Intuition behind neglecting higher order differentials in visual proofs of the Product Rule

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In most visual explanations of the product rule (like this video), the justification for the remaining square ($dfcdot dg$) being neglected is that it is "very small". But the whole point of the intuition is that the $df cdot g$ and $dg cdot f$ are very small too.



As such, I'm interested in either an intuitive explanation of this, or an other intuitive explanation for the product rule that takes this into account.



Thanks in advance




calculus intuition




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    up vote
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    favorite
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    In most visual explanations of the product rule (like this video), the justification for the remaining square ($dfcdot dg$) being neglected is that it is "very small". But the whole point of the intuition is that the $df cdot g$ and $dg cdot f$ are very small too.



    As such, I'm interested in either an intuitive explanation of this, or an other intuitive explanation for the product rule that takes this into account.



    Thanks in advance




    calculus intuition




    share|cite|improve this question























      up vote
      3
      down vote

      favorite
      1









      up vote
      3
      down vote

      favorite
      1






      1





      In most visual explanations of the product rule (like this video), the justification for the remaining square ($dfcdot dg$) being neglected is that it is "very small". But the whole point of the intuition is that the $df cdot g$ and $dg cdot f$ are very small too.



      As such, I'm interested in either an intuitive explanation of this, or an other intuitive explanation for the product rule that takes this into account.



      Thanks in advance




      calculus intuition




      share|cite|improve this question













      In most visual explanations of the product rule (like this video), the justification for the remaining square ($dfcdot dg$) being neglected is that it is "very small". But the whole point of the intuition is that the $df cdot g$ and $dg cdot f$ are very small too.



      As such, I'm interested in either an intuitive explanation of this, or an other intuitive explanation for the product rule that takes this into account.



      Thanks in advance





      calculus intuition





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 27 at 19:13









      Blue

      43.6k868141




      43.6k868141









      asked Jul 27 at 19:03









      Gabriel Alfour

      243




      243




















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          5
          down vote



          accepted










          It might help to compare the product rule against this (exact) equation for finite differences, where $Delta f(x)$ means $f(x + delta) - f(x)$:



          $$ Delta( fg(x) ) = g(x) cdot Delta f(x) + f(x+delta) cdot Delta g(x)$$



          If you find this equation helps you adequately intuit the product rule, then to answer the posted question observe the approximation



          $$ g(x) cdot Delta f(x) + f(x+delta) cdot Delta g(x)
          approx g(x) cdot Delta f(x) + f(x) cdot Delta g(x) $$



          ought to have negligible error; the difference between the two formulas is precisely $Delta f(x) Delta g(x)$.



          At some point you may be better off simply working through the algebra behind the proof of the product rule and basing your intuition off of the algebra directly, rather than looking for heuristic rationalizations.






          share|cite|improve this answer





















          • I already worked through the algebra. I'm looking for intuitions, so that I can manipulate the equations as mental concepts rather than going through computations everytime. But I think you are pointing at the right thing. These $df$/$dg$ are entailed by the $dx$ representation. I think I should be looking at something like the slope of a graph at a given point, but for function of $R^2$ to $R$. After checking Google, it's actually called the total derivative, and makes much more sense framed this way.
            – Gabriel Alfour
            Jul 27 at 20:33

















          up vote
          8
          down vote













          Indeed $dfcdot g$ and $gcdot df$ are very small but $dfcdot dg$ is even smaller.



          For the intuition, let assume for examaple



          • $Delta fapprox Delta gapprox10^-3$

          • $fapprox gapprox10$

          then



          $$Delta f cdot g approx10^-2ggDelta f cdot Delta g approx10^-6$$






          share|cite|improve this answer



















          • 1




            @AOrtiz Thanks for the edit and correction for the typo!
            – gimusi
            Jul 27 at 19:11

















          up vote
          8
          down vote













          The derivative is about the first order term in the expanded functions, higher order terms vanish in the limit:
          beginalign
          (f(x) , g(x))'
          &= lim_hto 0 fracf(x+h), g(x+h) - f(x), g(x)h \
          &= lim_hto 0 frac(f(x) + f'(x) h + O(h^2)), (g(x) + g'(x) h + O(h^2))) - f(x), g(x)h \
          &= lim_hto 0 fracf(x) , g(x) + f'(x) , g(x) h + f(x) , g'(x) h + O(h^2) - f(x), g(x)h \
          &= lim_hto 0 f'(x) , g(x) + f(x), g'(x) + O(h) \
          &= f'(x) , g(x) + f(x) , g'(x)
          endalign
          So too small in this context means being a term of order $k$ in $h$ with $k ge 2$.






