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Differentiation questions (mostly partial differentiation) with customised feedback, with the intention of catching common errors.

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

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Let $z=3xy+x^2-2y^3$. Find:

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

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"3y+3x+2x-6y^2\", \"The product rule is not needed to differentiate $3xy$. Remember, when partially differentiating with respect to $x$, treat $y$ as a number. This also goes for the last term: $2y^3$ is just treated as a number.\"],\n [\"(-3y-2x)/(3x-6y^2)\", \"It looks like you have used implicit differentiation rather than partial differentiation. Implicit differentiation is only used if you wish to find $\\\\frac{dy}{dx}$, not if you wish to find $\\\\frac{\\\\partial y}{\\\\partial x}$ or $\\\\frac{\\\\partial z}{\\\\partial x}$ etc.\"]\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 )))$\\displaystyle \\frac{\\partial z}{\\partial y} = $ [[0]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"3y+3x+2x-6y^2\", \"The product rule is not needed to differentiate $3xy$. Remember, when partially differentiating with respect to $y$, treat $x$ as a number. This also goes for the second term: $x^2$ is just treated as a number.\"],\n [\"(-3y-2x)/(3x-6y^2)\", \"It looks like you have used implicit differentiation rather than partial differentiation. Implicit differentiation is only used if you wish to find $\\\\frac{dy}{dx}$, not if you wish to find $\\\\frac{\\\\partial y}{\\\\partial x}$ or $\\\\frac{\\\\partial z}{\\\\partial y}$ etc.\"]\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 )))Now consider $40=3xy+x^2-2y^3$. Find $\\displaystyle \\frac{dy}{dx}$.

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

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "(3y+2x)/(6y^2-3x)", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}, {"name": "y", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "T6Q2 (custom feedback)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}], "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": "

$f_x(x,y)=$ [[0]]

", "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"vsetRange": [0, 1], "extendBaseMarkingAlgorithm": true, "scripts": {}, "checkVariableNames": false, "customMarkingAlgorithm": "malrules:\n [\n [\"(x^2+2*x)*e^x\", \"Be careful when differentiating the exponential function. The power on the exponential function never changes when you differentiate.\"],\n [\"2x*e^(x+y)\", \"Don't forget to use the product rule to differentiate the first term!\"], \n [\"2x*e^(x+y)+x^2*e^(x+y)+sin(y)\", \"Remember, when partially differentiating with respect to $x$ you treat $y$ as a number. Therefore anything to do with $y$ - such as $\\\\cos y$ - is treated as a number.\"], \n [\"x^2*e^x+2x*e^(x+y)\", \"Be careful when differentiating the exponential function. The power on the exponential function never changes when you differentiate.\"],\n [\"x^2*e^(x+y)*(dy)/(dx)+2x*e^(x+y)+sin(y)*(dy)/(dx)\", \"Remember with partial differentiation, $x$ and $y$ are independent of each other. In particular, $y$ does not depend on $x$, so when differentiating with respect to $x$, $y$ is treated like a number.\"],\n [\"(-x^2*e^(x+y)-2x*e^(x+y))/sin(y)\", \"It looks like you have tried to use implicit differentiation rather than partial differentiation. Implicit differentiation is only used if you wish to find $\\\\frac{dy}{dx}$, not if you wish to find $\\\\frac{\\\\partial y}{\\\\partial x}$ or $f_{x}$ etc.\"],\n [\"-2x*e^(x+y)/(x^2*e^(x+y)+sin(y))\", \"It looks like you have tried to use implicit differentiation rather than partial differentiation. Implicit differentiation is only used if you wish to find $\\\\frac{dy}{dx}$, not if you wish to find $\\\\frac{\\\\partial y}{\\\\partial x}$ or $f_{x}$ etc.\"]\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 )))$f_y(x,y)=$ [[0]]

