Answer to Question #178423 in Calculus for Anna Watson

Question

Evaluate dy/dx from the equation  when 3xy^2+4x/y-(y-5)/(2x^2-1)=11 when x=1

Answer 1

$Given \ 3xy^2 +\frac{4x}{y}-\frac{(y-5)}{(2x^2-1)}=11...(1)\\ To \ find \ \frac{dy}{dx} , when \ x=1.\\ We \ use \ implicit \ differentiation \\ Substituting \ x=1 \ in \ (1), we \ get \\ 3(1)y^2 +\frac{4(1)}{y}-\frac{(y-5)}{(2(1)^2-1)}=11\\ \Rightarrow 3y^2 +\frac{4}{y}-(y-5)=11\\ \Rightarrow \frac{3y^3 +4-y(y-5)}{y}=11\\ \Rightarrow 3y^3 +4-y(y-5)=11y \ (by\ cross \ multiplication)\\ \Rightarrow 3y^3 +4-y^2+5y-11y=0\\ \Rightarrow 3y^3 -y^2-6y+4=0\\$ $Since \ the \ sum \ of \ the \ coefficients \ is \ equal \ to \ zero \\ y=1 \ is \ a \ root \ of \ 3y^3-y^2-6y+4=0\\ By \ synthetic \ division \ we \ get \ 3y^2 +2y-4 \ as \ a \ factor\\ \Rightarrow y=\frac{-1+\sqrt(13)}{3} \ and \ y=\frac{-1-\sqrt(13)}{3}\\ \therefore y=1,\ \frac{-1+\sqrt(13)}{3} \ and \ \frac{-1-\sqrt(13)}{3}\\$$Differentiating (1), with \ respect \ to \ 'x',\ we \ get \\ \frac{d}{dx}(3x^2y +\frac{4x}{y}-\frac{(y-5)})=\frac{d}{dx}(11) \\ \Rightarrow \frac{d}{dx}(3x^2y)+\frac{d}{dx}(\frac{4x}{y})-\frac{d}{dx}(\frac{(y-5)}{(2x^2-1)})=\frac{d}{dx}(11)\\ \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}]\\ -[\frac{(2x^2-1)\frac{d}{dx}(y-5)-(y-5)\frac{d}{dx}(2x^2-1)}{(2x^2-1)^2}]=0\\ \Rightarrow [(3x^2)\frac{dy}{dx} +6xy]+[\frac{4y-4x\frac{dy}{dx}}{y^2}]\\ -[\frac{(2x^2-1)\frac{dy}{dx}-(y-5)(4x)}{(2x^2-1)^2}]=0\\ Regrouping \ the \ terms \ containing \ \frac{dy}{dx}, we get \\ [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}]\\ \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)}]}\\ Substituting \ x=1 \ and \ y=1 \ in \frac{dy}{dx}, we \ get \\ [\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)}]}\\ =\frac{[-6-4+16]}{[3-4-1]}\\ =\frac{(6)}{(-2)}=-3\\ \therefore [\frac{dy}{dx}]_{x=1,y=1}=-3$