{
  "index": "1988-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "A not uncommon calculus mistake is to believe that the product rule\nfor derivatives says that $(fg)' = f'g'$. If $f(x)=e^{x^2}$,\ndetermine, with proof, whether there exists an open interval $(a,b)$\nand a nonzero function $g$ defined on $(a,b)$ such that this wrong\nproduct rule is true for $x$ in $(a,b)$.",
  "solution": "Solution. We are asked for a solution \\( g \\) to \\( f g^{\\prime}+g f^{\\prime}=f^{\\prime} g^{\\prime} \\), which is equivalent to \\( g^{\\prime}+\\left(\\frac{f^{\\prime}}{f-f^{\\prime}}\\right) g=0 \\) if we avoid the zeros of \\( f-f^{\\prime}=(1-2 x) e^{x^{2}} \\), i.e., avoid \\( x=1 / 2 \\). By the existence and uniqueness theorem for first order linear ordinary differential equations [BD, Theorem 2.4.1], if \\( x_{0} \\neq 1 / 2 \\), and \\( y_{0} \\) is any real number, then there exists a unique solution \\( g(x) \\), defined in some neighborhood \\( (a, b) \\) of \\( x_{0} \\), with \\( g\\left(x_{0}\\right)=y_{0} \\). By taking \\( y_{0} \\) nonzero, we obtain a nonzero solution \\( g \\).\n\nRemark. One can solve the differential equation by separation of variables, to find all possible \\( g \\). If \\( g \\) is nonzero, the differential equation is equivalent to\n\\[\n\\begin{aligned}\n\\frac{g^{\\prime}}{g} & =\\frac{f^{\\prime}}{f^{\\prime}-f}=\\frac{2 x e^{x^{2}}}{(2 x-1) e^{x^{2}}}=1+\\frac{1}{2 x-1}, \\\\\n\\ln |g(x)| & =x+\\frac{1}{2} \\ln |2 x-1|+c\n\\end{aligned}\n\\]\nfrom which one finds that the nonzero solutions are of the form \\( g(x)=C e^{x}|2 x-1|^{1 / 2} \\) for any nonzero number \\( C \\), on any interval not containing \\( 1 / 2 \\).\n\nRelated question. What other pairs of functions \\( f \\) and \\( g \\) satisfy \\( (f g)^{\\prime}=f^{\\prime} g^{\\prime} \\) ? For some discussion, see [Hal, Problem 2G].",
  "vars": [
    "f",
    "g",
    "x",
    "x_0"
  ],
  "params": [
    "a",
    "b",
    "y_0",
    "C"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "firstfunc",
        "g": "secondfunc",
        "x": "varxval",
        "x_0": "initialx",
        "a": "lowerbdry",
        "b": "upperbdry",
        "y_0": "initialy",
        "C": "constcval"
      },
      "question": "A not uncommon calculus mistake is to believe that the product rule for derivatives says that $(firstfunc\\,secondfunc)' = firstfunc'\\,secondfunc'$. If $firstfunc(varxval)=e^{varxval^{2}}$, determine, with proof, whether there exists an open interval $(lowerbdry,upperbdry)$ and a nonzero function secondfunc defined on $(lowerbdry,upperbdry)$ such that this wrong product rule is true for varxval in $(lowerbdry,upperbdry)$.",
      "solution": "Solution. We are asked for a solution \\( secondfunc \\) to \\( firstfunc\\,secondfunc^{\\prime}+secondfunc\\,firstfunc^{\\prime}=firstfunc^{\\prime}\\,secondfunc^{\\prime} \\), which is equivalent to \\( secondfunc^{\\prime}+\\left(\\frac{firstfunc^{\\prime}}{firstfunc-firstfunc^{\\prime}}\\right) secondfunc=0 \\) if we avoid the zeros of \\( firstfunc-firstfunc^{\\prime}=(1-2\\,varxval) e^{varxval^{2}} \\), i.e., avoid \\( varxval=1/2 \\). By the existence and uniqueness theorem for first-order linear ordinary differential equations [BD, Theorem 2.4.1], if \\( initialx \\neq 1/2 \\) and \\( initialy \\) is any real number, then there exists a unique solution \\( secondfunc(varxval) \\), defined in some neighborhood \\( (lowerbdry, upperbdry) \\) of \\( initialx \\), with \\( secondfunc\\!\\left(initialx\\right)=initialy \\). By taking \\( initialy \\) nonzero, we obtain a nonzero solution \\( secondfunc \\).