// Numbas version: exam_results_page_options
{"name": "Harry's copy of Chain rule - trig and quadratic, ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Harry's copy of Chain rule - trig and quadratic, ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "ungrouped_variables": ["a", "s2", "c", "b", "s1"], "type": "question", "variables": {"c": {"name": "c", "group": "Ungrouped variables", "templateType": "anything", "definition": "s2*random(1..9)", "description": ""}, "a": {"name": "a", "group": "Ungrouped variables", "templateType": "anything", "definition": "random(2..9)", "description": ""}, "s2": {"name": "s2", "group": "Ungrouped variables", "templateType": "anything", "definition": "random(1,-1)", "description": ""}, "s1": {"name": "s1", "group": "Ungrouped variables", "templateType": "anything", "definition": "random(1,-1)", "description": ""}, "b": {"name": "b", "group": "Ungrouped variables", "templateType": "anything", "definition": "s1*random(1..9)", "description": ""}}, "functions": {}, "variablesTest": {"maxRuns": 100, "condition": ""}, "statement": "<p>Differentiate the following function $f(x)$ using the chain rule.</p>", "variable_groups": [], "question_groups": [{"name": "", "pickQuestions": 0, "pickingStrategy": "all-ordered", "questions": []}], "parts": [{"type": "gapfill", "stepsPenalty": 0, "scripts": {}, "prompt": "<p>\\[\\simplify[std]{f(x) = cos(e^({a}x) +{b}x^2+{c})}\\]</p>\n        <p>$\\displaystyle \\frac{df}{dx}=\\;$[[0]]</p>\n        <p>Click on Show steps for more information. You will not lose any marks by doing so.</p>", "showCorrectAnswer": true, "steps": [{"type": "information", "showCorrectAnswer": true, "marks": 0, "scripts": {}, "prompt": "\n            \n            \n            <p>The chain rule says that if $f(x)=g(h(x))$ then<br/>\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]<br/>One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.<br/>Then we use the chain rule in the form:<br/>\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df}{du}\\]<br/>Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.</p>\n            \n            "}], "gaps": [{"type": "jme", "checkingaccuracy": 0.001, "expectedvariablenames": [], "checkingtype": "absdiff", "vsetrangepoints": 5, "showCorrectAnswer": true, "vsetrange": [0, 1], "answersimplification": "std", "scripts": {}, "marks": 3, "answer": "-({a}e^({a}x)+{2*b}x)*sin(e^({a}x) +{b}x^2+{c})", "checkvariablenames": false, "showpreview": true}], "marks": 0}], "tags": ["Calculus", "MAS1601", "SFY0004", "Steps", "chain rule", "checked2015", "derivative of a function of a function", "differentiation", "function of a function"], "showQuestionGroupNames": false, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "notes": "<p><strong>1/08/2012:</strong></p>\n        <p>Added tags.</p>\n        <p>Added description.</p>\n        <p>Checked calculation. OK.</p>\n        <p>Added information about Show steps. Altered to 0 marks lost rather than 1.</p>\n        <p>Got rid of a redundant ruleset.</p>\n        <p>Improved display in prompt.</p>", "description": "<p>Differentiate $\\displaystyle \\cos(e^{ax}+bx^2+c)$</p>"}, "advice": "\n    \n    \n    <p>$\\simplify[std]{f(x) = cos(e^({a}x) +{b}x^2+{c})}$<br/>The chain rule says that if $f(x)=g(h(x))$ then<br/>\\[\\simplify[std]{f&#39;(x) = h&#39;(x)g&#39;(h(x))}\\]<br/>One way to find $f&#39;(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.<br/>Then we use the chain rule in the form:<br/>\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df(u)}{du}\\]<br/>Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.</p>\n    \n    \n    \n    <p>For this example, we let $u=\\simplify[std]{e^({a}x) +{b}x^2+{c}}$ and we have $f(u)=\\simplify[std]{cos(u)}$.<br/>This gives<br/>\\[\\begin{eqnarray*}\\frac{du}{dx} &#38;=&#38; \\simplify[std]{{a}e^({a}x) +{2*b}x}\\\\\n    \n    \\frac{df(u)}{du} &#38;=&#38; \\simplify[std]{-sin(u)} \\end{eqnarray*}\\]</p>\n    \n    \n    \n    <p>Hence on substituting into the chain rule above we get:</p>\n    \n    \n    \n    <p>\\[\\begin{eqnarray*}\\frac{df}{dx} &#38;=&#38; \\simplify[std]{({a}e^({a}x) +{2*b}x) * (-sin(u))}\\\\\n    \n    &#38;=&#38; \\simplify[std]{-({a}e^({a}x) +{2*b}x)*sin(e^({a}x) +{b}x^2+{c})}\n    \n    \\end{eqnarray*}\\]<br/>on replacing $u$ by $\\simplify[std]{e^({a}x) +{b}x^2+{c}}$.</p>\n    \n    ", "preamble": {"js": "", "css": ""}, "contributors": [{"name": "Harry Flynn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/976/"}]}]}], "contributors": [{"name": "Harry Flynn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/976/"}]}