// Numbas version: finer_feedback_settings {"name": "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": [{"statement": "

Differentiate the following function $f(x)$.

", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "ungrouped_variables": ["a", "s1", "b", "m", "n"], "rulesets": {}, "parts": [{"showFeedbackIcon": true, "type": "gapfill", "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]]

", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "customMarkingAlgorithm": "", "unitTests": [], "gaps": [{"checkingAccuracy": 0.001, "type": "jme", "variableReplacements": [], "showPreview": true, "vsetRangePoints": 5, "answer": "({((m * b) + (n * a))} + ({(n * b)} * x))", "showCorrectAnswer": true, "customMarkingAlgorithm": "", "unitTests": [], "checkingType": "absdiff", "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "vsetRange": [0, 1], "checkVariableNames": false, "scripts": {}, "expectedVariableNames": [], "showFeedbackIcon": true, "marks": "4", "answerSimplification": "all", "failureRate": 1}], "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst"}], "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"}, "name": "Nick's copy of Differentiation: product and chain rule, (a+bx)^m e^(nx), factorise answer", "preamble": {"css": "", "js": ""}, "functions": {}, "tags": [], "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, $\\simplify{g(x) = {m * b + n * a} + {n * b} * x}$.

", "variables": {"s1": {"description": "", "name": "s1", "group": "Ungrouped variables", "templateType": "anything", "definition": "random(1,-1)"}, "a": {"description": "", "name": "a", "group": "Ungrouped variables", "templateType": "anything", "definition": "random(1..4)"}, "b": {"description": "", "name": "b", "group": "Ungrouped variables", "templateType": "anything", "definition": "s1*random(1..5)"}, "m": {"description": "", "name": "m", "group": "Ungrouped variables", "templateType": "anything", "definition": "random(2..8)"}, "n": {"description": "", "name": "n", "group": "Ungrouped variables", "templateType": "anything", "definition": "random(2..6)"}}, "extensions": [], "type": "question", "contributors": [{"name": "Nick Walker", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2416/"}]}]}], "contributors": [{"name": "Nick Walker", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2416/"}]}