\n\nRemark. One can solve the differential equation by separation of variables to find all possible \\( secondfunc \\). If \\( secondfunc \\) is nonzero, the differential equation is equivalent to\n\\[\n\\begin{aligned}\n\\frac{secondfunc^{\\prime}}{secondfunc} &= \\frac{firstfunc^{\\prime}}{firstfunc^{\\prime}-firstfunc} = \\frac{2\\,varxval\\,e^{varxval^{2}}}{(2\\,varxval-1) e^{varxval^{2}}} = 1 + \\frac{1}{2\\,varxval-1},\\\\[6pt]\n\\ln |secondfunc(varxval)| &= varxval + \\frac{1}{2}\\,\\ln |2\\,varxval-1| + c.\n\\end{aligned}\n\\]\nFrom this one finds that the nonzero solutions are of the form \\( secondfunc(varxval)=constcval\\,e^{varxval}|2\\,varxval-1|^{1/2} \\) for any nonzero number \\( constcval \\), on any interval not containing \\( 1/2 \\).\n\nRelated question. What other pairs of functions \\( firstfunc \\) and \\( secondfunc \\) satisfy \\( (firstfunc\\,secondfunc)^{\\prime}=firstfunc^{\\prime}\\,secondfunc^{\\prime} \\)? For some discussion, see [Hal, Problem 2G]."
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "sunflower",
        "g": "pineapple",
        "x": "raincloud",
        "x_0": "meadowlark",
        "a": "trombone",
        "b": "cinnamon",
        "y_0": "watermelon",
        "C": "harmonica"
      },
      "question": "A not uncommon calculus mistake is to believe that the product rule\nfor derivatives says that $(sunflower pineapple)' = sunflower' pineapple'$. If $sunflower(raincloud)=e^{raincloud^{2}}$,\ndetermine, with proof, whether there exists an open interval $(trombone,cinnamon)$\nand a nonzero function $pineapple$ defined on $(trombone,cinnamon)$ such that this wrong\nproduct rule is true for $raincloud$ in $(trombone,cinnamon)$.",
      "solution": "Solution. We are asked for a solution \\( pineapple \\) to \\( sunflower\\,pineapple^{\\prime}+pineapple\\,sunflower^{\\prime}=sunflower^{\\prime}\\,pineapple^{\\prime} \\), which is equivalent to \\( pineapple^{\\prime}+\\left(\\frac{sunflower^{\\prime}}{sunflower-sunflower^{\\prime}}\\right) pineapple=0 \\) if we avoid the zeros of \\( sunflower-sunflower^{\\prime}=(1-2\\,raincloud) e^{raincloud^{2}} \\), i.e., avoid \\( raincloud=1 / 2 \\). By the existence and uniqueness theorem for first order linear ordinary differential equations [BD, Theorem 2.4.1], if \\( meadowlark \\neq 1 / 2 \\), and \\( watermelon \\) is any real number, then there exists a unique solution \\( pineapple(raincloud) \\), defined in some neighborhood \\( (trombone, cinnamon) \\) of \\( meadowlark \\), with \\( pineapple\\!\\left(meadowlark\\right)=watermelon \\). By taking \\( watermelon \\) nonzero, we obtain a nonzero solution \\( pineapple \\).\n\nRemark. One can solve the differential equation by separation of variables, to find all possible \\( pineapple \\). If \\( pineapple \\) is nonzero, the differential equation is equivalent to\n\\[\n\\begin{aligned}\n\\frac{pineapple^{\\prime}}{pineapple} & =\\frac{sunflower^{\\prime}}{sunflower^{\\prime}-sunflower}=\\frac{2\\,raincloud\\,e^{raincloud^{2}}}{(2\\,raincloud-1) e^{raincloud^{2}}}=1+\\frac{1}{2\\,raincloud-1}, \\\\\n\\ln |pineapple(raincloud)| & =raincloud+\\frac{1}{2} \\ln |2\\,raincloud-1|+c\n\\end{aligned}\n\\]\nfrom which one finds that the nonzero solutions are of the form \\( pineapple(raincloud)=harmonica\\,e^{raincloud}|2\\,raincloud-1|^{1 / 2} \\) for any nonzero number \\( harmonica \\), on any interval not containing \\( 1 / 2 \\).\n\nRelated question. What other pairs of functions \\( sunflower \\) and \\( pineapple \\) satisfy \\( (sunflower\\,pineapple)^{\\prime}=sunflower^{\\prime}\\,pineapple^{\\prime} \\) ? For some discussion, see [Hal, Problem 2G]."
