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Partial differentiation question with customised feedback to catch some common errors.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "parts": [{"scripts": {}, "showFeedbackIcon": true, "customMarkingAlgorithm": "", "unitTests": [], "prompt": "

$\\frac{\\partial z}{\\partial x} =$[[0]]

", "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"vsetRangePoints": 5, "type": "jme", "scripts": {}, "showFeedbackIcon": true, "customMarkingAlgorithm": "malrules:\n [\n [\"-1/x\", \"Don't forget the product rule!\"],\n [\"ln(y/x)+x^2\", \"Be careful when multiplying by $\\\\frac{\\\\partial}{\\\\partial x} \\\\left( \\\\frac{y}{x} \\\\right) = \\\\frac{\\\\partial}{\\\\partial x} (yx^{-1}) = ?$\"],\n [\"ln(y/x)+y^2\", \"The derivative of $\\\\ln \\\\left( \\\\frac{y}{x} \\\\right)$ is $\\\\frac{1}{\\\\left(\\\\frac{y}{x} \\\\right)} \\\\neq \\\\frac{y}{x}$. Also, be careful when multiplying by $\\\\frac{d}{dx} \\\\left( \\\\frac{y}{x} \\\\right) = \\\\frac{d}{dx} (yx^{-1}) = ?$\"],\n [\"ln(y/x)+x^2/y\", \"Don't forget to multiply by the derivative of $\\\\left( \\\\frac{y}{x} \\\\right)$ (chain rule). Remember $\\\\frac{\\\\partial}{\\\\partial x} \\\\ln x = \\\\frac{1}{x}$ while $\\\\frac{\\\\partial}{\\\\partial x} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right) = \\\\frac{1}{f(x)} \\\\times f'(x).$ Therefore, $\\\\frac{\\\\partial}{\\\\partial x} \\\\left( \\\\ln \\\\left(\\\\frac{y}{x} \\\\right) \\\\right) = \\\\frac{1}{\\\\frac{y}{x}} \\\\times \\\\frac{\\\\partial}{\\\\partial x} \\\\left( \\\\frac{y}{x} \\\\right) = \\\\frac{x}{y} \\\\times ?$\"],\n [\"ln(y/x)-y^2/x^2\", \"Look very closely at how you differentiated $\\\\ln \\\\left( \\\\frac{y}{x} \\\\right)$. Hint: $\\\\frac{1}{\\\\left(\\\\frac{y}{x} \\\\right)} \\\\neq \\\\frac{y}{x}$.\"]\n ]\n\n\nparsed_malrules: \n map(\n [\"expr\":parse(x[0]),\"feedback\":x[1]],\n x,\n malrules\n )\n\nagree_malrules (Do the student's answer and the expected answer agree on each of the sets of variable values?):\n map(\n len(filter(not x ,x,map(\n try(\n resultsEqual(unset(question_definitions,eval(studentexpr,vars)),unset(question_definitions,eval(malrule[\"expr\"],vars)),settings[\"checkingType\"],settings[\"checkingAccuracy\"]),\n message,\n false\n ),\n vars,\n vset\n )))$\\frac{\\partial z}{\\partial y}=$ [[0]]

", "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"vsetRangePoints": 5, "type": "jme", "scripts": {}, "showFeedbackIcon": true, "customMarkingAlgorithm": "malrules:\n [\n [\"y/x\", \"Look closely at how you have differentiated $\\\\ln \\\\left( \\\\frac{y}{x} \\\\right)$. Hint: $\\\\frac{1}{\\\\left(\\\\frac{y}{x} \\\\right)} \\\\neq \\\\frac{y}{x}$.\"],\n [\"y\", \"It looks like you have differentiated $\\\\ln \\\\left( \\\\frac{y}{x} \\\\right)$ incorrectly. Note that $\\\\frac{1}{\\\\left(\\\\frac{y}{x} \\\\right)} \\\\neq \\\\frac{y}{x}$. Also, don't forget to multiply by the derivative (with respect to $y$) of $\\\\left( \\\\frac{y}{x} \\\\right)$. Remember $\\\\frac{\\\\partial}{\\\\partial y} \\\\ln y = \\\\frac{1}{y}$ while $\\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\ln \\\\left( f(y) \\\\right) \\\\right) = \\\\frac{1}{f(y)} \\\\times f'(y).$ Therefore, $\\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\ln \\\\left(\\\\frac{y}{x} \\\\right) \\\\right) = \\\\frac{1}{\\\\frac{y}{x}} \\\\times \\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\frac{y}{x} \\\\right) = \\\\frac{x}{y} \\\\times ?$\"],\n [\"x^2/y\", \"Don't forget to multiply by the derivative (with respect to $y$) of $\\\\frac{y}{x}$ (chain rule). Remember $\\\\frac{\\\\partial}{\\\\partial y} \\\\ln y = \\\\frac{1}{y}$ while $\\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\ln \\\\left( f(y) \\\\right) \\\\right) = \\\\frac{1}{f(y)} \\\\times f'(y).$ Therefore, $\\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\ln \\\\left(\\\\frac{y}{x} \\\\right) \\\\right) = \\\\frac{1}{\\\\frac{y}{x}} \\\\times \\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\frac{y}{x} \\\\right) = \\\\frac{x}{y} \\\\times ?$\"]\n ]\n\n\nparsed_malrules: \n map(\n [\"expr\":parse(x[0]),\"feedback\":x[1]],\n x,\n malrules\n )\n\nagree_malrules (Do the student's answer and the expected answer agree on each of the sets of variable values?):\n map(\n len(filter(not x ,x,map(\n try(\n resultsEqual(unset(question_definitions,eval(studentexpr,vars)),unset(question_definitions,eval(malrule[\"expr\"],vars)),settings[\"checkingType\"],settings[\"checkingAccuracy\"]),\n message,\n false\n ),\n vars,\n vset\n )))$\\frac{\\partial^2 z}{\\partial x \\partial y} =$ [[0]]

