f : x, y to mathbb{R} \
frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}} = frac{partial f}{partial y}
$$



So that $frac{partial f}{partial y}$ in the denominator reduces $frac{partial^2 f}{partial y^2}$ to $frac{partial f}{partial y}$ like $frac{a^2}{a}=a$, without knowing the function $f$.



I would say no, that's not allowed, but I'm just wondering.



Thank you.










share|cite|improve this question













Is it possible to do some kind of simplifications on an expression like



$$
f : x, y to mathbb{R} \
frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}} = frac{partial f}{partial y}
$$



So that $frac{partial f}{partial y}$ in the denominator reduces $frac{partial^2 f}{partial y^2}$ to $frac{partial f}{partial y}$ like $frac{a^2}{a}=a$, without knowing the function $f$.



I would say no, that's not allowed, but I'm just wondering.



Thank you.







partial-derivative






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asked Nov 26 at 22:56









WolfgangP

1625




1625












  • I think that the closest you have to it is the chain rule: $$ frac{partial f}{partial u} = frac{partial x}{partial u} frac{partial f}{partial x} $$
    – rafa11111
    Nov 26 at 22:59












  • No, a second derivative has nothing to do with the square of the first derivative. This quotient cannot be simplified.
    – Yves Daoust
    Nov 26 at 23:00


















  • I think that the closest you have to it is the chain rule: $$ frac{partial f}{partial u} = frac{partial x}{partial u} frac{partial f}{partial x} $$
    – rafa11111
    Nov 26 at 22:59












  • No, a second derivative has nothing to do with the square of the first derivative. This quotient cannot be simplified.
    – Yves Daoust
    Nov 26 at 23:00
















I think that the closest you have to it is the chain rule: $$ frac{partial f}{partial u} = frac{partial x}{partial u} frac{partial f}{partial x} $$
– rafa11111
Nov 26 at 22:59






I think that the closest you have to it is the chain rule: $$ frac{partial f}{partial u} = frac{partial x}{partial u} frac{partial f}{partial x} $$
– rafa11111
Nov 26 at 22:59














No, a second derivative has nothing to do with the square of the first derivative. This quotient cannot be simplified.
– Yves Daoust
Nov 26 at 23:00




No, a second derivative has nothing to do with the square of the first derivative. This quotient cannot be simplified.
– Yves Daoust
Nov 26 at 23:00










4 Answers
4






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3
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accepted










$$frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}}=frac{frac{partial}{partial y}left(frac{partial f}{partial y}right)}{frac{partial f}{partial y}}nefrac{partial f}{partial y}$$



just as



$$ frac{y^{primeprime}}{y^prime}ne y^prime $$






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    2
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    That is indeed not allowed, even for single-variable derivatives. Take for example $f(x) = x^2$. Then
    $$ frac{f''}{f'} = frac{2}{2x} = frac{1}{x} not= 2x = f' $$






    share|cite|improve this answer

















    • 1




      Actually the case in the question is single-variable, as $x$ plays no role.
      – Yves Daoust
      Nov 26 at 23:01






    • 1




      @YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
      – MisterRiemann
      Nov 26 at 23:02




















    up vote
    1
    down vote













    No. Remember the second derivative is just the derivative of the first derivative, so using $g$ for the first derivative, your equation is equivalent to $frac{partial g}{partial y}=g^2$, which of course can't always be true.






    share|cite|improve this answer




























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      0
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      No, usually these kinds of analogies don't work. Take for example $f(x,y)=y$.






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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

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        active

        oldest

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        active

        oldest

        votes








        up vote
        3
        down vote



        accepted










        $$frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}}=frac{frac{partial}{partial y}left(frac{partial f}{partial y}right)}{frac{partial f}{partial y}}nefrac{partial f}{partial y}$$



        just as



        $$ frac{y^{primeprime}}{y^prime}ne y^prime $$






        share|cite|improve this answer

























          up vote
          3
          down vote



          accepted










          $$frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}}=frac{frac{partial}{partial y}left(frac{partial f}{partial y}right)}{frac{partial f}{partial y}}nefrac{partial f}{partial y}$$



          just as



          $$ frac{y^{primeprime}}{y^prime}ne y^prime $$






          share|cite|improve this answer























            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            $$frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}}=frac{frac{partial}{partial y}left(frac{partial f}{partial y}right)}{frac{partial f}{partial y}}nefrac{partial f}{partial y}$$



            just as



            $$ frac{y^{primeprime}}{y^prime}ne y^prime $$






            share|cite|improve this answer












            $$frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}}=frac{frac{partial}{partial y}left(frac{partial f}{partial y}right)}{frac{partial f}{partial y}}nefrac{partial f}{partial y}$$



            just as



            $$ frac{y^{primeprime}}{y^prime}ne y^prime $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 26 at 23:17









