// Numbas version: finer_feedback_settings
{"name": "cormac's copy of Chain rule - exponential of polynomial, ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [], "name": "cormac's copy of Chain rule - exponential of polynomial, ", "metadata": {"description": "<p>Differentiate $\\displaystyle e^{ax^{m} +bx^2+c}$</p>", "licence": "Creative Commons Attribution 4.0 International"}, "parts": [{"showCorrectAnswer": true, "showFeedbackIcon": true, "prompt": "<p>\\[\\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}\\]</p>\n<p>$\\displaystyle \\frac{df}{dx}=\\;$[[0]]</p>", "type": "gapfill", "gaps": [{"showCorrectAnswer": true, "answer": "({m*a}x^{m-1}+{2*b}x)*e^({a}x^{m} +{b}x^2+{c})", "showpreview": true, "marks": 3, "scripts": {}, "vsetrange": [0, 1], "variableReplacementStrategy": "originalfirst", "expectedvariablenames": [], "showFeedbackIcon": false, "type": "jme", "checkingtype": "absdiff", "answersimplification": "std", "checkingaccuracy": 0.001, "vsetrangepoints": 5, "checkvariablenames": false, "variableReplacements": []}], "marks": 0, "scripts": {}, "variableReplacementStrategy": "originalfirst", "variableReplacements": []}], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "statement": "<p>Differentiate the following function $f(x)$ using the chain rule.</p>", "functions": {}, "advice": "\n\t    \n\t    \n\t    <p>$\\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}$<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\t    \n\t    \n\t    \n\t    <p>For this example, we let $u=\\simplify[std]{{a}x^{m} +{b}x^2+{c}}$ and we have $f(u)=\\simplify[std]{e^u}$.<br/>This gives<br/>\\[\\begin{eqnarray*}\\frac{du}{dx} &#38;=&#38; \\simplify[std]{{a*m}x^{m-1} +{2*b}x}\\\\\n\t    \n\t    \\frac{df(u)}{du} &#38;=&#38; \\simplify[std]{e^u} \\end{eqnarray*}\\]</p>\n\t    \n\t    \n\t    \n\t    <p>Hence on substituting into the chain rule above we get:</p>\n\t    \n\t    \n\t    \n\t    <p>\\[\\begin{eqnarray*}\\frac{df}{dx} &#38;=&#38; \\simplify[std]{({a*m}x^{m-1} +{2*b}x) * (e^u)}\\\\\n\t    \n\t    &#38;=&#38; \\simplify[std]{({a*m}x^{m-1} +{2*b}x)*e^({a}x^{m} +{b}x^2+{c})}\n\t    \n\t    \\end{eqnarray*}\\]<br/>on replacing $u$ by $\\simplify[std]{{a}x^{m} +{b}x^2+{c}}$.</p>\n\t    \n\t    \n\t", "variablesTest": {"maxRuns": 100, "condition": ""}, "tags": [], "ungrouped_variables": ["a", "c", "b", "s2", "s1", "m"], "extensions": [], "preamble": {"css": "", "js": ""}, "variables": {"s2": {"description": "", "definition": "random(1,-1)", "group": "Ungrouped variables", "templateType": "anything", "name": "s2"}, "c": {"description": "", "definition": "s2*random(1..9)", "group": "Ungrouped variables", "templateType": "anything", "name": "c"}, "a": {"description": "", "definition": "random(2..9)", "group": "Ungrouped variables", "templateType": "anything", "name": "a"}, "s1": {"description": "", "definition": "random(1,-1)", "group": "Ungrouped variables", "templateType": "anything", "name": "s1"}, "m": {"description": "", "definition": "random(3..4)", "group": "Ungrouped variables", "templateType": "anything", "name": "m"}, "b": {"description": "", "definition": "s1*random(1..9)", "group": "Ungrouped variables", "templateType": "anything", "name": "b"}}, "type": "question", "contributors": [{"name": "cormac breen", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/306/"}]}]}], "contributors": [{"name": "cormac breen", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/306/"}]}