          share|cite|improve this answer

















          • 1




            Aside; this calculation assumes $f$ and $g$ are at least twice differentiable (which is not necessarily a defect). It can be made more general by using the little-oh bound $f(x+h) = f(x) + f'(x) h + o(h)$.
            – Hurkyl
            Jul 27 at 19:35











          • In the algebra I see that the higher order term vanish. The remaining term is a $f(x +h) cdot g(x+h) / h$, so one of them is a derivative, while the other is equal to zero. However, I don't see how it can be represented on those rectangles.
            – Gabriel Alfour
            Jul 27 at 20:38










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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          5
          down vote



          accepted










          It might help to compare the product rule against this (exact) equation for finite differences, where $Delta f(x)$ means $f(x + delta) - f(x)$:



          $$ Delta( fg(x) ) = g(x) cdot Delta f(x) + f(x+delta) cdot Delta g(x)$$



          If you find this equation helps you adequately intuit the product rule, then to answer the posted question observe the approximation



          $$ g(x) cdot Delta f(x) + f(x+delta) cdot Delta g(x)
          approx g(x) cdot Delta f(x) + f(x) cdot Delta g(x) $$



          ought to have negligible error; the difference between the two formulas is precisely $Delta f(x) Delta g(x)$.



          At some point you may be better off simply working through the algebra behind the proof of the product rule and basing your intuition off of the algebra directly, rather than looking for heuristic rationalizations.






          share|cite|improve this answer





















          • I already worked through the algebra. I'm looking for intuitions, so that I can manipulate the equations as mental concepts rather than going through computations everytime. But I think you are pointing at the right thing. These $df$/$dg$ are entailed by the $dx$ representation. I think I should be looking at something like the slope of a graph at a given point, but for function of $R^2$ to $R$. After checking Google, it's actually called the total derivative, and makes much more sense framed this way.
            – Gabriel Alfour
            Jul 27 at 20:33














          up vote
          5
          down vote



          accepted










          It might help to compare the product rule against this (exact) equation for finite differences, where $Delta f(x)$ means $f(x + delta) - f(x)$:



          $$ Delta( fg(x) ) = g(x) cdot Delta f(x) + f(x+delta) cdot Delta g(x)$$



          If you find this equation helps you adequately intuit the product rule, then to answer the posted question observe the approximation



          $$ g(x) cdot Delta f(x) + f(x+delta) cdot Delta g(x)
          approx g(x) cdot Delta f(x) + f(x) cdot Delta g(x) $$



          ought to have negligible error; the difference between the two formulas is precisely $Delta f(x) Delta g(x)$.



          At some point you may be better off simply working through the algebra behind the proof of the product rule and basing your intuition off of the algebra directly, rather than looking for heuristic rationalizations.






          share|cite|improve this answer





















          • I already worked through the algebra. I'm looking for intuitions, so that I can manipulate the equations as mental concepts rather than going through computations everytime. But I think you are pointing at the right thing. These $df$/$dg$ are entailed by the $dx$ representation. I think I should be looking at something like the slope of a graph at a given point, but for function of $R^2$ to $R$. After checking Google, it's actually called the total derivative, and makes much more sense framed this way.
            – Gabriel Alfour
            Jul 27 at 20:33












          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          It might help to compare the product rule against this (exact) equation for finite differences, where $Delta f(x)$ means $f(x + delta) - f(x)$:



          $$ Delta( fg(x) ) = g(x) cdot Delta f(x) + f(x+delta) cdot Delta g(x)$$



          If you find this equation helps you adequately intuit the product rule, then to answer the posted question observe the approximation



          $$ g(x) cdot Delta f(x) + f(x+delta) cdot Delta g(x)
          approx g(x) cdot Delta f(x) + f(x) cdot Delta g(x) $$



          ought to have negligible error; the difference between the two formulas is precisely $Delta f(x) Delta g(x)$.



