// Numbas version: finer_feedback_settings {"name": "Helge's copy of DIfferentiation: product rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"type": "question", "rulesets": {"dpoly": ["std", "fractionNumbers"], "std": ["all", "!collectNumbers"]}, "parts": [{"type": "gapfill", "gaps": [{"checkingtype": "absdiff", "type": "jme", "answersimplification": "dPoly", "expectedvariablenames": [], "answer": "({((m * b) + (n * a))} + ({(n * b)} * x))", "vsetrange": [0, 1], "showpreview": true, "showCorrectAnswer": true, "vsetrangepoints": 5, "checkvariablenames": false, "scripts": {}, "checkingaccuracy": 0.001, "marks": 3}], "showCorrectAnswer": true, "steps": [{"checkingtype": "absdiff", "type": "jme", "expectedvariablenames": [], "answer": "", "vsetrange": [0, 1], "showpreview": true, "showCorrectAnswer": true, "vsetrangepoints": 5, "checkvariablenames": false, "scripts": {}, "checkingaccuracy": 0.001, "prompt": "

The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]

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

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You are given that \\[\\simplify[dPoly]{Diff(f,x,1) = ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) * g(x)}\\]

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for a polynomial $g(x)$. You have to find $g(x)$.

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

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Click on Show steps to get more information, you will not lose any marks by doing so.

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Differentiate the following function $f(x)$ using the product rule.

", "metadata": {"description": "

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

", "licence": "Creative Commons Attribution 4.0 International", "notes": "\n \t\t

20/06/2012:

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Added tags.

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4/7/2012

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Added tags.

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31/07/2012:

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Added a tag.

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Allowed free look at Steps.

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The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]

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For this example:

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\\[\\simplify[dPoly]{u = ({a} + {b} * x) ^ {m}}\\Rightarrow \\simplify[dPoly]{Diff(u,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1}}\\]

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\\[\\simplify{v = e ^ ({n} * x)} \\Rightarrow \\simplify{Diff(v,x,1) = {n} * e ^ ({n} * x)}\\]

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Hence on substituting into the product rule above we get:

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\\[\\simplify[dPoly]{Diff(f,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) + {n} * ({a} + {b} * x) ^ {m} * e ^ ({n} * x) = ({a} + {b} * x) ^ {m -1} * ({m * b + n * a} + {n * b} * x) * e ^ ({n} * x)}\\]

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The last step was to take out the common term $\\simplify[dPoly]{({a} + {b} * x) ^ {m -1} * e ^ ({n} * x)}$.

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Hence \\[\\simplify[dPoly]{g(x) = {m * b + n * a} + {n * b} * x}\\].

\n \n \n ", "tags": ["algebraic manipulation", "Calculus", "calculus", "checked2015", "derivative", "derivative ", "deriving a function", "differentiate", "differentiating a function", "differentiating a product of functions", "differentiation", "exponential function", "functions", "mas1601", "MAS1601", "product rule", "Steps", "steps"], "contributors": [{"name": "Helge M\u00fcnnich", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1885/"}]}]}], "contributors": [{"name": "Helge M\u00fcnnich", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1885/"}]}