", "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"vsetRange": [0, 1], "extendBaseMarkingAlgorithm": true, "scripts": {}, "checkVariableNames": false, "customMarkingAlgorithm": "malrules:\n [\n [\"x^2*e^(x+y)+2x*e^(x+y)+sin(y)\", \"There is no need for the product rule when differentiating $x^2 e^{x+y}$ since $x$ is treated like a number when partially differentiating with respect to $y$.\"],\n [\"x^2*e^y+sin(y)\", \"Be careful when differentiating the exponential function. The power on the exponential function never changes when you differentiate.\"],\n [\"(-x^2*e^(x+y)-2x*e^(x+y))/cos(y)\", \"It looks like you have tried to use implicit differentiation rather than partial differentiation. Implicit differentiation is only used if you wish to find $\\\\frac{dy}{dx}$, not if you wish to find $\\\\frac{\\\\partial y}{\\\\partial x}$ or $\\\\frac{\\\\partial z}{\\\\partial y}$ or $f_{y}$ etc.\"],\n [\"-2x*e^(x+y)/(x^2*e^(x+y)+cos(y))\", \"It looks like you have tried to use implicit differentiation rather than partial differentiation. Implicit differentiation is only used if you wish to find $\\\\frac{dy}{dx}$, not if you wish to find $\\\\frac{\\\\partial y}{\\\\partial x}$ or $\\\\frac{\\\\partial z}{\\\\partial y}$ or $f_{y}$ etc.\"]\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 )))$f_{xy}(x,y)=$ [[0]]

", "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"vsetRange": [0, 1], "extendBaseMarkingAlgorithm": true, "scripts": {}, "checkVariableNames": false, "customMarkingAlgorithm": "malrules:\n [\n [\"x^2*e^(x+y)+4x*e^(x+y)+2e^(x+y)\", \"There is no need for the product rule when differentiating $x^2 e^{x+y}$ since $x$ is treated like a number when partially differentiating with respect to $y$. For the same reason, the product rule is not needed when differentiating $2xe^{x+y}$.\"],\n [\"x^2*e^y+2x*e^y\", \"Be careful when differentiating the exponential function. The power on the exponential function never changes when you differentiate.\"]\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 )))$f_{yy}(x,y)=$ [[0]]

", "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"vsetRange": [0, 1], "extendBaseMarkingAlgorithm": true, "scripts": {}, "checkVariableNames": false, "customMarkingAlgorithm": "malrules:\n [\n [\"x^2*e^(x+y)+2x*e^(x+y)+cos(y)\", \"There is no need for the product rule when differentiating $x^2 e^{x+y}$ since $x$ is treated like a number when partially differentiating with respect to $y$.\"],\n [\"x^2*e^y+cos(y)\", \"Be careful when differentiating the exponential function. The power on the exponential function never changes when you differentiate.\"]\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 )))If $f(x,y)=x^2e^{x+y}-\\cos y$, find the partial derivatives:

", "preamble": {"css": "", "js": ""}, "rulesets": {}, "ungrouped_variables": [], "advice": "", "variables": {}, "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "type": "question"}, {"name": "T6Q3 (custom feedback)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}], "tags": [], "metadata": {"description": "

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

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

If $x=t^5+7$ and $y=2t^3-4$, find $\\displaystyle \\frac{dy}{dx}$.

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

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"5t^2/6\", \"Double check you have used the right formula for $\\\\frac{dy}{dx}$. In particular what gets divided by what?\"],\n [\"(2t^3-4)/(t^5+7)\", \"You have not differentiated. How is $\\\\frac{dy}{dx}$ calculated in parametric differentiation?\"]\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 )))Logarithmic differentiation question with customised feedback to catch some common errors.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Find the gradient of the tangent to the curve $y=x^{x+4x^2}$ at the point $(1,1)$.