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "unvarying",
        "g": "stagnant",
        "x": "fixedpoint",
        "x_0": "fixedorigin",
        "a": "centerline",
        "b": "innercore",
        "y_0": "varybase",
        "C": "variable"
      },
      "question": "A not uncommon calculus mistake is to believe that the product rule for derivatives says that $(unvarying\\,stagnant)' = unvarying'\\,stagnant'$. If $unvarying(fixedpoint)=e^{fixedpoint^2}$, determine, with proof, whether there exists an open interval $(centerline,innercore)$ and a nonzero function $stagnant$ defined on $(centerline,innercore)$ such that this wrong product rule is true for $fixedpoint$ in $(centerline,innercore)$.",
      "solution": "Solution. We are asked for a solution \\( stagnant \\) to \\( unvarying\\,stagnant^{\\prime}+stagnant\\,unvarying^{\\prime}=unvarying^{\\prime}\\,stagnant^{\\prime} \\), which is equivalent to \\( stagnant^{\\prime}+\\left(\\frac{unvarying^{\\prime}}{unvarying-unvarying^{\\prime}}\\right) stagnant=0 \\) if we avoid the zeros of \\( unvarying-unvarying^{\\prime}=(1-2\\,fixedpoint) e^{fixedpoint^{2}} \\), i.e., avoid \\( fixedpoint=1 / 2 \\). By the existence and uniqueness theorem for first order linear ordinary differential equations [BD, Theorem 2.4.1], if \\( fixedorigin \\neq 1 / 2 \\), and \\( varybase \\) is any real number, then there exists a unique solution \\( stagnant(fixedpoint) \\), defined in some neighborhood \\( (centerline, innercore) \\) of \\( fixedorigin \\), with \\( stagnant\\left(fixedorigin\\right)=varybase \\). By taking \\( varybase \\) nonzero, we obtain a nonzero solution \\( stagnant \\).\n\nRemark. One can solve the differential equation by separation of variables, to find all possible \\( stagnant \\). If \\( stagnant \\) is nonzero, the differential equation is equivalent to\n\\[\n\\begin{aligned}\n\\frac{stagnant^{\\prime}}{stagnant} & =\\frac{unvarying^{\\prime}}{unvarying^{\\prime}-unvarying}=\\frac{2\\,fixedpoint\\, e^{fixedpoint^{2}}}{(2\\,fixedpoint-1) e^{fixedpoint^{2}}}=1+\\frac{1}{2\\,fixedpoint-1}, \\\\\n\\ln |stagnant(fixedpoint)| & =fixedpoint+\\frac{1}{2} \\ln |2\\,fixedpoint-1|+c\n\\end{aligned}\n\\]\nfrom which one finds that the nonzero solutions are of the form \\( stagnant(fixedpoint)=variable\\,e^{fixedpoint}|2\\,fixedpoint-1|^{1 / 2} \\) for any nonzero number \\( variable \\), on any interval not containing \\( 1 / 2 \\).\n\nRelated question. What other pairs of functions \\( unvarying \\) and \\( stagnant \\) satisfy \\( (unvarying\\,stagnant)^{\\prime}=unvarying^{\\prime}\\,stagnant^{\\prime} \\) ? For some discussion, see [Hal, Problem 2G]."