            John Wayland Bales

            13.8k21137




            13.8k21137






















                up vote
                2
                down vote













                That is indeed not allowed, even for single-variable derivatives. Take for example $f(x) = x^2$. Then
                $$ frac{f''}{f'} = frac{2}{2x} = frac{1}{x} not= 2x = f' $$






                share|cite|improve this answer

















                • 1




                  Actually the case in the question is single-variable, as $x$ plays no role.
                  – Yves Daoust
                  Nov 26 at 23:01






                • 1




                  @YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
                  – MisterRiemann
                  Nov 26 at 23:02

















                up vote
                2
                down vote













                That is indeed not allowed, even for single-variable derivatives. Take for example $f(x) = x^2$. Then
                $$ frac{f''}{f'} = frac{2}{2x} = frac{1}{x} not= 2x = f' $$






                share|cite|improve this answer

















                • 1




                  Actually the case in the question is single-variable, as $x$ plays no role.
                  – Yves Daoust
                  Nov 26 at 23:01






                • 1




                  @YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
                  – MisterRiemann
                  Nov 26 at 23:02















                up vote
                2
                down vote










                up vote
                2
                down vote









                That is indeed not allowed, even for single-variable derivatives. Take for example $f(x) = x^2$. Then
                $$ frac{f''}{f'} = frac{2}{2x} = frac{1}{x} not= 2x = f' $$






                share|cite|improve this answer












                That is indeed not allowed, even for single-variable derivatives. Take for example $f(x) = x^2$. Then
                $$ frac{f''}{f'} = frac{2}{2x} = frac{1}{x} not= 2x = f' $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 at 23:00









                MisterRiemann

                5,0731623




                5,0731623








                • 1




                  Actually the case in the question is single-variable, as $x$ plays no role.
                  – Yves Daoust
                  Nov 26 at 23:01






                • 1




                  @YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
                  – MisterRiemann
                  Nov 26 at 23:02
















                • 1




                  Actually the case in the question is single-variable, as $x$ plays no role.
                  – Yves Daoust
                  Nov 26 at 23:01






                • 1




                  @YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
                  – MisterRiemann
                  Nov 26 at 23:02










                1




                1




                Actually the case in the question is single-variable, as $x$ plays no role.
                – Yves Daoust
                Nov 26 at 23:01




                Actually the case in the question is single-variable, as $x$ plays no role.
                – Yves Daoust
                Nov 26 at 23:01




                1




                1




                @YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
                – MisterRiemann
                Nov 26 at 23:02






                @YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
                – MisterRiemann
                Nov 26 at 23:02












                up vote
                1
                down vote













                No. Remember the second derivative is just the derivative of the first derivative, so using $g$ for the first derivative, your equation is equivalent to $frac{partial g}{partial y}=g^2$, which of course can't always be true.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













                  No. Remember the second derivative is just the derivative of the first derivative, so using $g$ for the first derivative, your equation is equivalent to $frac{partial g}{partial y}=g^2$, which of course can't always be true.






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    No. Remember the second derivative is just the derivative of the first derivative, so using $g$ for the first derivative, your equation is equivalent to $frac{partial g}{partial y}=g^2$, which of course can't always be true.






                    share|cite|improve this answer












                    No. Remember the second derivative is just the derivative of the first derivative, so using $g$ for the first derivative, your equation is equivalent to $frac{partial g}{partial y}=g^2$, which of course can't always be true.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 26 at 23:00









                    Y. Forman

                    11.4k423




                    11.4k423






















                        up vote
                        0
                        down vote













                        No, usually these kinds of analogies don't work. Take for example $f(x,y)=y$.






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                          up vote
                          0
                          down vote













                          No, usually these kinds of analogies don't work. Take for example $f(x,y)=y$.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            No, usually these kinds of analogies don't work. Take for example $f(x,y)=y$.






                            share|cite|improve this answer












                            No, usually these kinds of analogies don't work. Take for example $f(x,y)=y$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 26 at 23:01









                            Scientifica

                            6,13141332




                            6,13141332






























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