First find $\\displaystyle \\frac{dy}{dx}$:

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

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"(x+4x^2)*x^(x+4x^2-1)\", \"You cannot use the power rule if there is a variable in the power. Whenever you have $x$'s to the power of $x$'s, what type of differentiation must you use?\"],\n [\"1+4x+(1+8x)*ln(x)\", \"Almost there. Did you remember to multiply both sides by $y$? Remember, when you take the natural log of both sides and then differentiate both sides, the derivative of the left hand side is $\\\\frac{d}{dx} \\\\left( \\\\ln y \\\\right) = \\\\frac{1}{y} \\\\frac{dy}{dx}$, not just $\\\\frac{dy}{dx}$.\"]\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 )))Now evaluate the derivative at the point $(1,1)$:

$\\displaystyle \\frac{dy}{dx} \\bigg|_{(1,1)} =$ [[0]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "5", "maxValue": "5", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "T6Q5 (custom feedback)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}], "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 )))$\\frac{\\partial z}{\\partial y}=$ [[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": "

$\\frac{\\partial^2 z}{\\partial y \\partial x} =$ [[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": {}, "variables": {}, "preamble": {"css": "", "js": ""}, "advice": "", "ungrouped_variables": [], "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "type": "question"}, {"name": "T6Q6 (custom feedback)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}], "tags": [], "metadata": {"description": "

Product rule differentiation question with customised feedback to catch some common errors.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Find $\\displaystyle \\frac{dy}{dx}$ if $y=\\sin x \\cos (x^2+4)$.

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

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"(cos(x^3+4x))^2-2x*(sin(x^3+4x))^2\", \"$\\\\sin A \\\\sin B \\\\neq \\\\sin^2 (AB).$ To see this, try entering $\\\\sin \\\\left(\\\\frac{\\\\pi}{2} \\\\right) \\\\times \\\\sin \\\\left( 2 \\\\right)$ into your calculator and compare to the answer you get if you enter $\\\\left( \\\\sin \\\\left( \\\\frac{\\\\pi}{2} \\\\times 2 \\\\right) \\\\right)^2$ into your calculator. Similarly $\\\\cos A \\\\cos B \\\\neq \\\\cos^2 (AB)$.\"],\n [\"cos(x^3+4x)-2x*sin(x^3+4x)\", \"$\\\\sin A \\\\sin B \\\\neq \\\\sin (AB).$ To see this, try entering $\\\\sin \\\\left(\\\\frac{\\\\pi}{2} \\\\right) \\\\times \\\\sin \\\\left( 2 \\\\right)$ into your calculator and compare to the answer you get if you enter $\\\\sin \\\\left( \\\\frac{\\\\pi}{2} \\\\times 2 \\\\right)$ into your calculator. Similarly $\\\\cos A \\\\cos B \\\\neq \\\\cos (AB)$.\"],\n [\"-2x*cos(x)*sin(x^2+4)\", \"Don't forget the product rule! Remember, you need the product rule whenever you have one function of $x$ multiplied by another function of $x$.\"],\n [\"cos(x)*cos(x^2+4)-sin(x)*sin(x^2+4)\", \"On the right track. Don't forget the derivative of $\\\\cos$ (angle) is $-\\\\sin$ (the same angle) $\\\\times$ (derivative of the angle).\"],\n [\"cos(x)*cos(x^2+4)-sin(x)*sin(2x)\", \"Be careful when differentiating $\\\\cos(x^2+4)$. Remember the angle never changes when you differentiate a trigonometric function. Instead, the derivative of $\\\\cos$ (angle) is $-\\\\sin$ (the same angle) $\\\\times$ (derivative of the angle).\"]\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 )))Logarithmic differentiation question with customised feedback to catch some common errors.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

If $y=x^{\\sqrt{x}}$, find $\\displaystyle \\frac{dy}{dx}$.