    },
    "garbled_string": {
      "map": {
        "f": "qzxwvtnp",
        "g": "hjgrksla",
        "x": "plmoknij",
        "x_0": "wsxedcrf",
        "a": "vfrtgbhu",
        "b": "yhnujmik",
        "y_0": "ikolpmaj",
        "C": "ujmnhygt"
      },
      "question": "A not uncommon calculus mistake is to believe that the product rule for derivatives says that $(qzxwvtnphjgrksla)' = qzxwvtnp'hjgrksla'$. If $qzxwvtnp(plmoknij)=e^{plmoknij^2}$, determine, with proof, whether there exists an open interval $(vfrtgbhu,yhnujmik)$ and a nonzero function $hjgrksla$ defined on $(vfrtgbhu,yhnujmik)$ such that this wrong product rule is true for $plmoknij$ in $(vfrtgbhu,yhnujmik)$.",
      "solution": "Solution. We are asked for a solution \\( hjgrksla \\) to \\( qzxwvtnp hjgrksla^{\\prime}+hjgrksla qzxwvtnp^{\\prime}=qzxwvtnp^{\\prime} hjgrksla^{\\prime} \\), which is equivalent to \\( hjgrksla^{\\prime}+\\left(\\frac{qzxwvtnp^{\\prime}}{qzxwvtnp-qzxwvtnp^{\\prime}}\\right) hjgrksla=0 \\) if we avoid the zeros of \\( qzxwvtnp-qzxwvtnp^{\\prime}=(1-2 plmoknij) e^{plmoknij^{2}} \\), i.e., avoid \\( plmoknij=1 / 2 \\). By the existence and uniqueness theorem for first order linear ordinary differential equations [BD, Theorem 2.4.1], if \\( wsxedcrf \\neq 1 / 2 \\), and \\( ikolpmaj \\) is any real number, then there exists a unique solution \\( hjgrksla(plmoknij) \\), defined in some neighborhood \\( (vfrtgbhu, yhnujmik) \\) of \\( wsxedcrf \\), with \\( hjgrksla\\left(wsxedcrf\\right)=ikolpmaj \\). By taking \\( ikolpmaj \\) nonzero, we obtain a nonzero solution \\( hjgrksla \\).\n\nRemark. One can solve the differential equation by separation of variables, to find all possible \\( hjgrksla \\). If \\( hjgrksla \\) is nonzero, the differential equation is equivalent to\n\\[\n\\begin{aligned}\n\\frac{hjgrksla^{\\prime}}{hjgrksla} & =\\frac{qzxwvtnp^{\\prime}}{qzxwvtnp^{\\prime}-qzxwvtnp}=\\frac{2 plmoknij e^{plmoknij^{2}}}{(2 plmoknij-1) e^{plmoknij^{2}}}=1+\\frac{1}{2 plmoknij-1}, \\\\\n\\ln |hjgrksla(plmoknij)| & =plmoknij+\\frac{1}{2} \\ln |2 plmoknij-1|+c\n\\end{aligned}\n\\]\nfrom which one finds that the nonzero solutions are of the form \\( hjgrksla(plmoknij)=ujmnhygt e^{plmoknij}|2 plmoknij-1|^{1 / 2} \\) for any nonzero number \\( ujmnhygt \\), on any interval not containing \\( 1 / 2 \\).\n\nRelated question. What other pairs of functions \\( qzxwvtnp \\) and \\( hjgrksla \\) satisfy \\( (qzxwvtnp hjgrksla)^{\\prime}=qzxwvtnp^{\\prime} hjgrksla^{\\prime} \\) ? For some discussion, see [Hal, Problem 2G]."