          At some point you may be better off simply working through the algebra behind the proof of the product rule and basing your intuition off of the algebra directly, rather than looking for heuristic rationalizations.






          share|cite|improve this answer













          It might help to compare the product rule against this (exact) equation for finite differences, where $Delta f(x)$ means $f(x + delta) - f(x)$:



          $$ Delta( fg(x) ) = g(x) cdot Delta f(x) + f(x+delta) cdot Delta g(x)$$



          If you find this equation helps you adequately intuit the product rule, then to answer the posted question observe the approximation



          $$ g(x) cdot Delta f(x) + f(x+delta) cdot Delta g(x)
          approx g(x) cdot Delta f(x) + f(x) cdot Delta g(x) $$



          ought to have negligible error; the difference between the two formulas is precisely $Delta f(x) Delta g(x)$.



          At some point you may be better off simply working through the algebra behind the proof of the product rule and basing your intuition off of the algebra directly, rather than looking for heuristic rationalizations.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 27 at 19:18









          Hurkyl

          107k9112253




          107k9112253











          • I already worked through the algebra. I'm looking for intuitions, so that I can manipulate the equations as mental concepts rather than going through computations everytime. But I think you are pointing at the right thing. These $df$/$dg$ are entailed by the $dx$ representation. I think I should be looking at something like the slope of a graph at a given point, but for function of $R^2$ to $R$. After checking Google, it's actually called the total derivative, and makes much more sense framed this way.
            – Gabriel Alfour
            Jul 27 at 20:33
















          • I already worked through the algebra. I'm looking for intuitions, so that I can manipulate the equations as mental concepts rather than going through computations everytime. But I think you are pointing at the right thing. These $df$/$dg$ are entailed by the $dx$ representation. I think I should be looking at something like the slope of a graph at a given point, but for function of $R^2$ to $R$. After checking Google, it's actually called the total derivative, and makes much more sense framed this way.
            – Gabriel Alfour
            Jul 27 at 20:33















          I already worked through the algebra. I'm looking for intuitions, so that I can manipulate the equations as mental concepts rather than going through computations everytime. But I think you are pointing at the right thing. These $df$/$dg$ are entailed by the $dx$ representation. I think I should be looking at something like the slope of a graph at a given point, but for function of $R^2$ to $R$. After checking Google, it's actually called the total derivative, and makes much more sense framed this way.
          – Gabriel Alfour
          Jul 27 at 20:33




          I already worked through the algebra. I'm looking for intuitions, so that I can manipulate the equations as mental concepts rather than going through computations everytime. But I think you are pointing at the right thing. These $df$/$dg$ are entailed by the $dx$ representation. I think I should be looking at something like the slope of a graph at a given point, but for function of $R^2$ to $R$. After checking Google, it's actually called the total derivative, and makes much more sense framed this way.
          – Gabriel Alfour
          Jul 27 at 20:33










          up vote
          8
          down vote













          Indeed $dfcdot g$ and $gcdot df$ are very small but $dfcdot dg$ is even smaller.



          For the intuition, let assume for examaple



          • $Delta fapprox Delta gapprox10^-3$

          • $fapprox gapprox10$

          then



          $$Delta f cdot g approx10^-2ggDelta f cdot Delta g approx10^-6$$






          share|cite|improve this answer



















          • 1




            @AOrtiz Thanks for the edit and correction for the typo!
            – gimusi
            Jul 27 at 19:11














          up vote
          8
          down vote













          Indeed $dfcdot g$ and $gcdot df$ are very small but $dfcdot dg$ is even smaller.



          For the intuition, let assume for examaple



          • $Delta fapprox Delta gapprox10^-3$

          • $fapprox gapprox10$

          then



          $$Delta f cdot g approx10^-2ggDelta f cdot Delta g approx10^-6$$






          share|cite|improve this answer



















          • 1




            @AOrtiz Thanks for the edit and correction for the typo!
            – gimusi
            Jul 27 at 19:11












          up vote
          8
          down vote










          up vote
          8
          down vote









          Indeed $dfcdot g$ and $gcdot df$ are very small but $dfcdot dg$ is even smaller.



