// Numbas version: finer_feedback_settings {"name": "Luis's copy of Quotient rule - differentiate exponential over exponential", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variablesTest": {"condition": "", "maxRuns": 100}, "parts": [{"gaps": [{"answer": "{c*(b^2-a^2)}", "vsetrange": [0, 1], "checkingtype": "absdiff", "scripts": {}, "type": "jme", "checkingaccuracy": 0.001, "showpreview": true, "showCorrectAnswer": true, "marks": 3, "checkvariablenames": false, "expectedvariablenames": [], "vsetrangepoints": 5}], "stepsPenalty": 0, "showCorrectAnswer": true, "marks": 0, "scripts": {}, "type": "gapfill", "prompt": "\n\t\t\t
\\[\\simplify[std]{f(x) = ({a}+{b}e^({c}x))/({b}+{a}e^({c}x))}\\]
\n\t\t\tYou are given that \\[\\simplify[std]{Diff(f,x,1) = (a*e^({c}x)) / ({b}+{a}e^({c}x))^2}\\]
\n\t\t\tfor a number $a$. You have to find $a$.
\n\t\t\t$a=\\;$[[0]]
\n\t\t\tYou can click on Show steps to get help. You will not lose any marks if you do so.
\n\t\t\t", "steps": [{"showCorrectAnswer": true, "marks": 0, "scripts": {}, "type": "information", "prompt": "The quotient rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\\]
Differentiate the following function $f(x)$ using the quotient rule.
", "metadata": {"description": "The derivative of $\\displaystyle \\frac{a+be^{cx}}{b+ae^{cx}}$ is $\\displaystyle \\frac{pe^{cx}} {(b+ae^{cx})^2}$. Find $p$.
", "notes": "\n\t\t1/08/2012:
\n\t\tAdded tags.
\n\t\tAdded description.
\n\t\tChecked calculation. OK.
\n\t\tAdded information about Show steps. Altered to 0 marks lost rather than 1.
\n\t\tChanged std rule set to include !noLeadingMinus, so expressions don't change order from that intended. Got rid of a redundant ruleset.
\n\t\t\n\t\t", "licence": "Creative Commons Attribution 4.0 International"}, "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "functions": {}, "variable_groups": [], "question_groups": [{"questions": [], "pickQuestions": 0, "name": "", "pickingStrategy": "all-ordered"}], "showQuestionGroupNames": false, "advice": "\n\t \n\t \n\t
The quotient rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\\]
For this example:
\n\t \n\t \n\t \n\t\\[\\simplify[std]{u = {a}+{b}e^({c}x)}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {b*c}e^({c}x)}\\]
\n\t \n\t \n\t \n\t\\[\\simplify[std]{v = {b}+{a}e^({c}x)} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {a*c}e^({c}x)}\\]
\n\t \n\t \n\t \n\tHence on substituting into the quotient rule above we get:
\n\t \n\t \n\t \n\t\\[\\begin{eqnarray*} \\frac{df}{dx}&=&\\simplify[std]{({b*c}e^({c}x)({b}+{a}e^({c}x))-{a*c}e^({c}x)({a}+{b}e^({c}x)))/({b}+{a}e^({c}x))^2}\\\\\n\t \n\t &=&\\simplify[std]{({b^2*c} e^({c}x)+{a*b*c}*e^({2*c}x)-{a^2*c}e^({c}x)-{a*b*c}*e^({2*c}x) )/({b}+{a}e^({c}x))^2}\\\\\n\t \n\t &=&\\simplify[std]{({b^2*c} e^({c}x)-{a^2*c}e^({c}x))/({b}+{a}e^({c}x))^2}\\\\\n\t \n\t &=&\\simplify[std]{({b^2*c-a^2*c} e^({c}x))/({b}+{a}e^({c}x))^2}\t\n\t \n\t \\end{eqnarray*}\\]
\n\t \n\t \n\t \n\tHence $a=\\var{c*(b^2-a^2)}$
\n\t \n\t \n\t", "type": "question", "tags": ["algebraic manipulation", "Calculus", "checked2015", "derivative of a quotient", "differentiation", "MAS1601", "quotient rule", "Steps"], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Luis Hernandez", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2870/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Luis Hernandez", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2870/"}]}