// Numbas version: finer_feedback_settings {"name": "Implicit differentiation 2 (Basic)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"preamble": {"js": "", "css": ""}, "parts": [{"useCustomName": false, "prompt": "

Hint: Type $2xy$ as 2x*y

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

", "extendBaseMarkingAlgorithm": true, "gaps": [{"vsetRangePoints": 5, "failureRate": 1, "useCustomName": false, "vsetRange": [0, 1], "extendBaseMarkingAlgorithm": true, "unitTests": [], "scripts": {}, "showFeedbackIcon": false, "answer": "-{b}y^2/({a}+{b}x*y)", "checkingType": "absdiff", "checkVariableNames": false, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": false, "checkingAccuracy": 0.001, "variableReplacements": [], "type": "jme", "showPreview": true, "customName": "", "valuegenerators": [{"value": "", "name": "x"}, {"value": "", "name": "y"}], "marks": 1, "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 )))$\\frac{dy}{dx} \\bigg|_{(\\var{c},1)}=$[[0]]

(Answer in fraction form if necessary)

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

"}, "name": "Implicit differentiation 2 (Basic)", "rulesets": {}, "statement": "

Find the gradient of the curve $\\ln(y^\\var{a})+\\var{b}xy=\\simplify{{c}*{b}}$ at the point $(\\var{c},1)$.

", "variable_groups": [], "advice": "

$\\frac{d}{dx}(\\ln(y^\\var{a})+\\var{b}xy)=\\frac{d}{dx}(\\simplify{{c}*{b}})$

$(\\frac{1}{y^\\var{a}}.\\simplify{{a}y^{a-1}}).\\frac{dy}{dx}+(\\var{b}.1.y+\\var{b}x.\\frac{dy}{dx})=0$

$(\\frac{\\var{a}}{y}+\\var{b}x)\\frac{dy}{dx}=-\\var{b}y$

$\\frac{dy}{dx}=\\frac{-\\var{b}y}{\\frac{\\var{a}}y+\\var{b}x}$

$\\frac{dy}{dx}=\\frac{-\\var{b}y}{\\frac{\\var{a}}y+\\var{b}x} \\times \\frac{y}{y}$

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

$\\ $

$\\frac{dy}{dx} \\bigg|_{(\\var{c},1)}=\\simplify{(-{b})/({a}+{b}*{c})}$

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