// Numbas version: exam_results_page_options {"name": "T6Q5 (custom feedback)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"tags": [], "metadata": {"description": "
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 )))$\\frac{\\partial^2 z}{\\partial y \\partial x} =$ [[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 )))