// Numbas version: exam_results_page_options {"name": "Praneetha's copy of Nick's copy of 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": [], "variables": {"n": {"description": "", "name": "n", "definition": "random(2..6)", "group": "Ungrouped variables", "templateType": "anything"}, "a": {"description": "", "name": "a", "definition": "random(1..4)", "group": "Ungrouped variables", "templateType": "anything"}, "m": {"description": "", "name": "m", "definition": "random(2..8)", "group": "Ungrouped variables", "templateType": "anything"}, "s1": {"description": "", "name": "s1", "definition": "random(1,-1)", "group": "Ungrouped variables", "templateType": "anything"}, "b": {"description": "", "name": "b", "definition": "s1*random(1..5)", "group": "Ungrouped variables", "templateType": "anything"}}, "functions": {}, "advice": "

\n

$f(x)$ 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.

\n

\n

Differentiating the first part, keeping the second half the same, gives the term: $\\simplify{{m} *{ b} * ({a} + {b} * x) ^ {m -1}} \\times \\simplify{e ^ ({n} * x)}$.  

\n

Note that that we needed the chain rule to do this differentiation.

\n

\n

\n

Differentiating the second part, keeping the first half the same, gives the term: $\\simplify{{n} * e ^ ({n} * x)} \\times \\simplify{({a} + {b}x)^{m}}$.

\n

Again, we needed the chain rule to do this differentiation.

\n

\n

Hence, $\\simplify{Diff(f,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) + {n} * ({a} + {b} * x) ^ {m} * e ^ ({n} * x)}$.

\n

           $= \\simplify{({a} + {b} * x) ^ {m -1} * ({m * b + n * a} + {n * b} * x) * e ^ ({n} * x)}$, (by doing some factorising)

\n

\n

Hence, $\\simplify{g(x) = {m * b + n * a} + {n * b} * x}$.

", "parts": [{"type": "gapfill", "showCorrectAnswer": true, "customMarkingAlgorithm": "", "showFeedbackIcon": true, "variableReplacements": [], "marks": 0, "scripts": {}, "unitTests": [], "extendBaseMarkingAlgorithm": true, "gaps": [{"type": "jme", "showCorrectAnswer": true, "customMarkingAlgorithm": "", "checkVariableNames": false, "checkingAccuracy": 0.001, "variableReplacements": [], "failureRate": 1, "showFeedbackIcon": true, "checkingType": "absdiff", "marks": "4", "scripts": {}, "unitTests": [], "vsetRange": [0, 1], "answer": "({((m * b) + (n * a))} + ({(n * b)} * x))", "vsetRangePoints": 5, "extendBaseMarkingAlgorithm": true, "showPreview": true, "variableReplacementStrategy": "originalfirst", "answerSimplification": "all", "expectedVariableNames": []}], "variableReplacementStrategy": "originalfirst", "prompt": "

$\\simplify{f(x) = ({a} + {b} * x) ^ {m} * e ^ ({n} * x)}$

\n

You are told that $\\simplify{Diff(f,x,1) = ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) * g(x)}$, for a polynomial $g(x)$.

\n

\n

You have to find $g(x)$.

\n

$g(x)=\\;$[[0]]

"}], "metadata": {"description": "

Differentiate the function $f(x)=(a + b x)^m  e ^ {n x}$ using the product and chain rule. Find $g(x)$ such that $f^{\\prime}(x)= (a + b x)^{m-1}  e ^ {n x}g(x)$. Non-calculator. Advice is given.

", "licence": "Creative Commons Attribution 4.0 International"}, "tags": [], "rulesets": {}, "preamble": {"css": "", "js": ""}, "extensions": [], "ungrouped_variables": ["a", "s1", "b", "m", "n"], "name": "Praneetha's copy of Nick's copy of Differentiation: product and chain rule, (a+bx)^m e^(nx), factorise answer", "statement": "

Differentiate the following function $f(x)$.

", "variablesTest": {"condition": "", "maxRuns": 100}, "type": "question", "contributors": [{"name": "Nick Walker", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2416/"}, {"name": "Vicki Fracarossi", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2477/"}, {"name": "Praneetha Singh", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2552/"}]}]}], "contributors": [{"name": "Nick Walker", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2416/"}, {"name": "Vicki Fracarossi", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2477/"}, {"name": "Praneetha Singh", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2552/"}]}