Answer to Question #178423 in Calculus for Anna Watson
Evaluate dy/dx from the equation when 3xy^2+4x/y-(y-5)/(2x^2-1)=11 when x=1
"Given \\ 3xy^2 +\\frac{4x}{y}-\\frac{(y-5)}{(2x^2-1)}=11...(1)\\\\\nTo \\ find \\ \\frac{dy}{dx} , when \\ x=1.\\\\\nWe \\ use \\ implicit \\ differentiation \\\\\nSubstituting \\ x=1 \\ in \\ (1), we \\ get \\\\\n3(1)y^2 +\\frac{4(1)}{y}-\\frac{(y-5)}{(2(1)^2-1)}=11\\\\\n\\Rightarrow 3y^2 +\\frac{4}{y}-(y-5)=11\\\\\n\\Rightarrow \\frac{3y^3 +4-y(y-5)}{y}=11\\\\\n\\Rightarrow 3y^3 +4-y(y-5)=11y \\ (by\\ cross \\ multiplication)\\\\\n\\Rightarrow 3y^3 +4-y^2+5y-11y=0\\\\\n\\Rightarrow 3y^3 -y^2-6y+4=0\\\\" "Since \\ the \\ sum \\ of \\ the \\ coefficients \\ is \\ equal \\ to \\ zero \\\\\ny=1 \\ is \\ a \\ root \\ of \\ 3y^3-y^2-6y+4=0\\\\\nBy \\ synthetic \\ division \\ we \\ get \\ 3y^2 +2y-4 \\ as \\ a \\ factor\\\\\n\\Rightarrow y=\\frac{-1+\\sqrt(13)}{3} \\ and \\ y=\\frac{-1-\\sqrt(13)}{3}\\\\\n\\therefore y=1,\\ \\frac{-1+\\sqrt(13)}{3} \\ and \\ \\frac{-1-\\sqrt(13)}{3}\\\\""Differentiating (1), with \\ respect \\ to \\ 'x',\\ we \\ get \\\\\n\\frac{d}{dx}(3x^2y +\\frac{4x}{y}-\\frac{(y-5)})=\\frac{d}{dx}(11) \\\\\n\\Rightarrow \n\n\\frac{d}{dx}(3x^2y)+\\frac{d}{dx}(\\frac{4x}{y})-\\frac{d}{dx}(\\frac{(y-5)}{(2x^2-1)})=\\frac{d}{dx}(11)\\\\\n\\Rightarrow [(3x^2)\\frac{d}{dx}(y) +3y\\frac{d}{dx}(x^2)]+[\\frac{y\\frac{d}{dx}(4x)-4x\\frac{d}{dx}(y)}{y^2}]\\\\\n-[\\frac{(2x^2-1)\\frac{d}{dx}(y-5)-(y-5)\\frac{d}{dx}(2x^2-1)}{(2x^2-1)^2}]=0\\\\\n\\Rightarrow [(3x^2)\\frac{dy}{dx} +6xy]+[\\frac{4y-4x\\frac{dy}{dx}}{y^2}]\\\\\n-[\\frac{(2x^2-1)\\frac{dy}{dx}-(y-5)(4x)}{(2x^2-1)^2}]=0\\\\\nRegrouping \\ the \\ terms \\ containing \\ \\frac{dy}{dx}, we get \\\\\n[3x^2-\\frac{4x}{y^2}-\\frac{1}{(2x^2-1)}]\\frac{dy}{dx}=[-6xy-\\frac{4y}{y^2}-\\frac{(y-5)(4x)}{(2x^2-1)^2}]\\\\\n\\Rightarrow \\frac{dy}{dx}=\\frac{[-6xy-\\frac{4y}{y^2}-\\frac{(y-5)(4x)}{(2x^2-1)^2}]}{[3x^2-\\frac{4x}{y^2}-\\frac{1}{(2x^2-1)}]}\\\\\nSubstituting \\ x=1 \\ and \\ y=1 \\ in \\frac{dy}{dx}, we \\ get \\\\\n[\\frac{dy}{dx}]_{x=1,y=1}=\\frac{[-6(1)(1)-\\frac{4(1)}{(1)^2}-\\frac{(1-5)(4(1))}{(2(1)^2-1)^2}]}{[3(1)^2-\\frac{4(1)}{(1)^2}-\\frac{1}{(2(1)^2-1)}]}\\\\\n=\\frac{[-6-4+16]}{[3-4-1]}\\\\\n=\\frac{(6)}{(-2)}=-3\\\\\n\\therefore [\\frac{dy}{dx}]_{x=1,y=1}=-3"
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