Question

Solve using implicit differentiation

Answer 1

the assumption being that "y" is encapsulating a function in x-terms, whilst "x" is a simple variable.

`3y^2+2x^2y-y=4\implies 3\stackrel{ \textit{chain rule} }{\left( 2y\cdot \cfrac{dy}{dx} \right)}+2\stackrel{ \textit{product rule} }{\left(2xy+x^2\cdot \cfrac{dy}{dx}\right)}-\cfrac{dy}{dx}=0 \\\\\\ 6y\cdot \cfrac{dy}{dx}+4xy+x^2\cdot \cfrac{dy}{dx}-\cfrac{dy}{dx}=0 \implies 6y\cdot \cfrac{dy}{dx}+x^2\cdot \cfrac{dy}{dx}-\cfrac{dy}{dx}=-4xy`

`\cfrac{dy}{dx}(6y+2x^2-1)=-4xy\implies \cfrac{dy}{dx}=\cfrac{-4xy}{6y+2x^2-1} \\\\[-0.35em] ~\dotfill\\\\ \left. \cfrac{dy}{dx} \right|_((x,y)=(1,1))\implies \cfrac{-4(1)(1)}{6(1)+2(1)^2-1}\implies \cfrac{-4}{7}`