    },
    "kernel_variant": {
      "question": "The following statement is a much more ambitious (and equally false) extension of the one–variable blunder  \n\\((fg)'=f'g'\\).\n\nFor a \\(C^{1}\\)-function \\(H:\\mathbb R^{2}\\to\\mathbb R\\) and a vector \\(v\\in\\mathbb R^{2}\\) write  \n\\[D_{v}H(x,y)=\\nabla H(x,y)\\!\\cdot\\! v\\]  \nfor the directional derivative in the direction \\(v\\).\n\nLet  \n\\[\nf(x,y)=e^{\\,x^{2}+3y^{2}} .\n\\]\n\nDecide, with proof, whether there exist  \n• an open set \\(U\\subset\\mathbb R^{2}\\) and  \n• a non–zero function \\(g\\in C^{1}(U)\\)  \n\nsuch that the following UNIVERSAL erroneous product rule holds simultaneously for every point \\((x,y)\\in U\\) **and for every vector \\(v\\in\\mathbb R^{2}\\)**:\n\\[\nD_{v}\\bigl(fg\\bigr)(x,y)=D_{v}f(x,y)\\,D_{v}g(x,y)\\qquad\\forall\\,v\\in\\mathbb R^{2}. \\tag{\\(*\\)}\n\\]\n\n(That is, the identity must be true not only in the \\(x\\)- and \\(y\\)-directions, but in \\emph{all} directions.)",
      "solution": "We show that no such non–zero \\(g\\) can exist.\n\n--------------------------------------------------------------------\n1.  Reformulating the condition\n--------------------------------------------------------------------\nBecause \\(D_{v}(fg)=f\\,D_{v}g+g\\,D_{v}f\\), the requirement (\\(*\\)) reads  \n\\[\nf\\,D_{v}g+g\\,D_{v}f=\\bigl(D_{v}f\\bigr)\\bigl(D_{v}g\\bigr)\\qquad\\forall v .\n\\]\nRe–arranging (and keeping in mind that \\(g\\not\\equiv 0\\)) gives\n\\[\n\\bigl(f-D_{v}f\\bigr)\\,D_{v}g=-g\\,D_{v}f\\qquad\\forall v . \\tag{1}\n\\]\n\nIntroduce the vector fields  \n\\[\na:=\\nabla f ,\\qquad b:=\\nabla(\\ln g)=\\frac{\\nabla g}{g}.\n\\]\nUsing \\(D_{v}g=g\\,D_{v}(\\ln g)=g\\,v\\!\\cdot\\! b\\) and \\(D_{v}f=v\\!\\cdot\\! a\\), equation (1) becomes\n\\[\n\\boxed{\\ (f-v\\!\\cdot\\! a)\\,v\\!\\cdot\\! b=-\\,v\\!\\cdot\\! a\\qquad\\forall v\\in\\mathbb R^{2}. \\ } \\tag{2}\n\\]\n\n--------------------------------------------------------------------\n2.  Consequences of holding for \\emph{all} directions\n--------------------------------------------------------------------\nFix a point \\((x,y)\\in U\\) and treat \\(a,b,f\\) as the corresponding constant vectors/numbers.\n\n(i)  Take \\(v\\) orthogonal to \\(a\\).  \nThen \\(v\\!\\cdot\\!a=0\\) and (2) collapses to  \n\\[\nf\\,(v\\!\\cdot\\! b)=0 .\n\\]\nSince \\(f=e^{x^{2}+3y^{2}}>0\\), we get \\(v\\!\\cdot\\! b=0\\) for every \\(v\\perp a\\); hence  \n\\[\nb \\ \\text{is parallel to}\\ a. \\tag{3}\n\\]\n\n(ii)  Therefore \\(b=\\lambda a\\) for some scalar function \\(\\lambda(x,y)\\).\n\n(iii)  Insert \\(b=\\lambda a\\) back into (2).  For an \\emph{arbitrary} vector \\(v\\),\n\\[\n(f-v\\!\\cdot\\!a)\\,\\lambda\\,(v\\!\\cdot\\!a)=-\\,v\\!\\cdot\\!a. \\tag{4}\n\\]\n\n--------------------------------------------------------------------\n3.  Determining the proportionality factor \\(\\lambda\\)\n--------------------------------------------------------------------\nChoose \\(v=a\\).  Then \\(v\\!