Question

If x²y - 3x = y³ – 3, then at the point (-1,2), dy/dx=

Answer 1

now, the assumption in this implicit differentiation is that "y" is really a function in x-terms, whilst "x" is just a simple variable

`x^2y-3x=y^3-3\implies \stackrel{product~rule}{\left( 2xy + x^2\cdot \cfrac{dy}{dx} \right)}-3=\stackrel{chain~rule}{3y^2\cdot \cfrac{dy}{dx}} \\\\\\ 2xy-3=3y^2\cdot \cfrac{dy}{dx}-x^2\cdot \cfrac{dy}{dx}\implies 2xy-3=\cfrac{dy}{dx}(3y^2-x^2) \\\\\\ \left. \cfrac{2xy-3}{3y^2-x^2}=\cfrac{dy}{dx} \right|_((-1,2))\implies \cfrac{2(-1)(2)-3}{3(2)^2-(-1)^2}\implies -\cfrac{7}{11}`