          For the intuition, let assume for examaple



          • $Delta fapprox Delta gapprox10^-3$

          • $fapprox gapprox10$

          then



          $$Delta f cdot g approx10^-2ggDelta f cdot Delta g approx10^-6$$






          share|cite|improve this answer















          Indeed $dfcdot g$ and $gcdot df$ are very small but $dfcdot dg$ is even smaller.



          For the intuition, let assume for examaple



          • $Delta fapprox Delta gapprox10^-3$

          • $fapprox gapprox10$

          then



          $$Delta f cdot g approx10^-2ggDelta f cdot Delta g approx10^-6$$







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 27 at 19:10









          AOrtiz

          8,84621238




          8,84621238











          answered Jul 27 at 19:09









          gimusi

          64.9k73483




          64.9k73483







          • 1




            @AOrtiz Thanks for the edit and correction for the typo!
            – gimusi
            Jul 27 at 19:11












          • 1




            @AOrtiz Thanks for the edit and correction for the typo!
            – gimusi
            Jul 27 at 19:11







          1




          1




          @AOrtiz Thanks for the edit and correction for the typo!
          – gimusi
          Jul 27 at 19:11




          @AOrtiz Thanks for the edit and correction for the typo!
          – gimusi
          Jul 27 at 19:11










          up vote
          8
          down vote













          The derivative is about the first order term in the expanded functions, higher order terms vanish in the limit:
          beginalign
          (f(x) , g(x))'
          &= lim_hto 0 fracf(x+h), g(x+h) - f(x), g(x)h \
          &= lim_hto 0 frac(f(x) + f'(x) h + O(h^2)), (g(x) + g'(x) h + O(h^2))) - f(x), g(x)h \
          &= lim_hto 0 fracf(x) , g(x) + f'(x) , g(x) h + f(x) , g'(x) h + O(h^2) - f(x), g(x)h \
          &= lim_hto 0 f'(x) , g(x) + f(x), g'(x) + O(h) \
          &= f'(x) , g(x) + f(x) , g'(x)
          endalign
          So too small in this context means being a term of order $k$ in $h$ with $k ge 2$.






          share|cite|improve this answer

















          • 1




            Aside; this calculation assumes $f$ and $g$ are at least twice differentiable (which is not necessarily a defect). It can be made more general by using the little-oh bound $f(x+h) = f(x) + f'(x) h + o(h)$.
            – Hurkyl
            Jul 27 at 19:35











          • In the algebra I see that the higher order term vanish. The remaining term is a $f(x +h) cdot g(x+h) / h$, so one of them is a derivative, while the other is equal to zero. However, I don't see how it can be represented on those rectangles.
            – Gabriel Alfour
            Jul 27 at 20:38














          up vote
          8
          down vote













          The derivative is about the first order term in the expanded functions, higher order terms vanish in the limit:
          beginalign
          (f(x) , g(x))'
          &= lim_hto 0 fracf(x+h), g(x+h) - f(x), g(x)h \
          &= lim_hto 0 frac(f(x) + f'(x) h + O(h^2)), (g(x) + g'(x) h + O(h^2))) - f(x), g(x)h \
          &= lim_hto 0 fracf(x) , g(x) + f'(x) , g(x) h + f(x) , g'(x) h + O(h^2) - f(x), g(x)h \
          &= lim_hto 0 f'(x) , g(x) + f(x), g'(x) + O(h) \
          &= f'(x) , g(x) + f(x) , g'(x)
          endalign
          So too small in this context means being a term of order $k$ in $h$ with $k ge 2$.






          share|cite|improve this answer

















          • 1




            Aside; this calculation assumes $f$ and $g$ are at least twice differentiable (which is not necessarily a defect). It can be made more general by using the little-oh bound $f(x+h) = f(x) + f'(x) h + o(h)$.
            – Hurkyl
            Jul 27 at 19:35