\\cdot\\!a=\\|a\\|^{2}\\ne 0\\) (because \\(\\nabla f\\neq 0\\) except at the isolated point \\((0,0)\\), which we may omit without loss of generality).  Equation (4) gives\n\\[\n\\bigl(f-\\|a\\|^{2}\\bigr)\\,\\lambda\\,\\|a\\|^{2}=-\\,\\|a\\|^{2}\n\\Longrightarrow\n\\boxed{\\ \\lambda=-\\frac{1}{\\,f-\\|a\\|^{2}\\,}\\ }. \\tag{5}\n\\]\n\n--------------------------------------------------------------------\n4.  Testing the identity for an arbitrary direction\n--------------------------------------------------------------------\nInsert (5) into (4):\n\\[\n(f-v\\!\\cdot\\!a)\\,\\Bigl(-\\frac{1}{\\,f-\\|a\\|^{2}\\,}\\Bigr)(v\\!\\cdot\\!a)=-\\,v\\!\\cdot\\!a .\n\\]\nCancel the factor \\(-v\\!\\cdot\\!a\\) (which may be \\(0\\) only for directions already covered) and obtain\n\\[\n\\frac{f-v\\!\\cdot\\!a}{\\,f-\\|a\\|^{2}\\,}=1\\qquad\\forall v .\n\\]\nThus we would need\n\\[\nf-v\\!\\cdot\\!a=f-\\|a\\|^{2}\\qquad\\forall v ,\n\\]\ni.e.  \n\\[\nv\\!\\cdot\\!a=\\|a\\|^{2}\\qquad\\forall v\\in\\mathbb R^{2},\n\\]\nwhich is impossible unless \\(a=0\\).  But \\(a=\\nabla f\\) vanishes only at \\((0,0)\\), not on any open set.\n\n--------------------------------------------------------------------\n5.  Conclusion\n--------------------------------------------------------------------\nNo point in any open set \\(U\\) can satisfy (\\(*\\)); hence no non–zero \\(C^{1}\\)-function \\(g\\) fulfilling the universal directional rule exists.\n\nTherefore the answer is:\n\n\\[\n\\boxed{\\text{Such a function }g\\text{ does not exist}.}\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.702583",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher-dimensional setting:  \n   The problem is promoted from one–variable calculus to a full two-variable (vector–calculus) context, incorporating \\emph{all} directional derivatives, not just partial derivatives along coordinate axes.\n\n2.  Universal quantification over directions:  \n   The identity must hold simultaneously for every vector \\(v\\in\\mathbb R^{2}\\), creating an infinite family of constraints that interact in a non-trivial way.\n\n3.  Vector–analytic reasoning:  \n   Solving the resulting system requires interpreting the condition in terms of the gradient fields \\(a=\\nabla f\\) and \\(b=\\nabla(\\ln g)\\), recognizing orthogonality constraints, and exploiting the geometry of \\(\\mathbb R^{2}\\).\n\n4.  Contradiction via parameter elimination:  \n   The proof eliminates all possibilities by analyzing how a scalar proportionality factor \\(\\lambda\\) must behave, leading to an impossibility argument that uses the full variability of the direction vector \\(v\\).\n\n5.  Deeper theoretical tools:  \n   The solution implicitly uses ideas from differential geometry (directional derivatives as covectors) and linear algebra (orthogonality spaces, universality over vectors) rather than elementary ODE techniques alone.\n\nThese layers of abstraction and the need to juggle infinitely many simultaneous differential identities make the enhanced variant substantially harder than both the original problem and the earlier kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "The following statement is a much more ambitious (and equally false) extension of the one–variable blunder  \n\\((fg)'=f'g'\\).