// Numbas version: exam_results_page_options {"name": "Implicit differentiation 1 (Basic)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variables": {"a": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "Random(-6..6 except 0)", "name": "a"}, "c": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "Random(-3..3)", "name": "c"}, "constant": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "{a}*{c}-{b}*{d}", "name": "constant"}, "d": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "if(c=0, random(-3..3),0)", "name": "d"}, "b": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "Random(-6..6 except 0 except a except -a)", "name": "b"}}, "preamble": {"css": "", "js": ""}, "functions": {}, "extensions": [], "rulesets": {}, "parts": [{"useCustomName": false, "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "sortAnswers": false, "scripts": {}, "variableReplacements": [], "gaps": [{"checkVariableNames": false, "useCustomName": false, "showPreview": true, "showCorrectAnswer": false, "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "checkingType": "absdiff", "answer": "-({a}+2x*y^2)/(2x^2*y-{b})", "scripts": {}, "checkingAccuracy": 0.001, "variableReplacements": [], "valuegenerators": [{"value": "", "name": "x"}, {"value": "", "name": "y"}], "vsetRange": [0, 1], "showFeedbackIcon": false, "unitTests": [], "failureRate": 1, "marks": 1, "type": "jme", "vsetRangePoints": 5, "customName": "", "customMarkingAlgorithm": "malrules:\n [\n [\"3/y+y+x\", \"This is an implicit differentiation question. Therefore you need to consider $y$ as having something to do with $x$. Therefore, whenever you differentiate part of this expression involving $y$ you need the chain rule and so need to include $\\\\frac{dy}{dx}$. For example, when using the product rule to differentiate $xy$ (which you spotted - well done!), you get $x \\\\cdot \\\\frac{dy}{dx} + y \\\\cdot 1 = x \\\\frac{dy}{dx}+y$. Similarly, when differentiating $\\\\ln \\\\left( y^3 \\\\right)$, recall that $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/(1+3/y)\", \"There are two main errors here. Firstly, since this is implicit differentiation, you are thinking of $y$ as having something to do with $x$. This means you need the product rule to differentiate $xy$, since $x \\\\cdot y$ is really $x \\\\times$ (something to do with $x$). Secondly, what is the derivative of the right hand side? Don't forget to differentiate $\\\\textbf{both}$ sides.\"],\n [\"-y/(x+1/y^3)\", \"Be careful when differentiating $\\\\ln \\\\left( y^3 \\\\right)$. Remember, $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/y^3+y+x\", \"This is an implicit differentiation question. Therefore you need to consider $y$ as having something to do with $x$. Therefore, whenever you differentiate part of this expression involving $y$ you need the chain rule and so need to include $\\\\frac{dy}{dx}$. For example, when using the product rule to differentiate $xy$ (which you spotted - well done!), you get $x \\\\cdot \\\\frac{dy}{dx} + y \\\\cdot 1 = x \\\\frac{dy}{dx}+y$. Similarly, when differentiating $\\\\ln \\\\left( y^3 \\\\right)$, recall that $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/(1/y^3+1)\", \"There are three things to watch here. Firstly, since this is implicit differentiation, you are thinking of $y$ as having something to do with $x$. This means you need the product rule to differentiate $xy$, since $x \\\\cdot y$ is really $x \\\\times$ (something to do with $x$). Secondly, be careful when differentiating $\\\\ln \\\\left( y^3 \\\\right)$. Remember, $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$. Finally, what is the derivative of the right hand side? Don't forget to differentiate $\\\\textbf{both}$ sides.\"]\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 )))Hint:  Type $2x^2y$  as  2x^2*y

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$\\frac{dy}{dx}=$ [[0]]

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$\\frac{dy}{dx} \\bigg|_{(\\var{c},\\var{d})}=$[[0]]

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(Answer in fraction form if necessary)

", "marks": 0, "type": "gapfill", "customName": "", "customMarkingAlgorithm": ""}], "name": "Implicit differentiation 1 (Basic)", "variable_groups": [], "ungrouped_variables": ["a", "b", "c", "d", "constant"], "metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": "

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

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$\\frac{d}{dx}(\\var{a}x+x^2y^2)=\\frac{d}{dx}(\\simplify{{constant}+{b}y})$

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$\\var{a}+2x.y^2+x^2.2y.\\frac{dy}{dx}=\\var{b}.\\frac{dy}{dx}$

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$(\\simplify{2x^2y-{b}}) \\frac{dy}{dx}=\\simplify{-{a}-2x*y^2}$

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$\\frac{dy}{dx}=\\frac{\\simplify{-{a}-2x*y^2}}{\\simplify{2x^2*y-{b}}} =\\simplify{-(2x*y^2+{a})/(2x^2*y-{b})}$

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$\\$

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$\\frac{dy}{dx} \\bigg|_{(\\var{c},\\var{d})}=\\simplify{{a}/{b}}$

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", "statement": "

Find the gradient of the curve   $\\simplify{{a}x}+x^2y^2=\\simplify{{constant}+{b}y}$   at the point   $(\\var{c},\\var{d})$.

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