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

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"sqrt(x)*x^(sqrt(x)-1)\", \"You cannot use the power rule when you have a variable in the power. What method of differentiation do you need when you have variables to the power of variables?\"],\n [\"1/sqrt(x)+ln(x)/(2*sqrt(x))\", \"Did you forget to multiply both sides by $y$? Remember, when you take the natural log of both sides and then differentiate both sides, the derivative of the left hand side is $\\\\frac{d}{dx} \\\\left( \\\\ln y \\\\right) = \\\\frac{1}{y} \\\\frac{dy}{dx}$, not just $\\\\frac{dy}{dx}$.\"],\n [\"y*(1/(2x*sqrt(x)))\", \"Take a closer look at what you were trying to differentiate on the right hand side - you should have $\\\\sqrt{x} \\\\ln x$ i.e. one function of $x$ multiplied by another function of $x$. You therefore need the product rule for this.\"],\n [\"x^(sqrt(x))*(1/(2x*sqrt(x)))\", \"Take a closer look at what you were trying to differentiate on the right hand side - you should have $\\\\sqrt{x} \\\\ln x$ i.e. one function of $x$ multiplied by another function of $x$. You therefore need the product rule for this.\"]\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 )))Implicit differentiation question with customised feedback to catch some common errors.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Find the gradient of the tangent to the curve $\\ln(y^3)+xy=1$ at the point $(1,1)$.

First find $\\displaystyle \\frac{dy}{dx}$:

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

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "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 )))Now, evaluate the derivative at the point $(1,1)$:

$\\displaystyle\\frac{dy}{dx} \\bigg|_{(1,1)}=$ [[0]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "-1/4", "maxValue": "-1/4", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "T6Q9 (custom feedback)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}], "parts": [{"variableReplacementStrategy": "originalfirst", "prompt": "

$\\frac{\\partial K}{\\partial r} =$ [[0]]

", "showCorrectAnswer": true, "sortAnswers": false, "type": "gapfill", "unitTests": [], "customMarkingAlgorithm": "", "scripts": {}, "variableReplacements": [], "showFeedbackIcon": true, "marks": 0, "gaps": [{"variableReplacementStrategy": "originalfirst", "failureRate": 1, "checkVariableNames": false, "showPreview": true, "vsetRange": [0, 1], "type": "jme", "showCorrectAnswer": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "unitTests": [], "customMarkingAlgorithm": "malrules:\n [\n [\"15(r+s)^2*(r-s)^4\", \"There are two things to watch here. Firstly, don't forget the product rule! Since this is one function of $s$ multiplied by another function of $s$, you need the product rule. Secondly, when differentiating the $(r-s)^5$ term, don't forget to multiply by $\\\\frac{\\\\partial}{\\\\partial s} \\\\left( r-s \\\\right)$.\"],\n [\"-15(r+s)^2*(r-s)^4\", \"Don't forget the product rule! Since this is one function of $s$ multiplied by another function of $s$, you need the product rule.\"],\n [\"3(r+s)^2*(r-s)^5+5(r+s)^3*(r-s)^4\", \"Almost there. When differentiating the $(r-s)^5$ term, don't forget to multiply by $\\\\frac{\\\\partial}{\\\\partial s} \\\\left( r-s \\\\right)$.\"]\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 )))Partial differentiation question with customised feedback to catch some common errors.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "ungrouped_variables": [], "variablesTest": {"condition": "", "maxRuns": 100}, "functions": {}, "variables": {}, "statement": "

Let $K=(r+s)^3(r-s)^5$. Find $\\frac{\\partial K}{\\partial r}$ and $\\frac{\\partial K}{\\partial s}$.

", "type": "question"}, {"name": "T6Q10 (custom feedback)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}], "tags": [], "metadata": {"description": "

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

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

The following parametric equations describe a curve: $x(t)=4 \\sin t$ and $y(t)=-\\cos t-1$. Find the slope of the tangent to the curve when $\\displaystyle t=\\frac{\\pi}{3}$.