\n\nFor a \\(C^{1}\\)-function \\(H:\\mathbb R^{2}\\to\\mathbb R\\) and a vector \\(v\\in\\mathbb R^{2}\\) write  \n\\[D_{v}H(x,y)=\\nabla H(x,y)\\!\\cdot\\! v\\]  \nfor the directional derivative in the direction \\(v\\).\n\nLet  \n\\[\nf(x,y)=e^{\\,x^{2}+3y^{2}} .\n\\]\n\nDecide, with proof, whether there exist  \n• an open set \\(U\\subset\\mathbb R^{2}\\) and  \n• a non–zero function \\(g\\in C^{1}(U)\\)  \n\nsuch that the following UNIVERSAL erroneous product rule holds simultaneously for every point \\((x,y)\\in U\\) **and for every vector \\(v\\in\\mathbb R^{2}\\)**:\n\\[\nD_{v}\\bigl(fg\\bigr)(x,y)=D_{v}f(x,y)\\,D_{v}g(x,y)\\qquad\\forall\\,v\\in\\mathbb R^{2}. \\tag{\\(*\\)}\n\\]\n\n(That is, the identity must be true not only in the \\(x\\)- and \\(y\\)-directions, but in \\emph{all} directions.)",
      "solution": "We show that no such non–zero \\(g\\) can exist.\n\n--------------------------------------------------------------------\n1.  Reformulating the condition\n--------------------------------------------------------------------\nBecause \\(D_{v}(fg)=f\\,D_{v}g+g\\,D_{v}f\\), the requirement (\\(*\\)) reads  \n\\[\nf\\,D_{v}g+g\\,D_{v}f=\\bigl(D_{v}f\\bigr)\\bigl(D_{v}g\\bigr)\\qquad\\forall v .\n\\]\nRe–arranging (and keeping in mind that \\(g\\not\\equiv 0\\)) gives\n\\[\n\\bigl(f-D_{v}f\\bigr)\\,D_{v}g=-g\\,D_{v}f\\qquad\\forall v . \\tag{1}\n\\]\n\nIntroduce the vector fields  \n\\[\na:=\\nabla f ,\\qquad b:=\\nabla(\\ln g)=\\frac{\\nabla g}{g}.\n\\]\nUsing \\(D_{v}g=g\\,D_{v}(\\ln g)=g\\,v\\!\\cdot\\! b\\) and \\(D_{v}f=v\\!\\cdot\\! a\\), equation (1) becomes\n\\[\n\\boxed{\\ (f-v\\!\\cdot\\! a)\\,v\\!\\cdot\\! b=-\\,v\\!\\cdot\\! a\\qquad\\forall v\\in\\mathbb R^{2}. \\ } \\tag{2}\n\\]\n\n--------------------------------------------------------------------\n2.  Consequences of holding for \\emph{all} directions\n--------------------------------------------------------------------\nFix a point \\((x,y)\\in U\\) and treat \\(a,b,f\\) as the corresponding constant vectors/numbers.\n\n(i)  Take \\(v\\) orthogonal to \\(a\\).  \nThen \\(v\\!\\cdot\\!a=0\\) and (2) collapses to  \n\\[\nf\\,(v\\!\\cdot\\! b)=0 .\n\\]\nSince \\(f=e^{x^{2}+3y^{2}}>0\\), we get \\(v\\!\\cdot\\! b=0\\) for every \\(v\\perp a\\); hence  \n\\[\nb \\ \\text{is parallel to}\\ a. \\tag{3}\n\\]\n\n(ii)  Therefore \\(b=\\lambda a\\) for some scalar function \\(\\lambda(x,y)\\).\n\n(iii)  Insert \\(b=\\lambda a\\) back into (2).  For an \\emph{arbitrary} vector \\(v\\),\n\\[\n(f-v\\!\\cdot\\!a)\\,\\lambda\\,(v\\!\\cdot\\!a)=-\\,v\\!\\cdot\\!a. \\tag{4}\n\\]\n\n--------------------------------------------------------------------\n3.  Determining the proportionality factor \\(\\lambda\\)\n--------------------------------------------------------------------\nChoose \\(v=a\\).  Then \\(v\\!\\cdot\\!a=\\|a\\|^{2}\\ne 0\\) (because \\(\\nabla f\\neq 0\\) except at the isolated point \\((0,0)\\), which we may omit without loss of generality).  