          • In the algebra I see that the higher order term vanish. The remaining term is a $f(x +h) cdot g(x+h) / h$, so one of them is a derivative, while the other is equal to zero. However, I don't see how it can be represented on those rectangles.
            – Gabriel Alfour
            Jul 27 at 20:38












          up vote
          8
          down vote










          up vote
          8
          down vote









          The derivative is about the first order term in the expanded functions, higher order terms vanish in the limit:
          beginalign
          (f(x) , g(x))'
          &= lim_hto 0 fracf(x+h), g(x+h) - f(x), g(x)h \
          &= lim_hto 0 frac(f(x) + f'(x) h + O(h^2)), (g(x) + g'(x) h + O(h^2))) - f(x), g(x)h \
          &= lim_hto 0 fracf(x) , g(x) + f'(x) , g(x) h + f(x) , g'(x) h + O(h^2) - f(x), g(x)h \
          &= lim_hto 0 f'(x) , g(x) + f(x), g'(x) + O(h) \
          &= f'(x) , g(x) + f(x) , g'(x)
          endalign
          So too small in this context means being a term of order $k$ in $h$ with $k ge 2$.






          share|cite|improve this answer













          The derivative is about the first order term in the expanded functions, higher order terms vanish in the limit:
          beginalign
          (f(x) , g(x))'
          &= lim_hto 0 fracf(x+h), g(x+h) - f(x), g(x)h \
          &= lim_hto 0 frac(f(x) + f'(x) h + O(h^2)), (g(x) + g'(x) h + O(h^2))) - f(x), g(x)h \
          &= lim_hto 0 fracf(x) , g(x) + f'(x) , g(x) h + f(x) , g'(x) h + O(h^2) - f(x), g(x)h \
          &= lim_hto 0 f'(x) , g(x) + f(x), g'(x) + O(h) \
          &= f'(x) , g(x) + f(x) , g'(x)
          endalign
          So too small in this context means being a term of order $k$ in $h$ with $k ge 2$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 27 at 19:26









          mvw

          30.2k22250




          30.2k22250







          • 1




            Aside; this calculation assumes $f$ and $g$ are at least twice differentiable (which is not necessarily a defect). It can be made more general by using the little-oh bound $f(x+h) = f(x) + f'(x) h + o(h)$.
            – Hurkyl
            Jul 27 at 19:35











          • In the algebra I see that the higher order term vanish. The remaining term is a $f(x +h) cdot g(x+h) / h$, so one of them is a derivative, while the other is equal to zero. However, I don't see how it can be represented on those rectangles.
            – Gabriel Alfour
            Jul 27 at 20:38












          • 1




            Aside; this calculation assumes $f$ and $g$ are at least twice differentiable (which is not necessarily a defect). It can be made more general by using the little-oh bound $f(x+h) = f(x) + f'(x) h + o(h)$.
            – Hurkyl
            Jul 27 at 19:35











          • In the algebra I see that the higher order term vanish. The remaining term is a $f(x +h) cdot g(x+h) / h$, so one of them is a derivative, while the other is equal to zero. However, I don't see how it can be represented on those rectangles.
            – Gabriel Alfour
            Jul 27 at 20:38







          1




          1




          Aside; this calculation assumes $f$ and $g$ are at least twice differentiable (which is not necessarily a defect). It can be made more general by using the little-oh bound $f(x+h) = f(x) + f'(x) h + o(h)$.
          – Hurkyl
          Jul 27 at 19:35





          Aside; this calculation assumes $f$ and $g$ are at least twice differentiable (which is not necessarily a defect). It can be made more general by using the little-oh bound $f(x+h) = f(x) + f'(x) h + o(h)$.
          – Hurkyl
          Jul 27 at 19:35













          In the algebra I see that the higher order term vanish. The remaining term is a $f(x +h) cdot g(x+h) / h$, so one of them is a derivative, while the other is equal to zero. However, I don't see how it can be represented on those rectangles.
          – Gabriel Alfour
          Jul 27 at 20:38




          In the algebra I see that the higher order term vanish. The remaining term is a $f(x +h) cdot g(x+h) / h$, so one of them is a derivative, while the other is equal to zero. However, I don't see how it can be represented on those rectangles.
          – Gabriel Alfour
          Jul 27 at 20:38












           

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          Color the edges and diagonals of a regular polygon

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          Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
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          Relationship between determinant of matrix and determinant of adjoint?

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          Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
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