// Numbas version: finer_feedback_settings
{"name": "Adrian's copy of Chain rule - power of a quadratic, ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "advice": "\n    \n    \n    <p>$\\simplify[std]{f(x) = f(x) = ({a} * x^{m}+{c}x^2+{b})^{n}}$<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 $u=\\simplify[std]{{a} * x^{m}+{c}*x^2+{b}}$ 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}+{2*c}x}\\\\\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}+{2*c}x) * ({n}*u^{n-1})}\\\\\n    \n    &#38;=&#38;\\simplify[std]{{n}*({m*a}x^{m-1}+{2*c}*x)u^{n-1}}\\\\\n    \n    &#38;=&#38; \\simplify[std]{{n}*({m*a}x^{m-1}+{2*c}*x)({a}*x^{m}+{c}*x^2+{b})^{n-1}}\n    \n    \\end{eqnarray*}\\]<br/>on replacing $u$ by $\\simplify[std]{{a}x^{m}+{c}x^2+{b}}$.</p>\n    \n    ", "extensions": [], "ungrouped_variables": ["a", "c", "b", "s2", "s1", "m", "n"], "functions": {}, "parts": [{"type": "gapfill", "gaps": [{"showpreview": true, "showCorrectAnswer": true, "vsetrange": [0, 1], "showFeedbackIcon": true, "marks": "2", "vsetrangepoints": 5, "answer": "{n}({a*m}x ^ {m-1}+{2*c}x) * ({a} * x^{m}+{c}x^2+{b})^{n-1}", "type": "jme", "checkingaccuracy": 0.001, "checkingtype": "absdiff", "checkvariablenames": false, "answersimplification": "std", "scripts": {}, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "expectedvariablenames": []}], "showCorrectAnswer": true, "variableReplacements": [], "marks": 0, "steps": [{"type": "information", "showCorrectAnswer": true, "variableReplacements": [], "marks": 0, "scripts": {}, "variableReplacementStrategy": "originalfirst", "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            ", "showFeedbackIcon": true}], "scripts": {}, "variableReplacementStrategy": "originalfirst", "prompt": "<p>\\[\\simplify[std]{f(x) = ({a} * x^{m}+{c}x^2+{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, "showFeedbackIcon": true}], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "<p>Differentiate $\\displaystyle (ax^m+bx^2+c)^{n}$.</p>"}, "preamble": {"js": "", "css": ""}, "variables": {"c": {"name": "c", "definition": "s2*random(1..9)", "group": "Ungrouped variables", "description": "", "templateType": "anything"}, "m": {"name": "m", "definition": "random(3..7)", "group": "Ungrouped variables", "description": "", "templateType": "anything"}, "b": {"name": "b", "definition": "s1*random(1..9)", "group": "Ungrouped variables", "description": "", "templateType": "anything"}, "n": {"name": "n", "definition": "random(3..5)", "group": "Ungrouped variables", "description": "", "templateType": "anything"}, "s1": {"name": "s1", "definition": "random(1,-1)", "group": "Ungrouped variables", "description": "", "templateType": "anything"}, "a": {"name": "a", "definition": "random(2..9)", "group": "Ungrouped variables", "description": "", "templateType": "anything"}, "s2": {"name": "s2", "definition": "random(1,-1)", "group": "Ungrouped variables", "description": "", "templateType": "anything"}}, "name": "Adrian's copy of Chain rule - power of a quadratic, ", "statement": "<p>Differentiate the following function $f(x)$ using the chain rule.&nbsp;</p>", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "tags": [], "type": "question", "contributors": [{"name": "Adrian Jannetta", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/164/"}]}]}], "contributors": [{"name": "Adrian Jannetta", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/164/"}]}