", "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"vsetRangePoints": 5, "type": "jme", "scripts": {}, "showFeedbackIcon": true, "customMarkingAlgorithm": "", "unitTests": [], "vsetRange": [0, 1], "showCorrectAnswer": true, "checkVariableNames": false, "checkingType": "absdiff", "answer": "1/y", "extendBaseMarkingAlgorithm": true, "showPreview": true, "expectedVariableNames": [], "failureRate": 1, "marks": 1, "checkingAccuracy": 0.001, "variableReplacements": [], "variableReplacementStrategy": "originalfirst"}], "extendBaseMarkingAlgorithm": true, "marks": 0, "type": "gapfill", "variableReplacements": [], "variableReplacementStrategy": "originalfirst"}, {"scripts": {}, "showFeedbackIcon": true, "customMarkingAlgorithm": "", "unitTests": [], "prompt": "

\n

$\\frac{\\partial^2 z}{\\partial y \\partial x} =$ [[0]]

", "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"vsetRangePoints": 5, "type": "jme", "scripts": {}, "showFeedbackIcon": true, "customMarkingAlgorithm": "malrules:\n [\n [\"y/x^2\", \"Look very closely at how you differentiated $\\\\ln \\\\left( \\\\frac{y}{x} \\\\right)$. Hint: $\\\\frac{1}{\\\\left(\\\\frac{y}{x} \\\\right)} \\\\neq \\\\frac{y}{x}$.\"],\n [\"x/y\", \"Don't forget to multiply by the derivative (with respect to $y$) of $\\\\frac{y}{x}$ (chain rule). Remember $\\\\frac{\\\\partial}{\\\\partial y} \\\\ln y = \\\\frac{1}{y}$ while $\\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\ln \\\\left( f(y) \\\\right) \\\\right) = \\\\frac{1}{f(y)} \\\\times f'(y).$ Therefore, $\\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\ln \\\\left(\\\\frac{y}{x} \\\\right) \\\\right) = \\\\frac{1}{\\\\frac{y}{x}} \\\\times \\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\frac{y}{x} \\\\right) = \\\\frac{x}{y} \\\\times ?$\"],\n [\"y/x\", \"It looks like you have differentiated $\\\\ln \\\\left( \\\\frac{y}{x} \\\\right)$ incorrectly. Hint: $\\\\frac{1}{\\\\left(\\\\frac{y}{x} \\\\right)} \\\\neq \\\\frac{y}{x}$. Also,don't forget to multiply by the derivative (with respect to $y$) of $\\\\frac{y}{x}$ (chain rule). Remember $\\\\frac{\\\\partial}{\\\\partial y} \\\\ln y = \\\\frac{1}{y}$ while $\\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\ln \\\\left( f(y) \\\\right) \\\\right) = \\\\frac{1}{f(y)} \\\\times f'(y).$ Therefore, $\\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\ln \\\\left(\\\\frac{y}{x} \\\\right) \\\\right) = \\\\frac{1}{\\\\frac{y}{x}} \\\\times \\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\frac{y}{x} \\\\right) = \\\\frac{x}{y} \\\\times ?$\"]\n ]\n\n\nparsed_malrules: \n map(\n [\"expr\":parse(x[0]),\"feedback\":x[1]],\n x,\n malrules\n )\n\nagree_malrules (Do the student's answer and the expected answer agree on each of the sets of variable values?):\n map(\n len(filter(not x ,x,map(\n try(\n resultsEqual(unset(question_definitions,eval(studentexpr,vars)),unset(question_definitions,eval(malrule[\"expr\"],vars)),settings[\"checkingType\"],settings[\"checkingAccuracy\"]),\n message,\n false\n ),\n vars,\n vset\n )))Let $z=x \\ln \\frac{y}{x}$. Show that $\\frac{\\partial^2 z}{\\partial x \\partial y} = \\frac{\\partial^2 z}{\\partial y \\partial x}$.

", "rulesets": {}, "name": "T6Q5 (custom feedback)", "variables": {}, "preamble": {"css": "", "js": ""}, "advice": "", "ungrouped_variables": [], "extensions": [], "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "type": "question", "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}]}]}], "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}]}