// Numbas version: finer_feedback_settings
{"name": "Quotient rule - differentiate exponential over exponential", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [], "variables": {"s1": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(1,-1)", "name": "s1", "description": ""}, "a": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(2..9)", "name": "a", "description": ""}, "c": {"group": "Ungrouped variables", "templateType": "anything", "definition": "s2*random(1..9)", "name": "c", "description": ""}, "b1": {"group": "Ungrouped variables", "templateType": "anything", "definition": "s1*random(2..8)", "name": "b1", "description": ""}, "b": {"group": "Ungrouped variables", "templateType": "anything", "definition": "if(a^2=b1^2,b1+1,b1)", "name": "b", "description": ""}, "s2": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(1,-1)", "name": "s2", "description": ""}}, "ungrouped_variables": ["a", "c", "b", "s2", "s1", "b1"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "name": "Quotient rule - differentiate exponential over exponential", "showQuestionGroupNames": false, "functions": {}, "parts": [{"stepsPenalty": 0, "scripts": {}, "gaps": [{"answer": "{c*(b^2-a^2)}", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "checkvariablenames": false, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "showCorrectAnswer": true, "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n\t\t\t<p>\\[\\simplify[std]{f(x) = ({a}+{b}e^({c}x))/({b}+{a}e^({c}x))}\\]</p>\n\t\t\t      <p>You are given that \\[\\simplify[std]{Diff(f,x,1) = (a*e^({c}x)) / ({b}+{a}e^({c}x))^2}\\]</p>\n\t\t\t      <p>for a number $a$. You have to find $a$.</p>\n\t\t\t      <p>$a=\\;$[[0]]</p>\n\t\t\t      <p>You can click on Show steps to get help. You will not lose any marks if you do so.</p>\n\t\t\t", "steps": [{"type": "information", "prompt": "<p>The quotient rule says that if $u$ and $v$ are functions of $x$ then<br/>\\[\\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\\]</p>", "showCorrectAnswer": true, "marks": 0, "scripts": {}}], "showCorrectAnswer": true, "marks": 0}], "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "<p>Differentiate the following function $f(x)$ using the quotient rule.</p>", "tags": ["algebraic manipulation", "Calculus", "checked2015", "derivative of a quotient", "differentiation", "MAS1601", "quotient rule", "Steps"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n\t\t<p><strong>1/08/2012:</strong></p>\n\t\t<p>Added tags.</p>\n\t\t<p>Added description.</p>\n\t\t<p>Checked calculation. OK.</p>\n\t\t<p>Added information about Show steps. Altered to 0 marks lost rather than 1.</p>\n\t\t<p>Changed std rule set to include !noLeadingMinus, so expressions don't change order from that intended. Got rid of a redundant ruleset.</p>\n\t\t<p>&nbsp;</p>\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "<p>The derivative of $\\displaystyle \\frac{a+be^{cx}}{b+ae^{cx}}$ is $\\displaystyle \\frac{pe^{cx}} {(b+ae^{cx})^2}$. Find $p$.</p>"}, "advice": "\n\t    \n\t    \n\t    <p>The quotient rule says that if $u$ and $v$ are functions of $x$ then<br/>\\[\\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\\]</p>\n\t    \n\t    \n\t    \n\t    <p>For this example:</p>\n\t    \n\t    \n\t    \n\t    <p>\\[\\simplify[std]{u = {a}+{b}e^({c}x)}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {b*c}e^({c}x)}\\]</p>\n\t    \n\t    \n\t    \n\t    <p>\\[\\simplify[std]{v = {b}+{a}e^({c}x)} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {a*c}e^({c}x)}\\]</p>\n\t    \n\t    \n\t    \n\t    <p>Hence on substituting into the quotient rule above we get:</p>\n\t    \n\t    \n\t    \n\t    <p>\\[\\begin{eqnarray*} \\frac{df}{dx}&#38;=&#38;\\simplify[std]{({b*c}e^({c}x)({b}+{a}e^({c}x))-{a*c}e^({c}x)({a}+{b}e^({c}x)))/({b}+{a}e^({c}x))^2}\\\\\n\t    \n\t    &#38;=&#38;\\simplify[std]{({b^2*c} e^({c}x)+{a*b*c}*e^({2*c}x)-{a^2*c}e^({c}x)-{a*b*c}*e^({2*c}x) )/({b}+{a}e^({c}x))^2}\\\\\n\t    \n\t    &#38;=&#38;\\simplify[std]{({b^2*c} e^({c}x)-{a^2*c}e^({c}x))/({b}+{a}e^({c}x))^2}\\\\\n\t    \n\t    &#38;=&#38;\\simplify[std]{({b^2*c-a^2*c} e^({c}x))/({b}+{a}e^({c}x))^2}\t\n\t    \n\t    \\end{eqnarray*}\\]</p>\n\t    \n\t    \n\t    \n\t    <p>Hence $a=\\var{c*(b^2-a^2)}$</p>\n\t    \n\t    \n\t", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}