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}\\]
a)
\nFor this example:
\n\\[\\simplify[std]{u = sin({a}x)}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {a}cos({a}x)}\\]
\n\\[\\simplify[std]{v = {b}sin({a}x)+{c}cos({a}x)} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {a*b}cos({a}x)+{-a*c}sin({a}x)}\\]
\nHence on substituting into the quotient rule above we get:
\n\\[\\begin{eqnarray*} \\frac{df}{dx}&=&\\simplify[std]{({a}cos({a}x)({b}sin({a}x)+{c}cos({a}x))-sin({a}x)({a*b}cos({a}x)+{-a*c}sin({a}x)))/({b}sin({a}x)+{c}cos({a}x))^2}\\\\ &=&\\simplify[std]{({a*b} cos({a}x) sin({a}x)+{a*c} cos({a}x)^2-{a*b} sin({a}x)cos({a}x)+{a*c}sin({a}x)^2)/({b}sin({a}x)+{c}cos({a}x))^2}\\\\ &=&\\simplify[std]{({a*c}cos({a}x)^2+{a*c}sin({a}x)^2)/({b}sin({a}x)+{c}cos({a}x))^2}\\\\ &=&\\simplify[std]{({a*c}(cos({a}x)^2+sin({a}x)^2))/({b}sin({a}x)+{c}cos({a}x))^2}\\\\ &=&\\simplify[std]{({a*c})/({b}sin({a}x)+{c}cos({a}x))^2} \\end{eqnarray*}\\]
\nHence $a=\\var{a*c}$
\nb)\\[\\simplify[std]{u = cos({a}x)}\\Rightarrow \\simplify[std]{Diff(u,x,1) = -{a}sin({a}x)}\\]
\n\\[\\simplify[std]{v = {b}sin({a}x)+{c}cos({a}x)} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {a*b}cos({a}x)+{-a*c}sin({a}x)}\\]
\nHence on substituting into the quotient rule above we get:
\n\\[\\begin{eqnarray*} \\frac{dg}{dx}&=&\\simplify[std]{({-a}sin({a}x)({b}sin({a}x)+{c}cos({a}x))-cos({a}x)({a*b}cos({a}x)+{-a*c}sin({a}x)))/({b}sin({a}x)+{c}cos({a}x))^2}\\\\ &=&\\simplify[std]{({-a*b}sin({a}x)^2-{a*c} sin({a}x)cos({a}x)-{a*b}cos({a}x)^2+{a*c}sin({a})cos({a}x))/({b}sin({a}x)+{c}cos({a}x))^2}\\\\ &=&\\simplify[std]{({-a*b}sin({a}x)^2-{a*b}cos({a}x)^2)/({b}sin({a}x)+{c}cos({a}x))^2}\\\\ &=&\\simplify[std]{({-a*b}(sin({a}x)^2+cos({a}x)^2))/({b}sin({a}x)+{c}cos({a}x))^2}\\\\ &=&\\simplify[std]{({-a*b})/({b}sin({a}x)+{c}cos({a}x))^2} \\end{eqnarray*}\\]
\nHence $b=\\var{-a*b}$
\nc)
\nWe have that $h(x)=\\simplify[std]{{m}f(x)+{n}g(x)}$
Hence \\[\\begin{eqnarray*}\\frac{dh}{dx} &=& \\simplify[std]{{m}*Diff(f,x,1)+{n}*Diff(f,x,1)}\\\\ &=&\\simplify[std]{{m}*({a*c}/({b}sin({a}x)+{c}cos({a}x))^2)+{n}({-a*b}/({b}sin({a}x)+{c}cos({a}x))^2)}\\\\ &=&\\simplify[std]{(({m}*{a*c})+({n}*{-a*b}))/({b}sin({a}x)+{c}cos({a}x))^2}\\\\ &=&\\simplify[std]{{res}/({b}sin({a}x)+{c}cos({a}x))^2} \\end{eqnarray*}\\]
Hence $c=\\var{res}$
\n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noleadingMinus"]}, "parts": [{"stepspenalty": 0.0, "prompt": "\n\\[\\simplify[std]{f(x) = (sin({a}x))/({b}sin({a}x)+{c}cos({a}x))}\\]
\nYou are given that \\[\\simplify[std]{Diff(f,x,1) = a / ({b}sin({a}x)+{c}cos({a}x))^2}\\]
\nfor a number $a$. You have to find $a$.
\n$a=\\;$[[0]]
\n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "marks": 3.0, "answer": "{a*c}", "type": "jme"}], "steps": [{"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}\\]
\\[\\simplify[std]{g(x) = (cos({a}x))/({b}sin({a}x)+{c}cos({a}x))}\\]
\n \n \n \nYou are given that \\[\\simplify[std]{Diff(g,x,1) = b / ({b}sin({a}x)+{c}cos({a}x))^2}\\]
\n \n \n \nfor a number $b$. You have to find $b$.
\n \n \n \n$b=\\;$[[0]]
\n \n \n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "marks": 3.0, "answer": "{-b*a}", "type": "jme"}], "type": "gapfill", "marks": 0.0}, {"prompt": "\n\\[\\simplify[std]{h(x) = ({m}sin({a}x)+{n}cos({a}x))/({b}sin({a}x)+{c}cos({a}x))}\\]
\nYou are given that \\[\\simplify[std]{Diff(h,x,1) = c / ({b}sin({a}x)+{c}cos({a}x))^2}\\]
\nfor a number $c$. You have to find $c$.
\n$c=\\;$[[0]]
\n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "marks": 3.0, "answer": "{res}", "type": "jme"}], "type": "gapfill", "marks": 0.0}], "extensions": [], "statement": "Differentiate the following functions using the quotient rule.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2..9)", "name": "a"}, "c": {"definition": "if(b^2=c1^2,c1+1,c1)", "name": "c"}, "b": {"definition": "random(1..9)", "name": "b"}, "s2": {"definition": "random(1,-1)", "name": "s2"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "m": {"definition": "random(2..9)", "name": "m"}, "n": {"definition": "if(m*c=n1*b,n1+1,n1)", "name": "n"}, "res": {"definition": "m*a*c-n*b*a", "name": "res"}, "n1": {"definition": "s2*random(2..9)", "name": "n1"}, "c1": {"definition": "s1*random(2..8)", "name": "c1"}}, "metadata": {"notes": "\n \t\t\n \t\t
1/08/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tChecked calculation. OK.
\n \t\tChanged std rule set to include !noLeadingMinus, so expressions don't change order. Got rid of a redundant ruleset.
\n \t\tImproved display in various places.
\n \t\tChanged to 0 penalty for accessing Show steps in first question.
\n \t\t\n \t\t
\n \t\t", "description": "
Find $\\displaystyle \\frac{d}{dx}\\left(\\frac{m\\sin(ax)+n\\cos(ax)}{b\\sin(ax)+c\\cos(ax)}\\right)$. Three part question.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}], "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}]}]}], "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}]}