// Numbas version: finer_feedback_settings
{"name": "Harry's copy of Chain rule - binomial, ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}], "rulesets": {"surdf": [{"pattern": "a/sqrt(b)", "result": "(sqrt(b)*a)/b"}], "std": ["all", "!collectNumbers", "fractionNumbers"]}, "parts": [{"gaps": [{"checkingaccuracy": 0.001, "marks": 3, "scripts": {}, "showCorrectAnswer": true, "vsetrangepoints": 5, "type": "jme", "answer": "{a*m*n}x ^ {m-1} * ({a} * x^{m}+{b})^{n-1}", "checkvariablenames": false, "checkingtype": "absdiff", "showpreview": true, "expectedvariablenames": [], "vsetrange": [0, 1], "answersimplification": "std"}], "prompt": "<p>\\[\\simplify[std]{f(x) = ({a} * x^{m}+{b})^{n}}\\]</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>", "stepsPenalty": 0, "marks": 0, "scripts": {}, "showCorrectAnswer": true, "type": "gapfill", "steps": [{"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            ", "showCorrectAnswer": true, "scripts": {}, "type": "information", "marks": 0}]}], "preamble": {"css": "", "js": ""}, "type": "question", "advice": "\n    \n    \n    <p>$\\simplify[std]{f(x) = ({a} * x^{m}+{b})^{n}}$<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 $u=\\simplify[std]{{a} * x^{m}+{b}}$ and we have $f(u)=\\simplify[std]{u^{n}}$.<br/>This gives<br/>\\[\\begin{eqnarray*}\\frac{du}{dx} &#38;=&#38; \\simplify[std]{{m*a}x ^ {m -1}}\\\\\n    \n    \\frac{df(u)}{du} &#38;=&#38; \\simplify[std]{{n}u^{n-1}} \\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]{{m*a}x ^ {m-1} * ({n}*u^{n-1})}\\\\\n    \n    &#38;=&#38;\\simplify[std]{{m*a*n}x^{m-1}u^{n-1}}\\\\\n    \n    &#38;=&#38; \\simplify[std]{{m*a*n}x^{m-1}({a}*x^{m}+{b})^{n-1}}\n    \n    \\end{eqnarray*}\\]<br/>on replacing $u$ by $\\simplify[std]{{a}x^{m}+{b}}$.</p>\n    \n    ", "tags": ["Calculus", "MAS1601", "SFY0004", "Steps", "chain rule", "checked2015", "derivative of a function of a function", "differentiation", "function of a function"], "name": "Harry's copy of Chain rule - binomial, ", "metadata": {"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>\n                                                                <p>&nbsp;</p>", "licence": "Creative Commons Attribution 4.0 International", "description": "<p>Differentiate $\\displaystyle (ax^m+b)^{n}$.</p>"}, "variables": {"n": {"definition": "random(5..9)", "name": "n", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "a": {"definition": "random(2..9)", "name": "a", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "b": {"definition": "s1*random(1..9)", "name": "b", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "m": {"definition": "random(2..9)", "name": "m", "group": "Ungrouped variables", "templateType": "anything", "description": ""}, "s1": {"definition": "random(1,-1)", "name": "s1", "group": "Ungrouped variables", "templateType": "anything", "description": ""}}, "functions": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "<p>Differentiate the following function $f(x)$ using the chain rule.</p>", "ungrouped_variables": ["a", "s1", "b", "m", "n"], "variable_groups": [], "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/"}]}