Question

y ​4 ​​ +3xy+x=2y, start superscript, 4, end superscript, plus, 3, x, y, plus, x, equals, 2y Find the value of d y d x ​dx ​ ​dy ​​ start fraction, d, y, divided by, d, x, end fraction at the point ( 2 , 0 ) (2,0)left parenthesis, 2, comma, 0, right parenthesis.

Answer 1

$\bf y^4+3xy+x=2\implies 4y^3\cfrac{dy}{dx}+3\left( 1\cdot y+x\cdot \cfrac{dy}{dx} \right)+1=0 \\\\\\ 4y^3\cfrac{dy}{dx}+3\left(y+x\cfrac{dy}{dx} \right)+1=0\implies 4y^3\cfrac{dy}{dx}+3y+3x\cfrac{dy}{dx}+1=0 \\\\\\ 4y^3\cfrac{dy}{dx}+3x\cfrac{dy}{dx}=-1-3y\implies \stackrel{common~factor}{\cfrac{dy}{dx}}(4y^3+3x)=-1-3y \\\\\\ \left. \cfrac{dy}{dx}=\cfrac{-1-3y}{4y^3+3x} \right|_{(2,0)}\implies \cfrac{-1-3(0)}{4(0)^3+3(2)}\implies \cfrac{-1}{6}$