Question

Find the rate of change of y with respect to x at the given values of x and y.


`xy^ 2-x^2y-2=0`; x = 1, y = −1

Answer 1

`\bf xy^2-x^2y-2=0\qquad (1,-1)\\\\ -------------------------------\\\\ \left( y^2+x2y\cfrac{dy}{dx} \right)-\left( 2xy+x^2\cfrac{dy}{dx} \right)-0=0 \\\\\\ 2xy\cfrac{dy}{dx}-x^2\cfrac{dy}{dx}=2xy-y^2\impliedby \textit{common factor} \\\\\\ \cfrac{dy}{dx}(2xy-x^2)=2xy-y^2\implies \boxed{\cfrac{dy}{dx}=\cfrac{2xy-y^2}{2xy-x^2}} \\\\\\ \left. \cfrac{2xy-y^2}{2xy-x^2} \right|_(1,-1)\implies \cfrac{-2-1}{-2-1}\implies 1\\\\ -------------------------------\\\\`


`\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-(-1)=1(x-1) \\ \left. \qquad \right. \uparrow\\ \textit{point-slope form} \\\\\\ y+1=x-1\implies \boxed{y=x-2}`