Equation (4) gives\n\\[\n\\bigl(f-\\|a\\|^{2}\\bigr)\\,\\lambda\\,\\|a\\|^{2}=-\\,\\|a\\|^{2}\n\\Longrightarrow\n\\boxed{\\ \\lambda=-\\frac{1}{\\,f-\\|a\\|^{2}\\,}\\ }. \\tag{5}\n\\]\n\n--------------------------------------------------------------------\n4.  Testing the identity for an arbitrary direction\n--------------------------------------------------------------------\nInsert (5) into (4):\n\\[\n(f-v\\!\\cdot\\!a)\\,\\Bigl(-\\frac{1}{\\,f-\\|a\\|^{2}\\,}\\Bigr)(v\\!\\cdot\\!a)=-\\,v\\!\\cdot\\!a .\n\\]\nCancel the factor \\(-v\\!\\cdot\\!a\\) (which may be \\(0\\) only for directions already covered) and obtain\n\\[\n\\frac{f-v\\!\\cdot\\!a}{\\,f-\\|a\\|^{2}\\,}=1\\qquad\\forall v .\n\\]\nThus we would need\n\\[\nf-v\\!\\cdot\\!a=f-\\|a\\|^{2}\\qquad\\forall v ,\n\\]\ni.e.  \n\\[\nv\\!\\cdot\\!a=\\|a\\|^{2}\\qquad\\forall v\\in\\mathbb R^{2},\n\\]\nwhich is impossible unless \\(a=0\\).  But \\(a=\\nabla f\\) vanishes only at \\((0,0)\\), not on any open set.\n\n--------------------------------------------------------------------\n5.  Conclusion\n--------------------------------------------------------------------\nNo point in any open set \\(U\\) can satisfy (\\(*\\)); hence no non–zero \\(C^{1}\\)-function \\(g\\) fulfilling the universal directional rule exists.\n\nTherefore the answer is:\n\n\\[\n\\boxed{\\text{Such a function }g\\text{ does not exist}.}\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.548589",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher-dimensional setting:  \n   The problem is promoted from one–variable calculus to a full two-variable (vector–calculus) context, incorporating \\emph{all} directional derivatives, not just partial derivatives along coordinate axes.\n\n2.  Universal quantification over directions:  \n   The identity must hold simultaneously for every vector \\(v\\in\\mathbb R^{2}\\), creating an infinite family of constraints that interact in a non-trivial way.\n\n3.  Vector–analytic reasoning:  \n   Solving the resulting system requires interpreting the condition in terms of the gradient fields \\(a=\\nabla f\\) and \\(b=\\nabla(\\ln g)\\), recognizing orthogonality constraints, and exploiting the geometry of \\(\\mathbb R^{2}\\).\n\n4.  Contradiction via parameter elimination:  \n   The proof eliminates all possibilities by analyzing how a scalar proportionality factor \\(\\lambda\\) must behave, leading to an impossibility argument that uses the full variability of the direction vector \\(v\\).\n\n5.  Deeper theoretical tools:  \n   The solution implicitly uses ideas from differential geometry (directional derivatives as covectors) and linear algebra (orthogonality spaces, universality over vectors) rather than elementary ODE techniques alone.\n\nThese layers of abstraction and the need to juggle infinitely many simultaneous differential identities make the enhanced variant substantially harder than both the original problem and the earlier kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}