First, find $\\displaystyle \\frac{dy}{dx}$:

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

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"cos(t)/(4sin(t))\", \"Have you put $\\\\frac{dx}{dt}$ and $\\\\frac{dy}{dt}$ together in the correct way?\"],\n [\"(sin(t)-1)/(4cos(t))\", \"Double check $\\\\frac{dy}{dt}$.\"],\n [\"-sin(t)/(4cos(t))\", \"Double check $\\\\frac{dy}{dt}$. What is $\\\\frac{d}{dt} \\\\left( \\\\cos t \\\\right)$? $\\\\Rightarrow $ $\\\\frac{d}{dt} \\\\left( - \\\\cos t \\\\right) = ?$\"],\n [\"sin(t)\", \"It looks like you have just found $\\\\frac{dy}{dt}$.\"],\n [\"(-cos(t)-1)/(4sin(t))\", \"You have not differentiated. How is $\\\\frac{dy}{dx}$ calculated for parametric differentiation?\"]\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 )))Next, evaluate the derivative when $\\displaystyle t = \\frac{\\pi}{3}$:

$\\displaystyle \\frac{dy}{dx} \\bigg|_{t=\\frac{\\pi}{3}}=$ [[0]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"0.005\", \"Is your calculator set to radians or degrees? Remember for this module, you should always be in radians.\"],\n [\"0.0046\", \"Is your calculator set to radians or degrees? Remember for this module, you should always be in radians.\"],\n [\"0.00457\", \"Is your calculator set to radians or degrees? Remember for this module, you should always be in radians.\"],\n [\"0.004570\", \"Is your calculator set to radians or degrees? Remember for this module, you should always be in radians.\"],\n [\"0.0045698\", \"Is your calculator set to radians or degrees? Remember for this module, you should always be in radians.\"],\n [\"0.00456977\", \"Is your calculator set to radians or degrees? Remember for this module, you should always be in radians.\"]\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 )))Differentiation question with customised feedback to catch some common errors.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

If $y=e^{x^2}$, find $\\displaystyle \\frac{1}{2} \\frac{d^2 y}{dx^2} + \\frac{dy}{dx}$.

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

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"x^2*e^(x^2-1)\", \"You cannot use the power rule when there is a variable in the power. What other rule could you use here? Look at page 25 of the log tables if you need to.\"],\n [\"e^(x^2)\", \"Don't forget to multiply by the derivative of the power. Remember, $\\\\frac{d}{dx} \\\\left( e^x \\\\right) = e^x$ but $\\\\frac{d}{dx} \\\\left( e^{f(x)} \\\\right) = e^{f(x)} \\\\cdot f'(x)$.\"],\n [\"2x*e^(x^2-1)\", \"Remember, the power on the exponential function never changes when you differentiate.\"],\n [\"e^(2x)\", \"Remember, the power on the exponential function never changes when you differentiate.\"]\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 )))$\\displaystyle \\frac{d^2 y}{dx^2}=$ [[0]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"4x*e^(x^2)\", \"Don't forget to use the product rule! The first derivative is one function of $x$ multiplied by another function of $x$ and so, to find the second derivative, you need the product rule.\"],\n [\"2e^(x^2)+2x*e^(2x)\", \"Remember: the power on the exponential function never changes when you differentiate.\"],\n [\"2e^(x^2)+2x*e^(x^2)\", \"Almost there! When differentiating $e^{x^2}$, don't forget to multiply by the derivative of the power. Remember, $\\\\frac{d}{dx} \\\\left( e^x \\\\right) = e^x$ but $\\\\frac{d}{dx} \\\\left( e^{f(x)} \\\\right) = e^{f(x)} \\\\cdot f'(x)$.\"],\n [\"2e^(2x)\", \"There are two things to note here. Firstly, don't forget to use the product rule! The first derivative is one function of $x$ multiplied by another function of $x$ and so, to find the second derivative, you need the product rule. Also, remember: the power on the exponential function never changes when you differentiate.\"],\n [\"2e^(x^2)\", \"Don't forget to use the product rule! The first derivative is one function of $x$ multiplied by another function of $x$ and so, to find the second derivative, you need the product rule.\"]\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 )))$\\displaystyle \\frac{1}{2} \\frac{d^2 y}{dx^2} + \\frac{dy}{dx}=$ [[0]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "e^(x^2)*(2x^2+2x+1)", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "T6Q12 (custom feedback)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}, {"name": "Catherine Palmer", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/423/"}], "tags": [], "metadata": {"description": "

