Practice problem
Find dydx\frac{dy}{dx}dxdy of x3y3−y=xx^3y^3-y=xx3y3−y=x
dydx=1−3x2y23y2x3−1\frac{dy}{dx}=\frac{1-3x^2y^2}{3y^2x^3-1}dxdy=3y2x3−11−3x2y2
dydx=1+3x2y23y2x3+1\frac{dy}{dx}=\frac{1+3x^2y^2}{3y^2x^3+1}dxdy=3y2x3+11+3x2y2
dydx=3x2y3+3y2x3−1\frac{dy}{dx}=3x^2y^3+3y^2x^3-1dxdy=3x2y3+3y2x3−1
dydx=3y2x3−11−3x2y2\frac{dy}{dx}=\frac{3y^2x^3-1} {1-3x^2y^2}dxdy=1−3x2y23y2x3−1
Type a concept, like “slope” or “photosynthesis”.