// Numbas version: exam_results_page_options {"name": "Sheet 2 Q4 - differentiation by rule with custom feedback", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Sheet 2 Q4 - differentiation by rule with custom feedback", "tags": [], "metadata": {"description": "Differentiation by rule question with feedback given for anticipated student errors.", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Let $x=2t^2+\\ln 2t$. 

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Find $\\displaystyle \\frac{dx}{dt}$.

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$\\displaystyle \\frac{dx}{dt} = $ [[0]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"4t+1/(2t)\", \"You're on the right track, but need to be careful when differentiating $\\\\ln 2t$. Remember that the derivative of $\\\\ln \\\\textbf{$t$}$ is $\\\\frac{1}{t}$. However, you need to differentiate $\\\\ln \\\\color{red}{\\\\textbf{$2$}}t$. What is the extra step you need to include in order to differentiate this correctly?\"],\n [\"4t+2/t\", \"Look back at how you differentiated $\\\\ln 2t$. Think of the $2t$ as being a single unit. If $\\\\frac{d}{dx} (\\\\ln x) = \\\\frac{1}{x}$, $\\\\frac{d}{dt} (\\\\ln t) = \\\\frac{1}{t}$ and $\\\\frac{d}{dA} (\\\\ln A) = \\\\frac{1}{A}$, what should you get for $\\\\frac{d}{dt} (\\\\ln 2t)$?\"]\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 )))Show that $\\displaystyle \\frac{d^2 x}{dt^2} + \\left( \\frac{dx}{dt} \\right)^2=16t^2+12$.

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$\\displaystyle \\frac{d^2 x}{dt^2}=$ [[0]]

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$\\displaystyle \\left( \\frac{dx}{dt} \\right)^2 = $ [[1]]

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$\\displaystyle \\frac{d^2 x}{dt^2} + \\left( \\frac{dx}{dt} \\right)^2= $ [[2]]

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