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

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

If $z=\\sin^{-1} \\frac{x}{y}$, find $\\frac{\\partial z}{\\partial x}$ and $\\frac{\\partial z}{\\partial y}$.

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

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"-1/sqrt(y^2-x^2)\", \"Check the rule for differentiating $\\\\sin^{-1} \\\\left( \\\\cdot \\\\right)$.\"],\n [\"1/sqrt(y^2+x^2)\", \"Check the rule for differentiating $\\\\sin^{-1} \\\\left( \\\\cdot \\\\right)$.\"],\n [\"y/(y^2+x^2)\", \"Check the rule for differentiating $\\\\sin^{-1} \\\\left( \\\\cdot \\\\right)$.\"]\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]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"1/sqrt(1-(x/y)^2)\", \"Don't forget to multiply by the derivative of the top. Recall that $\\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\sin^{-1} \\\\left( \\\\frac{y}{a} \\\\right) \\\\right) = \\\\frac{1}{\\\\sqrt{a^2-y^2}}$ but $\\\\frac{\\\\partial}{\\\\partial y} \\\\left( \\\\sin^{-1} \\\\left( \\\\frac{f(y)}{a} \\\\right) \\\\right) = \\\\frac{1}{\\\\sqrt{a^2-(f(y))^2}} \\\\cdot \\\\frac{\\\\partial}{\\\\partial y} \\\\left( f(y) \\\\right)$. i.e. $\\\\frac{1}{\\\\sqrt{(\\\\text{bottom})^2-(\\\\text{top})^2}} \\\\times \\\\left( \\\\text{derivative of the top} \\\\right)$\"],\n [\"1/sqrt(y^2-x^2)\", \"There are two things to note here. Firstly, you are looking for the partial derivative with respect to $y$ here. Remember that, in the rule given on page 25 of the log tables for differentiating $\\\\sin^{-1} \\\\left( \\\\frac{x}{a} \\\\right)$, the $x$ is something to do with the variable, while $a$ is a number. Can you rewrite $\\\\sin^{-1} \\\\left( \\\\frac{x}{y} \\\\right)$ so that it looks like $\\\\sin^{-1} \\\\left( \\\\frac{\\\\text{something to do with } y}{\\\\text{a number}} \\\\right)$? Secondly, once you have written $\\\\sin^{-1} \\\\left( \\\\frac{x}{y} \\\\right)$ in this way, don't forget to multiply by the derivative of the top, i.e. the derivative will be $\\\\frac{1}{\\\\sqrt{(\\\\text{bottom})^2-(\\\\text{top})^2}} \\\\times \\\\left( \\\\text{derivative of the top} \\\\right)$.\"],\n [\"-1/sqrt(y^2-x^2)*x/y^2\", \"You are looking for the partial derivative with respect to $y$ here. Remember that, in the rule given on page 25 of the log tables for differentiating $\\\\sin^{-1} \\\\left( \\\\frac{x}{a} \\\\right)$, the $x$ is something to do with the variable, while $a$ is a number. Can you rewrite $\\\\sin^{-1} \\\\left( \\\\frac{x}{y} \\\\right)$ so that it looks like $\\\\sin^{-1} \\\\left( \\\\frac{\\\\text{something to do with } y}{\\\\text{a number}} \\\\right)$?\"]\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 )))