// Numbas version: exam_results_page_options
{"name": "Differentiation: product and chain rule, (a+bx)^m e^(nx), factorise answer", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [], "preamble": {"js": "", "css": ""}, "ungrouped_variables": ["a", "s1", "b", "m", "n"], "functions": {}, "extensions": [], "variablesTest": {"condition": "", "maxRuns": 100}, "metadata": {"description": "<p>Differentiate the function $f(x)=(a + b x)^m &nbsp;e ^ {n x}$&nbsp;using the product and chain rule. Find $g(x)$ such that $f^{\\prime}(x)= (a + b x)^{m-1} &nbsp;e ^ {n x}g(x)$. Non-calculator. Advice is given.</p>", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "<p>Differentiate the following function $f(x)$.</p>", "variables": {"b": {"templateType": "anything", "description": "", "definition": "s1*random(1..5)", "group": "Ungrouped variables", "name": "b"}, "m": {"templateType": "anything", "description": "", "definition": "random(2..8)", "group": "Ungrouped variables", "name": "m"}, "a": {"templateType": "anything", "description": "", "definition": "random(1..4)", "group": "Ungrouped variables", "name": "a"}, "s1": {"templateType": "anything", "description": "", "definition": "random(1,-1)", "group": "Ungrouped variables", "name": "s1"}, "n": {"templateType": "anything", "description": "", "definition": "random(2..6)", "group": "Ungrouped variables", "name": "n"}}, "tags": [], "parts": [{"customName": "", "customMarkingAlgorithm": "", "sortAnswers": false, "scripts": {}, "prompt": "<p>$\\simplify{f(x) = ({a} + {b} * x) ^ {m} * e ^ ({n} * x)}$</p>\n<p>You are told that $\\simplify{Diff(f,x,1) = ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) * g(x)}$, for a polynomial $g(x)$.</p>\n<p></p>\n<p>You have to find $g(x)$.</p>\n<p>$g(x)=\\;$[[0]]</p>", "extendBaseMarkingAlgorithm": true, "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "type": "gapfill", "showFeedbackIcon": true, "marks": 0, "unitTests": [], "useCustomName": false, "gaps": [{"customName": "", "customMarkingAlgorithm": "", "failureRate": 1, "checkingType": "absdiff", "showCorrectAnswer": true, "checkVariableNames": false, "vsetRange": [0, 1], "showFeedbackIcon": true, "vsetRangePoints": 5, "checkingAccuracy": 0.001, "scripts": {}, "answer": "({((m * b) + (n * a))} + ({(n * b)} * x))", "valuegenerators": [{"value": "", "name": "x"}], "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "type": "jme", "answerSimplification": "all", "marks": "4", "unitTests": [], "showPreview": true, "useCustomName": false, "variableReplacements": []}], "variableReplacements": []}], "rulesets": {}, "advice": "<p></p>\n<p>$f(x)$ is the product of the two functions $\\simplify{({a} + {b}*x)^{m}}$ and $\\simplify{e ^ ({n} * x)}$, so we need to use the product rule.</p>\n<p></p>\n<p>Differentiating the first part, keeping the second half the same, gives the term: $\\simplify{{m} *{ b} * ({a} + {b} * x) ^ {m -1}} \\times \\simplify{e ^ ({n} * x)}$. &nbsp;</p>\n<p>Note that that we needed&nbsp;the chain rule to do this differentiation.</p>\n<p></p>\n<p></p>\n<p>Differentiating the second part, keeping the first half the same, gives the term: $\\simplify{{n} * e ^ ({n} * x)} \\times \\simplify{({a} + {b}x)^{m}}$.</p>\n<p>Again, we needed&nbsp;the chain rule to do this differentiation.</p>\n<p></p>\n<p>Hence, $\\simplify{Diff(f,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) + {n} * ({a} + {b} * x) ^ {m} * e ^ ({n} * x)}$.</p>\n<p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;$= \\simplify{({a} + {b} * x) ^ {m -1} * ({m * b + n * a} + {n * b} * x) * e ^ ({n} * x)}$, (by doing some factorising)</p>\n<p></p>\n<p>Hence, $\\simplify{g(x) = {m * b + n * a} + {n * b} * x}$.</p>", "name": "Differentiation: product and chain rule, (a+bx)^m e^(nx), factorise answer", "contributors": [{"name": "Lovkush Agarwal", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1358/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}]}], "contributors": [{"name": "Lovkush Agarwal", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1358/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}