Differentiate the following functions using the quotient rule.
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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.
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\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "
Find $\\displaystyle \\frac{d}{dx}\\left(\\frac{m\\sin(ax)+n\\cos(ax)}{b\\sin(ax)+c\\cos(ax)}\\right)$. Three part question.
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\\[\\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\\]
a)
\n\tFor this example:
\n\t\\[\\simplify[std]{u = sin({a}x)}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {a}cos({a}x)}\\]
\n\t\\[\\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)}\\]
\n\tHence on substituting into the quotient rule above we get:
\n\t\\[\\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*}\\]
\n\tHence $a=\\var{a*c}$
\n\tb)\\[\\simplify[std]{u = cos({a}x)}\\Rightarrow \\simplify[std]{Diff(u,x,1) = -{a}sin({a}x)}\\]
\n\t\\[\\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)}\\]
\n\tHence on substituting into the quotient rule above we get:
\n\t\\[\\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*}\\]
\n\tHence $b=\\var{-a*b}$
\n\tc)
\n\tWe 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 $c=\\var{res}$
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\\[\\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\\]
\\[\\simplify[std]{f(x) = (sin({a}x))/({b}sin({a}x)+{c}cos({a}x))}\\]
\n\t\t\tYou are given that \\[\\simplify[std]{Diff(f,x,1) = a / ({b}sin({a}x)+{c}cos({a}x))^2}\\]
\n\t\t\tfor a number $a$. You have to find $a$.
\n\t\t\t$a=\\;$[[0]]
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\n\t\t\t \n\t\t\t \n\t\t\t \n\t\t\tYou are given that \\[\\simplify[std]{Diff(g,x,1) = b / ({b}sin({a}x)+{c}cos({a}x))^2}\\]
\n\t\t\t \n\t\t\t \n\t\t\t \n\t\t\tfor a number $b$. You have to find $b$.
\n\t\t\t \n\t\t\t \n\t\t\t \n\t\t\t$b=\\;$[[0]]
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\n\t\t\tYou are given that \\[\\simplify[std]{Diff(h,x,1) = c / ({b}sin({a}x)+{c}cos({a}x))^2}\\]
\n\t\t\tfor a number $c$. You have to find $c$.
\n\t\t\t$c=\\;$[[0]]
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